Probability Distributions Class 12 Maths 2 Miscellaneous Exercise 7 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Probability Distributions Miscellaneous Exercise 7 Questions and Answers.

12th Maths Part 2 Probability Distributions Miscellaneous Exercise 7 Questions And Answers Maharashtra Board

(I) Choose the correct option from the given alternatives:

Question 1.
P.d.f. of a c.r.v. X is f(x) = 6x(1 – x), for 0 ≤ x ≤ 1 and = 0, otherwise (elsewhere) If P(X < a) = P(X > a), then a =
(a) 1
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{4}\)
Answer:
(b) \(\frac{1}{2}\)

Question 2.
If the p.d.f. of a c.r.v. X is f(x) = 3(1 – 2x2), for 0 < x < 1 and = 0, otherwise (elsewhere), then the c.d.f. of X is F(x) =
(a) 2x – 3x2
(b) 3x – 4x3
(c) 3x – 2x3
(d) 2x3 – 3x
Answer:
(c) 3x – 2x3

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7

Question 3.
If the p.d.f. of a c.r.v. X is f(x) = \(\frac{x^{2}}{18}\), for -3 < x < 3 and = 0, otherwise, then P(|X| < 1) =
(a) \(\frac{1}{27}\)
(b) \(\frac{1}{28}\)
(c) \(\frac{1}{29}\)
(d) \(\frac{1}{26}\)
Answer:
(a) \(\frac{1}{27}\)

Question 4.
If p.m.f. of a d.r.v. X takes values 0, 1, 2, 3, … which probability P(X = x) = k(x +1) . 5-x, where k is a constant, then P(X = 0) =
(a) \(\frac{7}{25}\)
(b) \(\frac{16}{25}\)
(c) \(\frac{18}{25}\)
(d) \(\frac{19}{25}\)
Answer:
(b) \(\frac{16}{25}\)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 I Q4

Question 5.
If p.m.f. of a d.r.v. X is P(X = x) = \(\frac{\left({ }^{5} \mathrm{C}_{x}\right)}{2^{5}}\), for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise. If a = P(X ≤ 2) and b = P(X ≥ 3), then
(a) a < b
(b) a > b
(c) a = b
(d) a + b
Answer:
(c) a = b

Question 6.
If p.m.f. of a d.r.v. X is P(X = x) = \(\frac{x}{n(n+1)}\), for x = 1, 2, 3, ……, n and = 0, otherwise, then E(X) =
(a) \(\frac{n}{1}+\frac{1}{2}\)
(b) \(\frac{n}{3}+\frac{1}{6}\)
(c) \(\frac{n}{2}+\frac{1}{5}\)
(d) \(\frac{n}{1}+\frac{1}{3}\)
Answer:
(b) \(\frac{n}{3}+\frac{1}{6}\)

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7

Question 7.
If p.m.f. of a d.r.v. X is P(x) = \(\frac{c}{x^{3}}\), for x = 1, 2, 3 and = 0, otherwise (elsewhere), then E(X) =
(a) \(\frac{343}{297}\)
(b) \(\frac{294}{251}\)
(c) \(\frac{297}{294}\)
(d) \(\frac{294}{297}\)
Answer:
(b) \(\frac{294}{251}\)

Question 8.
If the d.r.v. X has the following probability distribution:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 I Q8
then P(X = -1) =
(a) \(\frac{1}{10}\)
(b) \(\frac{2}{10}\)
(c) \(\frac{3}{10}\)
(d) \(\frac{4}{10}\)
Answer:
(a) \(\frac{1}{10}\)

Question 9.
If the d.r.v. X has the following probability distribution:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 I Q9
then k =
(a) \(\frac{1}{7}\)
(b) \(\frac{1}{8}\)
(c) \(\frac{1}{9}\)
(d) \(\frac{1}{10}\)
Answer:
(d) \(\frac{1}{10}\)

Question 10.
Find the expected value of X for the following p.m.f.
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 I Q10
(a) 0.85
(b) -0.35
(c) 0.15
(d) -0.15
Answer:
(b) -0.35

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7

(II) Solve the following:

Question 1.
Identify the random variable as either discrete or continuous in each of the following. If the random variable is discrete, list its possible values:
(i) An economist is interested in the number of unemployed graduates in the town of population 1 lakh.
(ii) Amount of syrup prescribed by a physician.
(iii) The person on a high protein diet is interesting to gain weight in a week.
(iv) 20 white rats are available for an experiment. Twelve rats are males. A scientist randomly selects 5 rats, the number of female rats selected on a specific day.
(v) A highway-safety group is interested in studying the speed (in km/hr) of a car at a checkpoint.
Solution:
(i) Let X = number of unemployed graduates in a town.
Since the population of the town is 1 lakh, X takes the finite values.
∴ random variable X is discrete.
Range = {0, 1, 2, …, 99999, 100000}.

(ii) Let X = amount of syrup prescribed by a physician.
Then X takes uncountable infinite values.
∴ random variable X is continuous.

(iii) Let X = gain of weight in a week
Then X takes uncountable infinite values
∴ random variable X is continuous.

(iv) Let X = number of female rats selected on a specific day.
Since the total number of rats is 20 which includes 12 males and 8 females, X takes the finite values.
∴ random variable X is discrete.
Range = {0, 1, 2, 3, 4, 5}

(v) Let X = speed of .the car in km/hr.
Then X takes uncountable infinite values
∴ random variable X is continuous.

Question 2.
The probability distribution of discrete r.v. X is as follows:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q2
(i) Determine the value of k.
(ii) Find P(X ≤ 4), P(2 < X < 4), P(X ≥ 3).
Solution:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q2.1
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q2.2

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7

Question 3.
The following is the probability distribution of X:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q3
Find the probability that
(i) X is positive
(ii) X is non-negative
(iii) X is odd
(iv) X is even.
Solution:
(i) P(X is positive) = P(X = 1) + P(X = 2) + P(X = 3)
= 0.25 + 0.15 + 0.1
= 0.50

(ii) P(X is non-negative)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.20 + 0.25 + 0.15 + 0.1
= 0.70

(iii) P(X is odd)
= P(X = -3) + P(X = -1) + P(X = 1) + P(X = 3)
= 0.05 + 0.15 + 0.25 + 0.1
= 0.55

(iv) P(X is even)
= P(X = -2) + P(X = 0) + P(X = 2)
= 0.10 + 0.20 + 0.15
= 0.45.

Question 4.
The p.m.f. of a r.v. X is given by P(X = x) = x = \(\frac{{ }^{5} \mathrm{C}_{\mathrm{x}}}{2^{5}}\), for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise. Then show that P(X ≤ 2) = P(X ≥ 3).
Solution:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= \(\frac{{ }^{5} \mathrm{C}_{0}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{1}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{2}}{2^{5}}\)
= \(\frac{{ }^{5} \mathrm{C}_{5}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{4}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{3}}{2^{5}}\) ………[latex]{ }^{n} \mathrm{C}_{r}={ }^{n} \mathrm{C}_{n-r}[/latex]
= P(X = 5) + P(X = 4) + P(X = 3)
= P(X ≥ 3)
∴ P(X ≤ 2) = P(X ≥ 3).

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7

Question 5.
In the p.m.f. of r.v. X
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q5
Find a and obtain c.d.f. of X.
Solution:
For p.m.f. of a r.v. X
\(\sum_{i=1}^{5} P(X=x)=1\)
∴ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 1
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q5.1
Let F(x) be the c.d.f. of X.
Then F(x) = P(X ≤ x)
∴ F(1) = P(X ≤ 1) = P(X = 1) = \(\frac{1}{20}\)
F(2) = P(X ≤ 2) = P(X = 1) + P (X = 2)
\(=\frac{1}{20}+\frac{3}{20}=\frac{4}{20}=\frac{1}{5}\)
P(3) = P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)
\(=\frac{1}{20}+\frac{3}{20}+\frac{5}{20}=\frac{9}{20}\)
F(4) = P(X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
\(=\frac{1}{20}+\frac{3}{20}+\frac{5}{20}+\frac{10}{20}=\frac{19}{20}\)
F(5) = P(X ≤ 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
\(=\frac{1}{20}+\frac{3}{20}+\frac{5}{20}+\frac{10}{20}+\frac{1}{20}=\frac{20}{20}=1\)
Hence, the c.d.f. of the random variable X is as follows:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q5.2

Question 6.
A fair coin is tossed 4 times. Let X denote the number of heads obtained. Write down the probability distribution of X. Also, find the formula for p.m.f. of X.
Solution:
When a fair coin is tossed 4 times then the sample space is
S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}
∴ n(S) = 16
X denotes the number of heads.
∴ X can take the value 0, 1, 2, 3, 4
When X = 0, then X = {TTTT}
∴ n (X) = 1
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{1}{16}=\frac{{ }^{4} \mathrm{C}_{0}}{16}\)
When X = 1, then
X = {HTTT, THTT, TTHT, TTTH}
∴ n(X) = 4
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{{ }^{4} C_{1}}{16}\)
When X = 2, then
X = {HHTT, HTHT, HTTH, THHT, THTH, TTHH}
∴ n(X) = 6
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{6}{16}=\frac{{ }^{4} \mathrm{C}_{2}}{16}\)
When X = 3, then
X = {HHHT, HHTH, HTHH, THHH}
∴ n(X) = 4
∴ P(X = 3) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{{ }^{4} C_{3}}{16}\)
When X = 4, then X = {HHHH}
∴ n(X) = 1
∴ P(X = 4) = \(\frac{n(X)}{n(S)}=\frac{1}{16}=\frac{{ }^{4} \mathrm{C}_{4}}{16}\)
∴ the probability distribution of X is as follows:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q6
Also, the formula for p.m.f. of X is
P(x) = \(\frac{{ }^{4} \mathrm{C}_{x}}{16}\), x = 0, 1, 2, 3, 4 and = 0, otherwise.

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7

Question 7.
Find the probability distribution of the number of successes in two tosses of a die, where success is defined as
(i) number greater than 4
(ii) six appear on at least one die.
Solution:
When a die is tossed two times, we obtain (6 × 6) = 36 number of observations.
Let X be the random variable, which represents the number of successes.
Here, success refers to the number greater than 4.
P(X = 0) = P(number less than or equal to 4 on both the tosses)
= \(\frac{4}{6} \times \frac{4}{6}=\frac{16}{36}=\frac{4}{9}\)
P(X = 1) = P(number less than or equal to 4 on first toss and greater than 4 on second toss) + P(number greater than 4 on first toss and less than or equal to 4 on second toss)
= \(\frac{4}{6} \times \frac{2}{6}+\frac{4}{6} \times \frac{2}{6}\)
= \(\frac{8}{36}+\frac{8}{36}\)
= \(\frac{16}{36}\)
= \(\frac{4}{9}\)
P(X = 2) = P(number greater than 4 on both the tosses)
= \(\frac{2}{6} \times \frac{2}{6}=\frac{4}{36}=\frac{1}{9}\)
Thus, the probability distribution is as follows:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q7
(ii) Here, success means six appears on at least one die.
P(Y = 0) = P(six appears on none of the dice) = \(\frac{5}{6} \times \frac{5}{6}=\frac{25}{36}\)
P(Y = 1) = P(six appears on none of the dice x six appears on at least one of the dice ) + P(six appears on none of the dice x six appears on at least one of the dice)
= \(\frac{1}{6} \times \frac{5}{6}+\frac{1}{6} \times \frac{5}{6}=\frac{5}{36}+\frac{5}{36}=\frac{10}{36}\)
P(Y = 2) = P(six appears on at least one of the dice) = \(\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}\)
Thus, the required probability distribution is as follows:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q7.1

Question 8.
A random variable X has the following probability distribution:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q8
Determine:
(i) k
(ii) P(X > 6)
(iii) P(0 < X < 3).

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7

Question 9.
The following is the c.d.f. of a r.v. X:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q9
Find
(i) p.m.f. of X
(ii) P( -1 ≤ X ≤ 2)
(iii) P(X ≤ X > 0).
Solution:
(i) From the given table
F(-3) = 0.1, F(-2) = 0.3, F(-1) = 0.5
F(0) = 0.65, f(1) = 0.75, F(2) = 0.85
F(3) = 0.9, F(4) = 1
P(X = -3) = F(-3) = 0.1
P(X = -2) = F(-2) – F(-3) = 0.3 – 0.1 = 0.2
P(X = -1) = F(-1) – F(-2) = 0.5 – 0.3 = 0.2
P(X = 0) = F(0) – F(-1) = 0.65 – 0.5 = 0.15
P(X = 1) = F(1) – F(0) = 0.75 – 0.65 = 0.1
P(X = 2) = F(2) – F(1) = 0.85 – 0.75 = 0.1
P(X = 3) = F(3) – F(2) = 0.9 – 0.85 = 0.1
P(X = 4) = F(4) – F(3) = 1 – 0.9 = 0.1
∴ the p.m.f of X is as follows:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q9.1
(ii) P(-1 ≤ X ≤ 2) = P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2)
= 0.2 + 0.15 + 0.1 + 0.1
= 0.55

(iii) (X ≤ 3) ∩ (X > 0)
= { -3, -2, -1, 0, 1, 2, 3} n {1, 2, 3, 4}
= {1, 2, 3}
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q9.2

Question 10.
Find the expected value, variance, and standard deviation of the random variable whose p.m.f’s are given below:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q10
Solution:
(i) We construct the following table to find the expected value, variance, and standard deviation:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q10.1
(ii) We construct the following table to find the expected value, variance, and standard deviation:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q10.2
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q10.3
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q10.4
(iii) We construct the following table to find the expected value, variance, and standard deviation:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q10.8
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q10.9
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q10.10
(iv) We construct the following table to find the expected value, variance, and standard deviation:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q10.5
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q10.6
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q10.7

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7

Question 11.
A player tosses two wins. He wins ₹ 10 if 2 heads appear, ₹ 5 if 1 head appears and ₹ 2 if no head appears. Find the expected winning amount and variance of the winning amount.
Solution:
When a coin is tossed twice, the sample space is
S = {HH, HT, TH, HH}
Let X denote the amount he wins.
Then X takes values 10, 5, 2.
P(X = 10) = P(2 heads appear) = \(\frac{1}{4}\)
P(X = 5) = P(1 head appears) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
P(X = 2) = P(no head appears) = \(\frac{1}{4}\)
We construct the following table to calculate the mean and the variance of X:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q11
From the table Σxi . P(xi) = 5.5, \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)\) = 38.5
E(X) = Σxi . P(xi) = 5.5
Var(X) = \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)\) – [E(X)]2
= 38.5 – (5.5)2
= 38.5 – 30.25
= 8.25
∴ Hence, expected winning amount = ₹ 5.5 and variance of winning amount = ₹ 8.25.

Question 12.
Let the p.m.f. of r.v. X be P(x) = \(\frac{3-x}{10}\), for x = -1, 0, 1, 2 and = 0, otherwise.
Calculate E(X) and Var(X).
Solution:
P(X) = \(\frac{3-x}{10}\)
X takes values -1, 0, 1, 2
P(X = -1) = P(-1) = \(\frac{3+1}{10}=\frac{4}{10}\)
P(X = 0) = P(0) = \(\frac{3-0}{10}=\frac{3}{10}\)
P(X = 1) = P(1) = \(\frac{3-1}{10}=\frac{2}{10}\)
P(X = 2) = P(2) = \(\frac{3-2}{10}=\frac{1}{10}\)
We construct the following table to calculate the mean and variance of X:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q12
From the table
ΣxiP(xi) = 0 and \(\Sigma x_{i}{ }^{2} \cdot P\left(x_{i}\right)\) = 1
E(X) = ΣxiP(xi) = 0
Var(X) = \(\Sigma x_{i}{ }^{2} \cdot P\left(x_{i}\right)\) – [E(X)]2
= 1 – 0
= 1
Hence, E(X) = 0, Var (X) = 1.

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7

Question 13.
Suppose the error involved in making a certain measurement is a continuous r.v. X with p.d.f.
f(x) = k(4 – x2), -2 ≤ x ≤ 2 and = 0 otherwise.
Compute
(i) P(X > 0)
(ii) P(-1 < X < 1)
(iii) P(X < -0.5 or X > 0.5).
Solution:
(i) P(X > 0)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q13
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q13.1
(ii) P(-1 < X < 1)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q13.2
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q13.3
(iii) P(X < -0.5 or X > 0.5)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q13.4
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q13.5
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q13.6
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q13.7
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q13.8

Question 14.
The p.d.f. of a continuous r.v. X is given by f(x) = \(\frac{1}{2 a}\), for 0 < x < 2a and = 0, otherwise. Show that P( X < \(\frac{a}{2}\)) = P(X > \(\frac{3a}{2}\))
Solution:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q14
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q14.1

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7

Question 15.
The p.d.f. of r.v. X is given by f(x) = \(\frac{k}{\sqrt{x}}\), for 0 < x < 4 and = 0, otherwise. Determine k. Determine c.d.f. of X and hence find P(X ≤ 2) and P(X ≤ 1).
Solution:
Since f is p.d.f. of the r.v. X,
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q15
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7 II Q15.1

Class 12 Maharashtra State Board Maths Solution 

Probability Distributions Class 12 Maths 2 Exercise 7.2 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Probability Distributions Ex 7.2 Questions and Answers.

12th Maths Part 2 Probability Distributions Exercise 7.2 Questions And Answers Maharashtra Board

Question 1.
Verify which of the following is p.d.f. of r.v. X:
(i) f(x) = sin x, for 0 ≤ x ≤ \(\frac{\pi}{2}\)
(ii) f(x) = x, for 0 ≤ x ≤ 1 and -2 – x for 1 < x < 2
(iii) fix) = 2, for 0 ≤ x ≤ 1.
Solution:
f(x) is the p.d.f. of r.v. X if
(a) f(x) ≥ 0 for all x ∈ R and
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q1
Hence, f(x) is the p.d.f. of X.

(ii) f(x) = x ≥ 0 if 0 ≤ x ≤ 1
For 1 < x < 2, -2 < -x < -1
-2 – 2 < -2 – x < -2 – 1
i.e. -4 < f(x) < -3 if 1 < x < 2
Hence, f(x) is not p.d.f. of X.

(iii) (a) f(x) = 2 ≥ 0 for 0 ≤ x ≤ 1
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q1.1
Hence, f(x) is not p.d.f. of X.

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 2.
The following is the p.d.f. of r.v. X:
f(x) = \(\frac{x}{8}\), for 0 < x < 4 and = 0 otherwise.
Find
(a) P(x < 1.5)
(b) P(1 < x < 2) (c) P(x > 2).
Solution:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q2
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q2.1

Question 3.
It is known that error in measurement of reaction temperature (in 0°C) in a certain experiment is continuous r.v. given by
f(x) = \(\frac{x^{2}}{3}\) for -1 < x < 2
= 0. otherwise.
(i) Verify whether f(x) is p.d.f. of r.v. X
(ii) Find P(0 < x ≤ 1)
(iii) Find the probability that X is negative.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q3

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 4.
Find k if the following function represents p.d.f. of r.v. X
(i) f(x) = kx. for 0 < x < 2 and = 0 otherwise.
Also find P(\(\frac{1}{4}\) < x < \(\frac{3}{2}\)).
(ii) f(x) = kx(1 – x), for 0 < x < 1 and = 0 otherwise.
Also find P(\(\frac{1}{4}\) < x < \(\frac{1}{2}\)), P(x < \(\frac{1}{2}\)).
Solution:
(i) Since, the function f is p.d.f. of X
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q4
(ii) Since, the function f is the p.d.f. of X,
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q4.1
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q4.2
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q4.3

Question 5.
Let X be the amount of time for which a book is taken out of the library by a randomly selected students and suppose X has p.d.f.
f(x) = 0.5x, for 0 ≤ x ≤ 2 and = 0 otherwise.
Calculate:
(i) P(x ≤ 1)
(ii) P(0.5 ≤ x ≤ 1.5)
(iii) P(x ≥ 1.5).
Solution:
(i) P(x ≤ 1)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q5
(ii) P(0.5 ≤ x ≤ 1.5)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q5.1
(iii) P(x ≥ 1.5)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q5.2
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q5.3

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 6.
Suppose that X is waiting time in minutes for a bus and its p.d.f. is given by f(x) = \(\frac{1}{5}\), for 0 ≤ x ≤ 5 and = 0 otherwise. Find the probability that
(i) waiting time is between 1 and 3
(ii) waiting time is more than 4 minutes.
Solution:
(i) Required probability = P(1 < X < 3)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q6
(ii) Required probability = P(X > 4)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q6.1

Question 7.
Suppose the error involved in making a certain measurement is a continuous r.v. X with p.d.f.
f(x) = k(4 – x2), -2 ≤ x ≤ 2 and 0 otherwise.
Compute:
(i) P(X > 0)
(ii) P(-1 < X < 1)
(iii) P(-0.5 < X or X > 0.5).
Solution:
Since, f is the p.d.f. of X,
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q7
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q7.1
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q7.2
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q7.3
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q7.4
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q7.5

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 8.
The following is the p.d.f. of continuous r.v. X
f(x) = \(\frac{x}{8}\), for 0 < x < 4 and = 0 otherwise.
(i) Find expression for c.d.f. of X.
(ii) Find F(x) at x = 0.5, 1.7 and 5.
Solution:
(i) Let F(x) be the c.d.f. of X
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q8
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q8.1

Question 9.
Given the p.d.f. of a continuous random r.v. X, f(x) = \(\frac{x^{2}}{3}\), for -1 < x < 2 and = 0 otherwise. Determine c.d.f. of X and hence find P(X < 1); P(X < -2), P(X > 0), P(1 < X < 2).
Solution:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q9
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q9.1

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 10.
If a r.v. X has p.d.f.
f(x) = \(\frac{c}{x}\) for 1 < x < 3, c > 0. Find c, E(X), Var (X).
Solution:
Since f(x) is p.d.f of r.v. X
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q10
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2 Q10.1

Class 12 Maharashtra State Board Maths Solution 

Probability Distributions Class 12 Maths 2 Exercise 7.1 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Probability Distributions Ex 7.1 Questions and Answers.

12th Maths Part 2 Probability Distributions Exercise 7.1 Questions And Answers Maharashtra Board

Question 1.
Let X represent the difference between a number of heads and the number of tails when a coin is tossed 6 times. What are the possible values of X?
Solution:
When a coin is tossed 6 times, the number of heads can be 0, 1, 2, 3, 4, 5, 6.
The corresponding number of tails will be 6, 5, 4, 3, 2, 1, 0.
∴ X can take values 0 – 6, 1 – 5, 2 – 4, 3 – 3, 4 – 2, 5 – 1, 6 – 0
i.e. -6, -4, -2, 0, 2, 4, 6.
∴ X = {-6, -4, -2, 0, 2, 4, 6}.

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 2.
An urn contains 5 red and 2 black balls. Two balls are drawn at random. X denotes the number of black balls drawn. What are the possible values of X?
Solution:
The urn contains 5 red and 2 black balls.
If two balls are drawn from the urn, it contains either 0 or 1 or 2 black balls.
X can take values 0, 1, 2.
∴ X = {0, 1, 2}.

Question 3.
State which of the following are not the probability mass function of a random variable. Give reasons for your answer.
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3
Solution:
P.m.f. of random variable should satisfy the following conditions:
(a) 0 ≤ pi ≤ 1
(b) Σpi = 1.

(i)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3.1
(a) Here 0 ≤ pi ≤ 1
(b) Σpi = 0.4 + 0.4 + 0.2 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

(ii)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3.2
P(X = 3) = -0.1, i.e. Pi < 0 which does not satisfy 0 ≤ Pi ≤ 1
Hence, P(X) cannot be regarded as p.m.f. of the random variable X.

(iii)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3.3
(a) Here 0 ≤ pi ≤ 1
(b) ∑pi = 0.1 + 0.6 + 0.3 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

(iv)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3.4
Here ∑pi = 0.3 + 0.2 + 0.4 + 0 + 0.05 = 0.95 ≠ 1
Hence, P(Z) cannot be regarded as p.m.f. of the random variable Z.

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1

(v)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3.5
Here ∑pi = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Hence, P(Y) cannot be regarded as p.m.f. of the random variable Y.

(vi)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3.6
(a) Here 0 ≤ pi ≤ 1
(b) ∑pi = 0.3 + 0.4 + 0.3 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

Question 4.
Find the probability distribution of
(i) number of heads in two tosses of a coin.
(ii) number of tails in the simultaneous tosses of three coins.
(iii) number of heads in four tosses of a coin.
Solution:
(i) For two tosses of a coin the sample space is {HH, HT, TH, TT}
Let X denote the number of heads in two tosses of a coin.
Then X can take values 0, 1, 2.
∴ P[X = 0] = P(0) = \(\frac{1}{4}\)
P[X = 1] = P(1) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
P[X = 2] = P(2) = \(\frac{1}{4}\)
∴ the required probability distribution is
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q4

(ii) When three coins are tossed simultaneously, then the sample space is
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let X denotes the number of tails.
Then X can take the value 0, 1, 2, 3.
∴ P[X = 0] = P(0) = \(\frac{1}{8}\)
P[X = 1] = P(1) = \(\frac{3}{8}\)
P[X = 2] = P(2) = \(\frac{3}{8}\)
P[X = 3] = P(3) = \(\frac{1}{8}\)
∴ the required probability distribution is
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q4.1

(iii) When a fair coin is tossed 4 times, then the sample space is
S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}
∴ n(S) = 16
Let X denotes the number of heads.
Then X can take the value 0, 1, 2, 3, 4
When X = 0, then X = {TTTT}
∴ n(X) = 1
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{1}{16}\)
When X = 1, then
X = {HTTT, THTT, TTHT, TTTH}
∴ n(X) = 4
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{1}{4}\)
When X = 2, then
X = {HHTT, HTHT, HTTH, THHT, THTH, TTHH}
∴ n(X) = 6
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{6}{16}=\frac{3}{8}\)
When X = 3, then
X = {HHHT, HHTH, HTHH, THHH}
∴ n(X) = 4
∴ P(X = 3) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{1}{4}\)
When X = 4, then X = {HHHH}
∴ n(X) = 1
∴ P(X = 4) = \(\frac{n(X)}{n(S)}=\frac{1}{16}\)
∴ the probability distribution of X is as follows:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q4.2

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 5.
Find the probability distribution of a number of successes in two tosses of a die, where success is defined as a number greater than 4 appearing on at least one die.
Solution:
When a die is tossed twice, the sample space s has 6 × 6 = 36 sample points.
∴ n(S) = 36
The trial will be a success if the number on at least one die is 5 or 6.
Let X denote the number of dice on which 5 or 6 appears.
Then X can take values 0, 1, 2.
When X = 0 i.e., 5 or 6 do not appear on any of the dice, then
X = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}
∴ n(X) = 16.
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{16}{36}=\frac{4}{9}\)
When X = 1, i.e. 5 or 6 appear on exactly one of the dice, then
X = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4)}
∴ n(X) = 16
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{16}{36}=\frac{4}{9}\)
When X = 2, i.e. 5 or 6 appear on both of the dice, then
X = {(5, 5), (5, 6), (6, 5), (6, 6)}
∴ n(X) = 4
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)
∴ the required probability distribution is
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q5

Question 6.
From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Solution:
Here, the number of defective bulbs is the random variable.
Let the number of defective bulbs be denoted by X.
∴ X can take the value 0, 1, 2, 3, 4.
Since the draws are done with replacement, therefore the four draws are independent experiments.
Total number of bulbs is 30 which include 6 defectives.
∴ P(X = 0) = P(0) = P(all 4 non-defective bulbs)
= \(\frac{24}{30} \times \frac{24}{30} \times \frac{24}{30} \times \frac{24}{30}\)
= \(\frac{256}{625}\)
P(X = 1) = P (1) = P (1 defective and 3 non-defective bulbs)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q6
P(X = 2) = P(2) = P(2 defective and 2 non-defective)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q6.1
P(X = 3) = P(3) = P(3 defectives and 1 non-defective)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q6.2
P(X = 4) = P(4) = P(all 4 defectives)
= \(\frac{6}{30} \times \frac{6}{30} \times \frac{6}{30} \times \frac{6}{30}\)
= \(\frac{1}{625}\)
∴ the required probability distribution is
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q6.3

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 7.
A coin is biased so that the head is 3 times as likely to occur as the tail. If the coin is tossed twice. Find the probability distribution of a number of tails.
Solution:
Given a biased coin such that heads is 3 times as likely as tails.
∴ P(H) = \(\frac{3}{4}\) and P(T) = \(\frac{1}{4}\)
The coin is tossed twice.
Let X can be the random variable for the number of tails.
Then X can take the value 0, 1, 2.
∴ P(X = 0) = P(HH) = \(\frac{3}{4} \times \frac{3}{4}=\frac{9}{16}\)
P(X = 1) = P(HT, TH) = \(\frac{3}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{3}{4}=\frac{6}{16}=\frac{3}{8}\)
P(X = 2) = P(TT) = \(\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}\)
∴ the required probability distribution is
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q7

Question 8.
A random variable X has the following probability distribution:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q8
Determine:
(i) k
(ii) P(X < 3) (iii) P(X > 4)
Solution:
(i) Since P (x) is a probability distribution of x,
\(\sum_{x=0}^{7} P(x)=1\)
⇒ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) = 1
⇒ 0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
⇒ 10k2 + 9k – 1 = 0
⇒ 10k2 + 10k – k – 1 = 0
⇒ 10k(k + 1) – 1(k + 1) = 0
⇒ (k + 1)(10k – 1) = 0
⇒ 10k – 1 = 0 ……..[∵ k ≠ -1]
⇒ k = \(\frac{1}{10}\)

(ii) P(X< 3) = P(0) + P(1) + P(2)
= 0 + k + 2k
= 3k
= 3(\(\frac{1}{10}\))
= \(\frac{3}{10}\)

(iii) P(0 < X < 3) = P (1) + P (2)
= k + 2k
= 3k
= 3(\(\frac{1}{10}\))
= \(\frac{3}{10}\)

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 9.
Find expected value and variance of X for the following p.m.f.:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q9
Solution:
We construct the following table to calculate E(X) and V(X):
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q9.1
From the table,
Σxipi = -0.05 and \(\Sigma x_{i}^{2} \cdot p_{i}\) = 2.25
∴ E(X) = Σxipi = -0.05
and V(X) = \(\Sigma x_{i}^{2}+p_{i}-\left(\sum x_{i}+p_{i}\right)^{2}\)
= 2.25 – (-0.05)2
= 2.25 – 0.0025
= 2.2475
Hence, E(X) = -0.05 and V(X) = 2.2475.

Question 10.
Find expected value and variance of X, where X is the number obtained on the uppermost face when a fair die is thrown.
Solution:
If a die is tossed, then the sample space for the random variable X is
S = {1, 2, 3, 4, 5, 6}
∴ P(X) = \(\frac{1}{6}\); X = 1, 2, 3, 4, 5, 6.
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q10
Hence, E(X) = 3.5 and V(X) = 2.9167.

Question 11.
Find the mean number of heads in three tosses of a fair coin.
Solution:
When three coins are tossed the sample space is {HHH, HHT, THH, HTH, HTT, THT, TTH, TTT}
∴ n(S) = 8
Let X denote the number of heads when three coins are tossed.
Then X can take values 0, 1, 2, 3
P(X = 0) = P(0) = \(\frac{1}{8}\)
P(X = 1) = P(1) = \(\frac{3}{8}\)
P(X = 2) = P(2) = \(\frac{3}{8}\)
P(X = 3) = P(3) = \(\frac{1}{8}\)
∴ mean = E(X) = ΣxiP(xi)
= \(0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8}\)
= \(0+\frac{3}{8}+\frac{6}{8}+\frac{3}{8}\)
= \(\frac{12}{8}\)
= 1.5

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 12.
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Solution:
When two dice are thrown, the sample space S has 6 × 6 = 36 sample points.
∴ n(S) = 36
Let X denote the number of sixes when two dice are thrown.
Then X can take values 0, 1, 2
When X = 0, then
X = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5)}
∴ n(X) = 25
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{25}{36}\)
When X = 1, then
X = {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
∴ n(X) = 10
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{10}{36}\)
When X = 2, then X = {(6, 6)}
∴ n(X) = 1
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{1}{36}\)
∴ E(X) = ΣxiP(xi)
= \(0 \times \frac{25}{36}+1 \times \frac{10}{36}+2 \times \frac{1}{36}\)
= \(0+\frac{10}{36}+\frac{2}{36}\)
= \(\frac{1}{3}\)

Question 13.
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers. Find E(X).
Solution:
Two numbers are chosen from the first 6 positive integers.
∴ n(S) = \({ }^{6} C_{2}=\frac{6 \times 5}{1 \times 2}\) = 15
Let X denote the larger of the two numbers.
Then X can take values 2, 3, 4, 5, 6.
When X = 2, the other positive number which is less than 2 is 1.
∴ n(X) = 1
∴ P(X = 2) = P(2) = \(\frac{n(X)}{n(S)}=\frac{1}{15}\)
When X = 3, the other positive number less than 3 can be 1 or 2 and hence can be chosen in 2 ways.
∴ n(X) = 2
P(X = 3) = P(3) = \(\frac{n(X)}{n(S)}=\frac{2}{15}\)
Similarly, P(X = 4) = P(4) = \(\frac{3}{15}\)
P(X = 5) = P(5) = \(\frac{4}{15}\)
P(X = 6) = P(6) = \(\frac{5}{15}\)
∴ E(X) = ΣxiP(xi)
= \(2 \times \frac{1}{15}+3 \times \frac{2}{15}+4 \times \frac{3}{15}+5 \times \frac{4}{15}+6 \times \frac{5}{15}\)
= \(\frac{2+6+12+20+30}{15}\)
= \(\frac{70}{15}\)
= \(\frac{14}{3}\)

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 14.
Let X denote the sum of numbers obtained when two fair dice are rolled. Find the standard deviation of X.
Solution:
If two fair dice are rolled then the sample space S of this experiment is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let X denote the sum of the numbers on uppermost faces.
Then X can take the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q14
∴ the probability distribution of X is given by
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q14.1
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q14.2
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q14.3

Question 15.
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the student is recorded. What is the probability distribution of the random variable X? Find mean, variance, and standard deviation of X.
Solution:
Let X denote the age of the chosen student. Then X can take values 14, 15, 16, 17, 18, 19, 20, 21.
We make a frequency table to find the number of students with age X:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q15
The chances of any student selected are equally likely.
If there are m students with age X, then P(X) = \(\frac{m}{15}\)
Using this, the following is the probability distribution of X:
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q15.1
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q15.2
Variance = V(X) = \(\Sigma x_{i}^{2}\) . P(xi) – [E(X)]2
= 312.2 – (17.53)2
= 312.2 – 307.3
= 4.9
Standard deviation = √V(X) = √4.9 = 2.21
Hence, mean = 17.53, variance = 4.9 and standard deviation = 2.21.

Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 16.
In a meeting, 70% of the member’s favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed and X = 1 if he is in favour. Find E(X) and Var(X).
Solution:
X takes values 0 and 1.
It is given that
P(X = 0) = P(0) = 30% = \(\frac{30}{100}\) = 0.3
P(X = 1) = P(1) = 70% = \(\frac{70}{100}\) = 0.7
∴ E(X) = Σxi . P(xi) = 0 × 0.3 + 1 × 0.7 = 0.7
Also, \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)\) = 0 × 0.3 + 1 × 0.7 = 0.7
∴ Variance = V(X) = \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)-[E(X)]^{2}\)
= 0.7 – (0.7)2
= 0.7 – 0.49
= 0.21
Hence, E(X) = 0.7 and Var(X) = 0.21.

Class 12 Maharashtra State Board Maths Solution 

Differential Equations Class 12 Maths 2 Miscellaneous Exercise 6 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Miscellaneous Exercise 6 Questions and Answers.

12th Maths Part 2 Differential Equations Miscellaneous Exercise 6 Questions And Answers Maharashtra Board

(I) Choose the correct option from the given alternatives:

Question 1.
The order and degree of the differential equation \(\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=\left(\frac{d^{2} y}{d x^{2}}\right)^{\frac{3}{2}}\) are respectively……..
(a) 2, 1
(b) 1, 2
(c) 3, 2
(d) 2, 3
Answer:
(d) 2, 3

Question 2.
The differential equation of y = c2 + \(\frac{c}{x}\) is…….
(a) \(x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}=y\)
(b) \(\frac{d y}{d x^{2}}+x \frac{d y}{d x}+y=0\)
(c) \(x^{3}\left(\frac{d y}{d x}\right)^{2}+x \frac{d y}{d x}=y\)
(d) \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-y=0\)
Answer:
(a) \(x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}=y\)

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

Question 3.
x2 + y2 = a2 is a solution of ………
(a) \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-y=0\)
(b) \(y=x \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}+a^{2} y\)
(c) \(y=x \frac{d y}{d x}+a \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\)
(d) \(\frac{d^{2} y}{d x^{2}}=(x+1) \frac{d y}{d x}\)
Answer:
(c) \(y=x \frac{d y}{d x}+a \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 I Q3

Question 4.
The differential equation of all circles having their centres on the line y = 5 and touching the X-axis is
(a) \(y^{2}\left(1+\frac{d y}{d x}\right)=25\)
(b) \((y-5)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)
(c) \((y-5)^{2}+\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)
(d) \((y-5)^{2}\left[1-\left(\frac{d y}{d x}\right)^{2}\right]=25\)
Answer:
(b) \((y-5)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 I Q4

Question 5.
The differential equation y \(\frac{d y}{d x}\) + x = 0 represents family of ………
(a) circles
(b) parabolas
(c) ellipses
(d) hyperbolas
Answer:
(a) circles

Hint:
y \(\frac{d y}{d x}\) + x = 0
∴ ∫y dy + ∫x dx = c
∴ \(\frac{y^{2}}{2}+\frac{x^{2}}{2}=c\)
∴ x2 + y2 = 2c which is a circle.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

Question 6.
The solution of \(\frac{1}{x} \cdot \frac{d y}{d x}=\tan ^{-1} x\) is……
(a) \(\frac{x^{2} \tan ^{-1} x}{2}+c=0\)
(b) x tan-1x + c = 0
(c) x – tan-1x = c
(d) \(y=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c\)
Answer:
(d) \(y=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c\)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 I Q6

Question 7.
The solution of (x + y)2 \(\frac{d y}{d x}\) = 1 is…….
(a) x = tan-1(x + y) + c
(b) y tan-1(\(\frac{x}{y}\)) = c
(c) y = tan-1(x + y) + c
(d) y + tan-1(x + y) = c
Answer:
(c) y = tan-1(x + y) + c
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 I Q7
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 I Q7.1

Question 8.
The Solution of \(\frac{d y}{d x}=\frac{y+\sqrt{x^{2}-y^{2}}}{2}\) is……
(a) sin-1(\(\frac{y}{x}\)) = 2 log |x| + c
(b) sin-1(\(\frac{y}{x}\)) = log |x| + c
(c) sin(\(\frac{x}{y}\)) = log |x| + c
(d) sin(\(\frac{y}{x}\)) = log |y| + c
Answer:
(b) sin-1(\(\frac{y}{x}\)) = log |x| + c
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 I Q8

Question 9.
The solution of \(\frac{d y}{d x}\) + y = cos x – sin x is……
(a) y ex = cos x + c
(b) y ex + ex cos x = c
(c) y ex = ex cos x + c
(d) y2 ex = ex cos x + c
Answer:
(c) y ex = ex cos x + c
Hint:
\(\frac{d y}{d x}\) + y = cos x – sin x
I.F. = \(e^{\int 1 d x}=e^{x}\)
∴ the solution is y . ex = ∫(cos x – sin x) ex + c
∴ y . ex = ex cos x + c

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

Question 10.
The integrating factor of linear differential equation x \(\frac{d y}{d x}\) + 2y = x2 log x is……..
(a) \(\frac{1}{x}\)
(b) k
(c) \(\frac{1}{n^{2}}\)
(d) x2
Answer:
(d) x2
Hint:
I.F. = \(e^{\int \frac{2}{x} d x}\)
= e2 log x
= x2

Question 11.
The solution of the differential equation \(\frac{d y}{d x}\) = sec x – y tan x is…….
(a) y sec x + tan x = c
(b) y sec x = tan x + c
(c) sec x + y tan x = c
(d) sec x = y tan x + c
Answer:
(b) y sec x = tan x + c

Hint:
\(\frac{d y}{d x}\) = sec x – y tan x
∴ \(\frac{d y}{d x}\) + y tan x = sec x
I.F. = \(e^{\int \tan x d x}=e^{\log \sec x}\) = sec x
∴ the solution is
y . sec x = ∫sec x . sec x dx + c
∴ y sec x = tan x + c

Question 12.
The particular solution of \(\frac{d y}{d x}=x e^{y-x}\), when x = y = 0 is……
(a) ex-y = x + 1
(b) ex+y = x + 1
(c) ex + ey = x + 1
(d) ey-x = x – 1
Answer:
(a) ex-y = x + 1
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 I Q12

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

Question 13.
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is a solution of……..
(a) \(\frac{d^{2} y}{d x^{2}}+y x+\left(\frac{d y}{d x}\right)^{2}=0\)
(b) \(x y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0\)
(c) \(y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}+y=0\)
(d) \(x y \frac{d y}{d x}+y \frac{d^{2} y}{d x^{2}}=0\)
Answer:
(b) \(x y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0\)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 I Q13

Question 14.
The decay rate of certain substances is directly proportional to the amount present at that instant. Initially, there are 27 grams of substance and 3 hours later it is found that 8 grams left. The amount left after one more hour is……
(a) 5\(\frac{2}{3}\) grams
(b) 5\(\frac{1}{3}\) grams
(c) 5.1 grams
(d) 5 grams
Answer:
(b) 5\(\frac{1}{3}\) grams

Question 15.
If the surrounding air is kept at 20°C and the body cools from 80°C to 70°C in 5 minutes, the temperature of the body after 15 minutes will be…..
(a) 51.7°C
(b) 54.7°C
(c) 52.7°C
(d) 50.7°C
Answer:
(b) 54.7°C

(II) Solve the following:

Question 1.
Determine the order and degree of the following differential equations:
(i) \(\frac{d^{2} y}{d x^{2}}+5 \frac{d y}{d x}+y=x^{3}\)
Solution:
The given D.E. is \(\frac{d^{2} y}{d x^{2}}+5 \frac{d y}{d x}+y=x^{3}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. is of order 2 and degree 1.

(ii) \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=\sqrt[5]{1+\frac{d y}{d x}}\)
Solution:
The given D.E. is \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=\sqrt[5]{1+\frac{d y}{d x}}\)
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{2 \times 5}=1+\frac{d y}{d x}\)
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{10}=1+\frac{d y}{d x}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 10.
∴ the given D.E. is of order 3 and degree 10.

(iii) \(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
Solution:
The given D.E. is \(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
On cubing both sides, we get
\(1+\left(\frac{d y}{d x}\right)^{2}=\left(\frac{d^{2} y}{d x^{2}}\right)^{3}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 3.
∴ the given D.E. is of order 2 and degree 3.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

(iv) \(\frac{d y}{d x}=3 y+\sqrt[4]{1+5\left(\frac{d y}{d x}\right)^{2}}\)
Solution:
The given D.E. is
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q1 (iv)
This D.E. has the highest order derivative \(\frac{d y}{d x}\) with power 4.
∴ the given D.E. is of order 1 and degree 4.

(v) \(\frac{d^{4} y}{d x^{4}}+\sin \left(\frac{d y}{d x}\right)=0\)
Solution:
The given D.E. is \(\frac{d^{4} y}{d x^{4}}+\sin \left(\frac{d y}{d x}\right)=0\)
This D.E. has highest order derivative \(\frac{d^{4} y}{d x^{4}}\).
∴ order = 4
Since this D.E. cannot be expressed as a polynomial in differential coefficient, the degree is not defined.

Question 2.
In each of the following examples verify that the given function is a solution of the differential equation.
(i) \(x^{2}+y^{2}=r^{2} ; x \frac{d y}{d x}+r \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=y\)
Solution:
x2 + y2 = r2 ……. (1)
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q2 (i)
Hence, x2 + y2 = r2 is a solution of the D.E.
\(x \frac{d y}{d x}+r \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=y\)

(ii) y = eax sin bx; \(\frac{d^{2} y}{d x^{2}}-2 a \frac{d y}{d x}+\left(a^{2}+b^{2}\right) y=0\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q2 (ii)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q2 (ii).1

(iii) y = 3 cos(log x) + 4 sin(log x); \(x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0\)
Solution:
y = 3 cos(log x) + 4 sin (log x) …… (1)
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q2 (iii)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q2 (iii).1

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

(iv) xy = aex + be-x + x2; \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+x^{2}=x y+2\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q2 (iv)

(v) x2 = 2y2 log y, x2 + y2 = xy \(\frac{d x}{d y}\)
Solution:
x2 = 2y2 log y ……(1)
Differentiating both sides w.r.t. y, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q2 (v)
∴ x2 + y2 = xy \(\frac{d x}{d y}\)
Hence, x2 = 2y2 log y is a solution of the D.E.
x2 + y2 = xy \(\frac{d x}{d y}\)

Question 3.
Obtain the differential equation by eliminating the arbitrary constants from the following equations:
(i) y2 = a(b – x)(b + x)
Solution:
y2 = a(b – x)(b + x) = a(b2 – x2)
Differentiating both sides w.r.t. x, we get
2y \(\frac{d y}{d x}\) = a(0 – 2x) = -2ax
∴ y \(\frac{d y}{d x}\) = -ax …….(1)
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q3 (i)
This is the required D.E.

(ii) y = a sin(x + b)
Solution:
y = a sin(x + b)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q3 (ii)
This is the required D.E.

(iii) (y – a)2 = b(x + 4)
Solution:
(y – a)2 = b(x + 4) …….(1)
Differentiating both sides w.r.t. x, we get
\(2(y-a) \cdot \frac{d}{d x}(y-a)=b \frac{d}{d x}(x+4)\)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q3 (iii)

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

(iv) y = \(\sqrt{a \cos (\log x)+b \sin (\log x)}\)
Solution:
y = \(\sqrt{a \cos (\log x)+b \sin (\log x)}\)
∴ y2 = a cos (log x) + b sin (log x) …….(1)
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q3 (iv)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q3 (iv).1

(v) y = Ae3x+1 + Be-3x+1
Solution:
y = Ae3x+1 + Be-3x+1 …… (1)
Differentiating twice w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q3 (v)
This is the required D.E.

Question 4.
Form the differential equation of:
(i) all circles which pass through the origin and whose centres lie on X-axis.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q4 (i)
Let C (h, 0) be the centre of the circle which pass through the origin. Then radius of the circle is h.
∴ equation of the circle is (x – h)2 + (y – 0)2 = h2
∴ x2 – 2hx + h2 + y2 = h2
∴ x2 + y2 = 2hx ……..(1)
Differentiating both sides w.r.t. x, we get
2x + 2y \(\frac{d y}{d x}\) = 2h
Substituting the value of 2h in equation (1), we get
x2 + y2 = (2x + 2y \(\frac{d y}{d x}\)) x
∴ x2 + y2 = 2x2 + 2xy \(\frac{d y}{d x}\)
∴ 2xy \(\frac{d y}{d x}\) + x2 – y2 = 0
This is the required D.E.

(ii) all parabolas which have 4b as latus rectum and whose axis is parallel to Y-axis.
Solution:
Let A(h, k) be the vertex of the parabola which has 4b as latus rectum and whose axis is parallel to the Y-axis.
Then equation of the parabola is
(x – h)2 = 4b(y – k) ……. (1)
where h and k are arbitrary constants.
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q4 (ii)
Differentiating both sides of (1) w.r.t. x, we get
2(x – h). \(\frac{d}{d x}\)(x – h) = 4b . \(\frac{d}{d x}\)(y – k)
∴ 2(x – h) x (1 – 0) = 4b(\(\frac{d y}{d x}\) – 0)
∴ (x – h) = 2b \(\frac{d y}{d x}\)
Differentiating again w.r.t. x, we get
1 – 0 = 2b \(\frac{d^{2} y}{d x^{2}}\)
∴ 2b \(\frac{d^{2} y}{d x^{2}}\) – 1 = 0
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

(iii) an ellipse whose major axis is twice its minor axis.
Solution:
Let 2a and 2b be lengths of the major axis and minor axis of the ellipse.
Then 2a = 2(2b)
∴ a = 2b
∴ equation of the ellipse is
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ \(\frac{x^{2}}{(2 b)^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ \(\frac{x^{2}}{4 b^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ x2 + 4y2 = 4b2
Differentiating w.r.t. x, we get
2x + 4 × 2y \(\frac{d y}{d x}\) = 0
∴ x + 4y \(\frac{d y}{d x}\) = 0
This is the required D.E.

(iv) all the lines which are normal to the line 3x + 2y + 7 = 0.
Solution:
Slope of the line 3x – 2y + 7 = 0 is \(\frac{-3}{-2}=\frac{3}{2}\).
∴ slope of normal to this line is \(-\frac{2}{3}\)
Then the equation of the normal is
y = \(-\frac{2}{3}\)x + k, where k is an arbitrary constant.
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=-\frac{2}{3} \times 1+0\)
∴ 3\(\frac{d y}{d x}\) + 2 = 0
This is the required D.E.

(v) the hyperbola whose length of transverse and conjugate axes are half of that of the given hyperbola \(\frac{x^{2}}{16}-\frac{y^{2}}{36}=k\).
Solution:
The equation of the hyperbola is \(\frac{x^{2}}{16}-\frac{y^{2}}{36}=k\)
i.e., \(\frac{x^{2}}{16 k}-\frac{y^{2}}{36 k}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a2 = 16k, b2 = 36k
∴ a = 4√k, b = 6√k
∴ l(transverse axis) = 2a = 8√k
and l(conjugate axis) = 2b = 12√k
Let 2A and 2B be the lengths of the transverse and conjugate axes of the required hyperbola.
Then according to the given condition
2A = a = 4√k and 2B = b = 6√k
∴ A = 2√k and B = 3√k
∴ equation of the required hyperbola is
\(\frac{x^{2}}{A^{2}}-\frac{y^{2}}{B^{2}}=1\)
i.e., \(\frac{x^{2}}{4 k}-\frac{y^{2}}{9 k}=1\)
∴ 9x2 – 4y2 = 36k, where k is an arbitrary constant.
Differentiating w.r.t. x, we get
9 × 2x – 4 × 2y \(\frac{d y}{d x}\) = 0
∴ 9x – 4y \(\frac{d y}{d x}\) = 0
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

Question 5.
Solve the following differential equations:
(i) log(\(\frac{d y}{d x}\)) = 2x + 3y
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q5 (i)

(ii) \(\frac{d y}{d x}\) = x2y + y
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q5 (ii)

(iii) \(\frac{d y}{d x}=\frac{2 y-x}{2 y+x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q5 (iii)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q5 (iii).1
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q5 (iii).2

(iv) x dy = (x + y + 1) dx
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q5 (iv)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q5 (iv).1

(v) \(\frac{d y}{d x}\) + y cot x = x2 cot x + 2x
Solution:
\(\frac{d y}{d x}\) + y cot x = x cot x + 2x ……..(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q, where P = cot x and Q = x2 cot x + 2x
∴ I.F. = \(e^{\int P d x}\)
= \(e^{\int \cot x d x}\)
= \(e^{\log (\sin x)}\)
= sin x
∴ the solution of (1) is given by
y(I.F.) = ∫Q . (I.F.) dx + c
∴ y sin x = ∫(x2 cot x + 2x) sin x dx + c
∴ y sinx = ∫(x2 cot x . sin x + 2x sin x) dx + c
∴ y sinx = ∫x2 cos x dx + 2∫x sin x dx + c
∴ y sinx = x2 ∫cos x dx – ∫[\(\frac{d}{d x}\left(x^{2}\right)\) ∫cos x dx] dx + 2∫x sin x dx + c
∴ y sin x = x2 (sin x) – ∫2x(sin x) dx + 2∫x sin x dx + c
∴ y sin x = x2 sin x – 2∫x sin x dx + 2∫x sin x dx + c
∴ y sin x = x2 sin x + c
∴ y = x2 + c cosec x
This is the general solution.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

(vi) y log y = (log y2 – x) \(\frac{d y}{d x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q5 (vi)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q5 (vi).1

(vii) 4 \(\frac{d x}{d y}\) + 8x = 5e-3y
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q5 (vii)

Question 6.
Find the particular solution of the following differential equations:
(i) y(1 + log x) = (log xx) \(\frac{d y}{d x}\), when y(e) = e2
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (i)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (i).1
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (i).2

(ii) (x + 2y2) \(\frac{d y}{d x}\) = y, when x = 2, y = 1
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (ii)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (ii).1
This is the general solution.
When x = 2, y = 1, we have
2 = 2(1)2 + c(1)
∴ c = 0
∴ the particular solution is x = 2y2.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

(iii) \(\frac{d y}{d x}\) – 3y cot x = sin 2x, when y(\(\frac{\pi}{2}\)) = 2
Solution:
\(\frac{d y}{d x}\) – 3y cot x = sin 2x
\(\frac{d y}{d x}\) = (3 cot x) y = sin 2x ……..(1)
This is the linear differential equation of the form
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (iii)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (iii).1
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (iii).2

(iv) (x + y) dy + (x – y) dx = 0; when x = 1 = y
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (iv)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (iv).1
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (iv).2

(v) \(2 e^{\frac{x}{y}} d x+\left(y-2 x e^{\frac{x}{y}}\right) d y=0\), when y(0) = 1
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (v)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (v).1
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q6 (v).2

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

Question 7.
Show that the general solution of defferential equation \(\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0\) is given by (x + y + 1) = c(1 – x – y – 2xy).
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q7
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q7.1
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q7.2

Question 8.
The normal lines to a given curve at each point (x, y) on the curve pass through (2, 0). The curve passes through (2, 3). Find the equation of the curve.
Solution:
Let P(x, y) be a point on the curve y = f(x).
Then slope of the normal to the curve is \(-\frac{1}{\left(\frac{d y}{d x}\right)}\)
∴ equation of the normal is
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q8
This is the general equation of the curve.
Since, the required curve passed through the point (2, 3), we get
22 + 32 = 4(2) + c
∴ c = 5
∴ equation of the required curve is x2 + y2 = 4x + 5.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

Question 9.
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.
Solution:
Let r be the radius and V be the volume of the spherical balloon at any time t.
Then the rate of change in volume of the spherical balloon is \(\frac{d V}{d t}\) which is a constant.
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q9
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6 II Q9.1
Hence, the radius of the spherical balloon after t seconds is \((63 t+27)^{\frac{1}{3}}\) units.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

Question 10.
A person’s assets start reducing in such a way that the rate of reduction of assets is proportional to the square root of the assets existing at that moment. If the assets at the beginning are ₹ 10 lakhs and they dwindle down to ₹ 10,000 after 2 years, show that the person will be bankrupt in 2\(\frac{2}{9}\) years from the start.
Solution:
Let x be the assets of the presort at time t years.
Then the rate of reduction is \(\frac{d x}{d t}\) which is proportional to √x.
∴ \(\frac{d x}{d t}\) ∝ √x
∴ \(\frac{d x}{d t}\) = -k√x, where k > 0
∴ \(\frac{d x}{\sqrt{x}}\) = -k dt
Integrating both sides, we get
\(\int x^{-\frac{1}{2}} d x\) = -k∫dt
∴ \(\frac{x^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}\) = -kt + c
∴ 2√x = -kt + c
At the beginning, i.e. at t = 0, x = 10,00,000
2√10,00,000 = -k(0) + c
∴ c = 2000
∴ 2√x = -kt + 2000 ……..(1)
Also, when t = 2, x = 10,000
∴ 2√10000 = -k × 2 + 2000
∴ 2k = 1800
∴ k = 900
∴ (1) becomes,
∴ 2√x = -900t + 2000
When the person will be bankrupt, x = 0
∴ 0 = -900t + 2000
∴ 900t = 2000
∴ t = \(\frac{20}{9}=2 \frac{2}{9}\)
Hence, the person will be bankrupt in \(2 \frac{2}{9}\) years.

Class 12 Maharashtra State Board Maths Solution 

Differential Equations Class 12 Maths 2 Exercise 6.6 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.6 Questions and Answers.

12th Maths Part 2 Differential Equations Exercise 6.6 Questions And Answers Maharashtra Board

Question 1.
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.
Solution:
Let x be the number of bacteria in the culture at time t.
Then the rate of increase is \(\frac{d x}{d t}\) which is proportional to x.
∴ \(\frac{d x}{d t}\) ∝ x
∴ \(\frac{d x}{d t}\) = kx, where k is a constant
∴ \(\frac{d x}{x}\) = k dt
On integrating, we get
\(\int \frac{d x}{x}\) = k∫dt + c
∴ log x = kt + c
Initially, i.e. when t = 0, let x = x0
log x0 = k × 0 + c
∴ c = log x0
∴ log x = kt + log x0
∴ log x – log x0 = kt
∴ log(\(\frac{x}{x_{0}}\)) = kt ………(1)
Since the number doubles in 4 hours, i.e. when t = 4, x = 2x0
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q1
∴ the number of bacteria will be 8 times the original number in 12 hours.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6

Question 2.
If the population of a country doubles in 60 years; in how many years will it be triple (treble) under the assumption that the rate of increase is proportional to the number of inhabitants?
[Given log 2 = 0.6912, log 3 = 1.0986]
Solution:
Let P be the population at time t years.
Then \(\frac{d P}{d t}\), the rate of increase of population is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k is a constant
∴ \(\frac{d P}{P}\) = k dt
On integrating, we get
\(\int \frac{d P}{P}\) = k∫dt + c
∴ log P = kt + c
Initially i.e. when t = 0, let P = P0
∴ log P0 = k x 0 + c
∴ c = log P0
∴ log P = kt + log P0
∴ log P – log P0 = kt
∴ log(\(\frac{P}{P_{0}}\)) = kt ……(1)
Since, the population doubles in 60 years, i.e. when t = 60, P = 2P0
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q2
∴ the population becomes triple in 95.4 years (approximately).

Question 3.
If a body cools from 80°C to 50°C at room temperature of 25°C in 30 minutes, find the temperature of the body after 1 hour.
Solution:
Let θ°C be the temperature of the body at time t minutes.
The room temperature is given to be 25°C.
Then by Newton’s law of cooling, \(\frac{d \theta}{d t}\), the rate of change of temperature, is proportional to (θ – 25).
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q3
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q3.1
∴ the temperature of the body will be 36.36°C after 1 hour.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6

Question 4.
The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after 2½ hours. [Take √2 = 1.414]
Solution:
Let x be the number of bacteria at time t.
Then the rate of increase is \(\frac{d x}{d t}\) which is proportional to x.
∴ \(\frac{d x}{d t}\) ∝ x
∴ \(\frac{d x}{d t}\) = kx, where k is a constant
∴ \(\frac{d x}{x}\) = k dt
On integrating, we get
\(\int \frac{d x}{x}\) = k∫dt + c
∴ log x = kt + c
Initially, i.e. when t = 0, x = 1000
∴ log 1000 = k × 0 + c
∴ c = log 1000
∴ log x = kt + log 1000
∴ log x – log 1000 = kt
∴ log(\(\frac{x}{1000}\)) = kt ……(1)
Now, when t = 1, x = 2 × 1000 = 2000
∴ log(\(\frac{2000}{1000}\)) = k
∴ k = log 2
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q4
∴ the number of bacteria after 2½ hours = 5656.

Question 5.
The rate of disintegration of a radioactive element at any time t is proportional to its mass at that time. Find the time during which the original mass of 1.5 gm will disintegrate into its mass of 0.5 gm.
Solution:
Let m be the mass of the radioactive element at time t.
Then the rate of disintegration is \(\frac{d m}{d t}\) which is proportional to m.
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q5
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q5.1
∴ log(3)-1 = -kt
∴ -log 3 = -kt
∴ t = \(\frac{1}{k}\) log 3
∴ the original mass will disintegrate to 0.5 gm when t = \(\frac{1}{k}\) log 3

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6

Question 6.
The rate of decay of certain substances is directly proportional to the amount present at that instant. Initially, there is 25 gm of certain substance and two hours later it is found that 9 gm are left. Find the amount left after one more hour.
Solution:
Let x gm be the amount of the substance left at time t.
Then the rate of decay is \(\frac{d x}{d t}\), which is proportional to x.
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q6
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q6.1
∴ \(\frac{x}{25}=\frac{27}{125}\)
∴ x = \(\frac{27}{5}\)
∴ the amount left after 3 hours \(\frac{27}{5}\) gm.

Question 7.
Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years, the population increased from 30,000 to 40,000.
Solution:
Let P be the population of the city at time t.
Then \(\frac{d P}{d t}\), the rate of increase of population is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k is a constant.
∴ \(\frac{d P}{P}\) = k dt
On integrating, we get
\(\int \frac{1}{P} d P\) = k∫dt + c
∴ log P = kt + c
Initially, i.e. when t = 0, P = 30000
∴ log 30000 = k × 0 + c
∴ c = log 30000
∴ log P = kt + log 30000
∴ log P – log 30000 = kt
∴ log(\(\frac{P}{30000}\)) = kt …….(1)
Now, when t = 40, P = 40000
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q7
∴ the population of the city at time t = 30000\(\left(\frac{4}{3}\right)^{\frac{t}{40}}\).

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6

Question 8.
A body cools according to Newton’s law from 100°C to 60°C in 20 minutes. The temperature of the surroundings is 20°C. How long will it take to cool down to 30°C?
Solution:
Let θ°C be the temperature of the body at time t.
The temperature of the surrounding is given to be 20°C.
According to Newton’s law of cooling
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q8
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q8.1
∴ the body will cool down to 30°C in 60 minutes, i.e. in 1 hour.

Question 9.
A right circular cone has a height of 9 cm and a radius of the base of 5 cm. It is inverted and water is poured into it. If at any instant the water level rises at the rate of \(\left(\frac{\pi}{A}\right)\) cm/sec, where A is the area of the water surface
at that instant, show that the vessel will be full in 75 seconds.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q9
Let r be the radius of the water surface and h be the height of the water at time t.
∴ area of the water surface A = πr2 sq cm.
Since height of the right circular cone is 9 cm and radius of the base is 5 cm.
\(\frac{r}{h}=\frac{5}{9}\)
∴ r = \(\frac{5}{9} h\)
∴ area of water surface, i.e. A = \(\pi\left(\frac{5}{9} h\right)^{2}\)
∴ A = \(\frac{25 \pi h^{2}}{81}\) ……..(1)
The water level, i.e. the rate of change of h is \(\frac{d h}{d t}\) rises at the rate of \(\left(\frac{\pi}{A}\right)\) cm/sec.
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q9.1
∴ t = \(\frac{81 \times 9 \times 25}{3 \times 81}\) = 75
Hence, the vessel will be full in 75 seconds.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6

Question 10.
Assume that a spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is 3 mm and 1 hour later has been reduced to 2 mm, find an expression for the radius of the raindrop at any time t.
Solution:
Let r be the radius, V be the volume and S be the surface area of the spherical raindrop at time t.
Then V = \(\frac{4}{3}\)πr3 and S = 4πr2
The rate at which the raindrop evaporates is \(\frac{d V}{d t}\) which is proportional to the surface area.
∴ \(\frac{d V}{d t}\) ∝ S
∴ \(\frac{d V}{d t}\) = -kS, where k > 0 ………(1)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q10
On integrating, we get
∫dr = -k∫dt + c
∴ r = -kt + c
Initially, i.e. when t = 0, r = 3
∴ 3 = -k × 0 + c
∴ c = 3
∴ r = -kt + 3
When t = 1, r = 2
∴ 2 = -k × 1 + 3
∴ k = 1
∴ r = -t + 3
∴ r = 3 – t, where 0 ≤ t ≤ 3.
This is the required expression for the radius of the raindrop at any time t.

Question 11.
The rate of growth of the population of a city at any time t is proportional to the size of the population. For a certain city, it is found that the constant of proportionality is 0.04. Find the population of the city after 25 years, if the initial population is 10,000. [Take e = 2.7182]
Solution:
Let P be the population of the city at time t.
Then the rate of growth of population is \(\frac{d P}{d t}\) which is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k = 0.04
∴ \(\frac{d P}{d t}\) = (0.04)P
∴ \(\frac{1}{P}\) dP = (0.04)dt
On integrating, we get
\(\int \frac{1}{P} d P\) = (0.04) ∫dt + c
∴ log P = (0.04)t + c
Initially, i.e., when t = 0, P = 10000
∴ log 10000 = (0.04) × 0 + c
∴ c = log 10000
∴ log P = (0.04)t + log 10000
∴ log P – log 10000 = (0.04)t
∴ log(\(\frac{P}{10000}\)) = (0.04)t
When t = 25, then
∴ log(\(\frac{P}{10000}\)) = 0.04 × 25 = 1
∴ log(\(\frac{P}{10000}\)) = log e ……[∵ log e = 1]
∴ \(\frac{P}{10000}\) = e = 2.7182
∴ P = 2.7182 × 10000 = 27182
∴ the population of the city after 25 years will be 27,182.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6

Question 12.
Radium decomposes at a rate proportional to the amount present at any time. If p percent of the amount disappears in one year, what percent of the amount of radium will be left after 2 years?
Solution:
Let x be the amount of the radium at time t.
Then the rate of decomposition is \(\frac{d x}{d t}\) which is proportional to x.
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q12
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6 Q12.1
Hence, \(\left(10-\frac{p}{10}\right)^{2} \%\) of the amount will be left after 2 years.

Class 12 Maharashtra State Board Maths Solution 

Differential Equations Class 12 Maths 2 Exercise 6.5 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.5 Questions and Answers.

12th Maths Part 2 Differential Equations Exercise 6.5 Questions And Answers Maharashtra Board

Question 1.
Solve the following differential equations:
(i) \(\frac{d y}{d x}+\frac{y}{x}=x^{3}-3\)
Solution:
\(\frac{d y}{d x}+\frac{y}{x}=x^{3}-3\) …….(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + P . y = Q, where P = \(\frac{1}{x}\) and Q = x3 – 3
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (i)
This is the general solution.

(ii) cos2x . \(\frac{d y}{d x}\) + y = tan x
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (ii)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (ii).1

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5

(iii) (x + 2y3) \(\frac{d y}{d x}\) = y
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (iii)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (iii).1

(iv) \(\frac{d y}{d x}\) + y . sec x = tan x
Solution:
\(\frac{d y}{d x}\) + y sec x = tan x
∴ \(\frac{d y}{d x}\) + (sec x) . y = tan x ……..(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + P . y = Q, where P = sec x and Q = tan x
∴ I.F. = \(e^{\int P d x}\)
= \(e^{\int \sec x d x}\)
= \(e^{\log (\sec x+\tan x)}\)
= sec x + tan x
∴ the solution of (1) is given by
y (I.F.) = ∫Q . (I.F.) dx + c
∴ y(sec x + tan x) = ∫tan x (sec x + tan x) dx + c
∴ (sec x + tan x) . y = ∫(sec x tan x + tan2x) dx + c
∴ (sec x + tan x) . y = ∫(sec x tan x + sec2x – 1) dx + c
∴ (sec x + tan x) . y = sec x + tan x – x + c
∴ y(sec x + tan x) = sec x + tan x – x + c
This is the general solution.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5

(v) x \(\frac{d y}{d x}\) + 2y = x2 . log x
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (v)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (v).1

(vi) (x + y) \(\frac{d y}{d x}\) = 1
Solution:
(x + y) \(\frac{d y}{d x}\) = 1
∴ \(\frac{d x}{d y}\) = x + y
∴ \(\frac{d x}{d y}\) – x = y
∴ \(\frac{d x}{d y}\) + (-1) x = y ……….(1)
This is the linear differential equation of the form
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (vi)
This is the general solution.

(vii) (x + a) \(\frac{d y}{d x}\) – 3y = (x + a)5
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (vii)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (vii).1

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5

(viii) dr + (2r cot θ + sin 2θ) dθ = 0
Solution:
dr + (2r cot θ + sin 2θ) dθ = 0
∴ \(\frac{d r}{d \theta}\) + (2r cot θ + sin 2θ) = 0
∴ \(\frac{d r}{d \theta}\) + (2 cot θ)r = -sin 2θ ………(1)
This is the linear differential equation of the form dr
\(\frac{d r}{d \theta}\) + P . r = Q, where P = 2 cot θ and Q = -sin 2θ
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (viii)
This is the general solution.

(ix) y dx + (x – y2) dy = 0
Solution:
y dx + (x – y2) dy = 0
∴ y dx = -(x – y2) dy
∴ \(\frac{d x}{d y}=-\frac{\left(x-y^{2}\right)}{y}=-\frac{x}{y}+y\)
∴ \(\frac{d x}{d y}+\left(\frac{1}{y}\right) \cdot x=y\) ………(1)
This is the linear differential equation of the form
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (ix)
This is the general solution.

(x) \(\left(1-x^{2}\right) \frac{d y}{d x}+2 x y=x\left(1-x^{2}\right)^{\frac{1}{2}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (x)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (x).1

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5

(xi) \(\left(1+x^{2}\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (xi)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q1 (xi).1

Question 2.
Find the equation of the curve which passes through the origin and has the slope x + 3y – 1 at any point (x, y) on it.
Solution:
Let A(x, y) be the point on the curve y = f(x).
Then slope of the tangent to the curve at the point A is \(\frac{d y}{d x}\).
According to the given condition,
\(\frac{d y}{d x}\) = x + 3y – 1
∴ \(\frac{d y}{d x}\) – 3y = x – 1 ………(1)
This is the linear differential equation of the form
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q2
This is the general equation of the curve.
But the required curve is passing through the origin (0, 0).
∴ by putting x = 0 and y = 0 in (2), we get
0 = 2 + c
∴ c = -2
∴ from (2), the equation of the required curve is 3(x + 3y) = 2 – 2e3x i.e. 3(x + 3y) = 2 (1 – e3x).

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5

Question 3.
Find the equation of the curve passing through the point \(\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)\) having slope of the tangent to the curve at any point (x, y) is \(-\frac{4 x}{9 y}\).
Solution:
Let A(x, y) be the point on the curve y = f(x).
Then the slope of the tangent to the curve at point A is \(\frac{d y}{d x}\).
According to the given condition
\(\frac{d y}{d x}=-\frac{4 x}{9 y}\)
∴ y dy = \(-\frac{4}{9}\) x dx
Integrating both sides, we get
∫y dy= \(-\frac{4}{9}\) ∫x dx
∴ \(\frac{y^{2}}{2}=-\frac{4}{9} \cdot \frac{x^{2}}{2}+c_{1}\)
∴ 9y2 = -4x2 + 18c1
∴ 4x2 + 9y2 = c where c = 18c1
This is the general equation of the curve.
But the required curve is passing through the point \(\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)\).
∴ by putting x = \(\frac{3}{\sqrt{2}}\) and y = √2 in (1), we get
\(4\left(\frac{3}{\sqrt{2}}\right)^{2}+9(\sqrt{2})^{2}=c\)
∴ 18 + 18 = c
∴ c = 36
∴ from (1), the equation of the required curve is 4x2 + 9y2 = 36.

Question 4.
The curve passes through the point (0, 2). The sum of the coordinates of any point on the curve exceeds the slope of the tangent to the curve at any point by 5. Find the equation of the curve.
Solution:
Let A(x, y) be any point on the curve.
Then slope of the tangent to the curve at the point A is \(\frac{d y}{d x}\).
According to the given condition
x + y = \(\frac{d y}{d x}\) + 5
∴ \(\frac{d y}{d x}\) – y = x – 5 ………(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + P . y = Q, where P = -1 and Q = x – 5
∴ I.F. = \(e^{\int P d x}=e^{\int-1 d x}=e^{-x}\)
∴ the solution of (1) is given by
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q4
This is the general equation of the curve.
But the required curve is passing through the point (0, 2).
∴ by putting x = 0, y = 2 in (2), we get
2 = 4 – 0 + c
∴ c = -2
∴ from (2), the equation of the required curve is y = 4 – x – 2ex.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5

Question 5.
If the slope of the tangent to the curve at each of its point is equal to the sum of abscissa and the product of the abscissa and ordinate of the point. Also, the curve passes through the point (0, 1). Find the equation of the curve.
Solution:
Let A(x, y) be the point on the curve y = f(x).
Then slope of the tangent to the curve at the point A is \(\frac{d y}{d x}\).
According to the given condition
\(\frac{d y}{d x}\) = x + xy
∴ \(\frac{d y}{d x}\) – xy = x ……….. (1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q, where P = -x and Q = x
∴ I.F. = \(e^{\int P d x}=e^{\int-x d x}=e^{-\frac{x^{2}}{2}}\)
∴ the solution of (1) is given by
y . (I.F.) = ∫Q . (I.F.) dx + c
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5 Q5
This is the general equation of the curve.
But the required curve is passing through the point (0, 1).
∴ by putting x = 0 and y = 1 in (2), we get
1 + 1 = c
∴ c = 2
∴ from (2), the equation of the required curve is 1 + y = \(2 e^{\frac{x^{2}}{2}}\).

Class 12 Maharashtra State Board Maths Solution 

Differential Equations Class 12 Maths 2 Exercise 6.4 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.4 Questions and Answers.

12th Maths Part 2 Differential Equations Exercise 6.4 Questions And Answers Maharashtra Board

I. Solve the following differential equations:

Question 1.
\(x \sin \left(\frac{y}{x}\right) d y=\left[y \sin \left(\frac{y}{x}\right)-x\right] d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q1

Question 2.
(x2 + y2) dx – 2xy . dy = 0
Solution:
(x2 + y2) dx – 2xy dy = 0
∴ 2xy dy = (x2 + y2) dx
∴ \(\frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}\) ………(1)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q2
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q2.1
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q2.2

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4

Question 3.
\(\left(1+2 e^{\frac{x}{y}}\right)+2 e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) \frac{d y}{d x}=0\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q3
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q3.1

Question 4.
y2 dx + (xy + x2) dy = 0
Solution:
y2 dx + (xy + x2) dy = 0
∴ (xy + x2) dy = -y2 dx
∴ \(\frac{d y}{d x}=\frac{-y^{2}}{x y+x^{2}}\) ……..(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
Substituting these values in (1), we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q4
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q4.1

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4

Question 5.
(x2 – y2) dx + 2xy dy = 0
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q5
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q5.1
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q5.2

Question 6.
\(\frac{d y}{d x}+\frac{x-2 y}{2 x-y}=0\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q6
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q6.1
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q6.2

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4

Question 7.
\(x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q7
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q7.1

Question 8.
\(\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{X}{y}\right) d y=0\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q8
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q8.1

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4

Question 9.
\(y^{2}-x^{2} \frac{d y}{d x}=x y \frac{d y}{d x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q9
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q9.1

Question 10.
xy \(\frac{d y}{d x}\) = x2 + 2y2, y(1) = 0
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q10
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q10.1

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4

Question 11.
x dy + 2y · dx = 0, when x = 2, y = 1
Solution:
∴ x dy + 2y · dx = 0
∴ x dy = -2y dx
∴ \(\frac{1}{y} d y=\frac{-2}{x} d x\)
Integrating, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q11
This is the general solution.
When x = 2, y = 1, we get
4(1) = c
∴ c = 4
∴ the particular solution is x2y = 4.

Question 12.
x2 \(\frac{d y}{d x}\) = x2 + xy + y2
Solution:
x2 \(\frac{d y}{d x}\) = x2 + xy + y2
∴ \(\frac{d y}{d x}=\frac{x^{2}+x y+y^{2}}{x^{2}}\) ………(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q12

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4

Question 13.
(9x + 5y) dy + (15x + 11y) dx = 0
Solution:
(9x + 5y) dy + (15x + 11y) dx = 0
∴ (9x + 5y) dy = -(15x + 11y) dx
∴ \(\frac{d y}{d x}=\frac{-(15 x+11 y)}{9 x+5 y}\) ………(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q13
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q13.1
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q13.2

Question 14.
(x2 + 3xy + y2) dx – x2 dy = 0
Solution:
(x2 + 3xy + y2) dx – x2 dy = 0
∴ x2 dy = (x2 + 3xy + y2) dx
∴ \(\frac{d y}{d x}=\frac{x^{2}+3 x y+y^{2}}{x^{2}}\) ………(1)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q14
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q14.1

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4

Question 15.
(x2 + y2) dx – 2xy dy = 0.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q15
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q15.1
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4 Q15.2

Class 12 Maharashtra State Board Maths Solution 

Differential Equations Class 12 Maths 2 Exercise 6.3 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.3 Questions and Answers.

12th Maths Part 2 Differential Equations Exercise 6.3 Questions And Answers Maharashtra Board

Question 1.
In each of the following examples verify that the given expression is a solution of the corresponding differential equation.
(i) xy = log y + c; \(\frac{d y}{d x}=\frac{y^{2}}{1-x y}\)
Solution:
xy = log y + c
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q1 (i)
Hence, xy = log y + c is a solution of the D.E.
\(\frac{d y}{d x}=\frac{y^{2}}{1-x y^{\prime}}, x y \neq 1\)

(ii) y = (sin-1x)2 + c; (1 – x2) \(\frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=2\)
Solution:
y = (sin-1 x)2 + c …….(1)
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q1 (ii)
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q1 (ii).1

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(iii) y = e-x + Ax + B; \(e^{x} \frac{d^{2} y}{d x^{2}}=1\)
Solution:
y = e-x + Ax + B
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q1 (iii)
∴ \(e^{x} \frac{d^{2} y}{d x^{2}}=1\)
Hence, y = e-x + Ax + B is a solution of the D.E.
\(e^{x} \frac{d^{2} y}{d x^{2}}=1\)

(iv) y = xm; \(x^{2} \frac{d^{2} y}{d x^{2}}-m x \frac{d y}{d x}+m y=0\)
Solution:
y = xm
Differentiating twice w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q1 (iv)
This shows that y = xm is a solution of the D.E.
\(x^{2} \frac{d^{2} y}{d x^{2}}-m x \frac{d y}{d x}+m y=0\)

(v) y = a + \(\frac{b}{x}\); \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)
Solution:
y = a + \(\frac{b}{x}\)
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q1 (v)
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q1 (v).1
Hence, y = a + \(\frac{b}{x}\) is a solution of the D.E.
\(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(vi) y = eax; x \(\frac{d y}{d x}\) = y log y
Solution:
y = eax
log y = log eax = ax log e
log y = ax …….(1) ……..[∵ log e = 1]
Differentiating w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}\) = a × 1
∴ \(\frac{d y}{d x}\) = ay
∴ x \(\frac{d y}{d x}\) = (ax)y
∴ x \(\frac{d y}{d x}\) = y log y ………[By (1)]
Hence, y = eax is a solution of the D.E.
x \(\frac{d y}{d x}\) = y log y.

Question 2.
Solve the following differential equations.
(i) \(\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q2 (i)

(ii) log(\(\frac{d y}{d x}\)) = 2x + 3y
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q2 (ii)

(iii) y – x \(\frac{d y}{d x}\) = 0
Solution:
y – x \(\frac{d y}{d x}\) = 0
∴ x \(\frac{d y}{d x}\) = y
∴ \(\frac{1}{x} d x=\frac{1}{y} d y\)
Integrating both sides, we get
\(\int \frac{1}{x} d x=\int \frac{1}{y} d y\)
∴ log |x| = log |y| + log c
∴ log |x| = log |cy|
∴ x = cy
This is the general solution.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(iv) sec2x . tan y dx + sec2y . tan x dy = 0
Solution:
sec2x . tan y dx + sec2y . tan x dy = 0
∴ \(\frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0\)
Integrating both sides, we get
\(\int \frac{\sec ^{2} x}{\tan x} d x+\int \frac{\sec ^{2} y}{\tan y} d y=c_{1}\)
Each of these integrals is of the type
\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log |f(x)| + c
∴ the general solution is
∴ log|tan x| + log|tan y | = log c, where c1 = log c
∴ log |tan x . tan y| = log c
∴ tan x . tan y = c
This is the general solution.

(v) cos x . cos y dy – sin x . sin y dx = 0
Solution:
cos x . cos y dy – sin x . sin y dx = 0
\(\frac{\cos y}{\sin y} d y-\frac{\sin x}{\cos x} d x=0\)
Integrating both sides, we get
∫cot y dy – ∫tan x dx = c1
∴ log|sin y| – [-log|cos x|] = log c, where c1 = log c
∴ log |sin y| + log|cos x| = log c
∴ log|sin y . cos x| = log c
∴ sin y . cos x = c
This is the general solution.

(vi) \(\frac{d y}{d x}\) = -k, where k is a constant.
Solution:
\(\frac{d y}{d x}\) = -k
∴ dy = -k dx
Integrating both sides, we get
∫dy = -k∫dx
∴ y = -kx + c
This is the general solution.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(vii) \(\frac{\cos ^{2} y}{x} d y+\frac{\cos ^{2} x}{y} d x=0\)
Solution:
\(\frac{\cos ^{2} y}{x} d y+\frac{\cos ^{2} x}{y} d x=0\)
∴ y cos2y dy + x cos2x dx = 0
∴ \(x\left(\frac{1+\cos 2 x}{2}\right) d x+y\left(1+\frac{\cos 2 y}{2}\right) d y=0\)
∴ x(1 + cos 2x) dx + y(1 + cos 2y) dy = 0
∴ x dx + x cos 2x dx + y dy+ y cos 2y dy = 0
Integrating both sides, we get
∫x dx + ∫y dy + ∫x cos 2x dx + ∫y cos 2y dy = c1 ……..(1)
Using integration by parts
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q2 (vii)
Multiplying throughout by 4, this becomes
2x2 + 2y2 + 2x sin 2x + cos 2x + 2y sin 2y + cos 2y = 4c1
∴ 2(x2 + y2) + 2(x sin 2x + y sin 2y) + cos 2y + cos 2x + c = 0, where c = -4c1
This is the general solution.

(viii) \(y^{3}-\frac{d y}{d x}=x^{2} \frac{d y}{d x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q2 (viii)

(ix) 2ex+2y dx – 3 dy = 0
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q2 (ix)

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(x) \(\frac{d y}{d x}\) = ex+y + x2 ey
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q2 (x)
∴ 3ex + 3e-y + x3 = -3c1
∴ 3ex + 3e-y + x3 = c, where c = -3c1
This is the general solution.

Question 3.
For each of the following differential equations, find the particular solution satisfying the given condition:
(i) 3ex tan y dx + (1 + ex) sec2y dy = 0, when x = 0, y = π
Solution:
3ex tan y dx + (1 + ex) sec2y dy = 0
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q3 (i)

(ii) (x – y2x) dx – (y + x2y) dy = 0, when x = 2, y = 0
Solution:
(x – y2x) dx – (y + x2y) dy = 0
∴ x(1 – y2) dx – y(1 + x2) dy = 0
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q3 (ii)
When x = 2, y = 0, we have
(1 + 4)(1 – 0) = c
∴ c = 5
∴ the particular solution is (1 + x2)(1 – y2) = 5.

(iii) y(1 + log x) \(\frac{d x}{d y}\) – x log x = 0, y = e2, when x = e
Solution:
y(1 + log x) \(\frac{d x}{d y}\) – x log x = 0
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q3 (iii)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q3 (iii).1

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(iv) (ey + 1) cos x + ey sin x \(\frac{d y}{d x}\) = 0, when x = \(\frac{\pi}{6}\), y = 0
Solution:
(ey + 1) cos x + ey sin x \(\frac{d y}{d x}\) = 0
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q3 (iv)
\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log|f(x)| + c
∴ from (1), the general solution is
log|sin x| + log|ey + 1| = log c, where c1 = log c
∴ log|sin x . (ey + 1)| = log c
∴ sin x . (ey + 1) = c
When x = \(\frac{\pi}{4}\), y = 0, we get
\(\left(\sin \frac{\pi}{4}\right)\left(e^{0}+1\right)=c\)
∴ c = \(\frac{1}{\sqrt{2}}\)(1 + 1) = √2
∴ the particular solution is sin x . (ey + 1) = √2

(v) (x + 1) \(\frac{d y}{d x}\) – 1 = 2e-y, y = 0, when x = 1
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q3 (v)
This is the general solution.
Now, y = 0, when x = 1
∴ 2 + e0 = c(1 + 1)
∴ 3 = 2c
∴ c = \(\frac{3}{2}\)
∴ the particular solution is 2 + ey = \(\frac{3}{2}\) (x + 1)
∴ 2(2 + ey) = 3(x + 1).

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(vi) cos(\(\frac{d y}{d x}\)) = a, a ∈ R, y (0) = 2
Solution:
cos(\(\frac{d y}{d x}\)) = a
∴ \(\frac{d y}{d x}\) = cos-1 a
∴ dy = (cos-1 a) dx
Integrating both sides, we get
∫dy = (cos-1 a) ∫dx
∴ y = (cos-1 a) x + c
∴ y = x cos-1 a + c
This is the general solution.
Now, y(0) = 2, i.e. y = 2,
when x = 0, 2 = 0 + c
∴ c = 2
∴ the particular solution is
∴ y = x cos-1 a + 2
∴ y – 2 = x cos-1 a
∴ \(\frac{y-2}{x}\) = cos-1a
∴ cos(\(\frac{y-2}{x}\)) = a

Question 4.
Reduce each of the following differential equations to the variable separable form and hence solve:
(i) \(\frac{d y}{d x}\) = cos(x + y)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q4 (i)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q4 (i).1

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(ii) (x – y)2 \(\frac{d y}{d x}\) = a2
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q4 (ii)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q4 (ii).1

(iii) x + y \(\frac{d y}{d x}\) = sec(x2 + y2)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q4 (iii)
Integrating both sides, we get
∫cos u du = 2 ∫dx
∴ sin u = 2x + c
∴ sin(x2 + y2) = 2x + c
This is the general solution.

(iv) cos2(x – 2y) = 1 – 2 \(\frac{d y}{d x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q4 (iv)
Integrating both sides, we get
∫dx = ∫sec2u du
∴ x = tan u + c
∴ x = tan(x – 2y) + c
This is the general solution.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(v) (2x – 2y + 3) dx – (x – y + 1) dy = 0, when x = 0, y = 1
Solution:
(2x – 2y + 3) dx – (x – y + 1) dy = 0
∴ (x – y + 1) dy = (2x – 2y + 3) dx
∴ \(\frac{d y}{d x}=\frac{2(x-y)+3}{(x-y)+1}\) ………(1)
Put x – y = u, Then \(1-\frac{d y}{d x}=\frac{d u}{d x}\)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q4 (v)
∴ u – log|u + 2| = -x + c
∴ x – y – log|x – y + 2| = -x + c
∴ (2x – y) – log|x – y + 2| = c
This is the general solution.
Now, y = 1, when x = 0.
∴ (0 – 1) – log|0 – 1 + 2| = c
∴ -1 – o = c
∴ c = -1
∴ the particular solution is
(2x – y) – log|x – y + 2| = -1
∴ (2x – y) – log|x – y + 2| + 1 = 0

Class 12 Maharashtra State Board Maths Solution 

Differential Equations Class 12 Maths 2 Exercise 6.2 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.2 Questions and Answers.

12th Maths Part 2 Differential Equations Exercise 6.2 Questions And Answers Maharashtra Board

Question 1.
Obtain the differential equation by eliminating the arbitrary constants from the following equations:
(i) x3 + y3 = 4ax
Solution:
x3 + y3 = 4ax ……..(1)
Differentiating both sides w.r.t. x, we get
3x2 + 3y2 \(\frac{d y}{d x}\) = 4a × 1
∴ 3x2 + 3y2 \(\frac{d y}{d x}\) = 4a
Substituting the value of 4a in (1), we get
x3 + y3 = (3x2 + 3y2 \(\frac{d y}{d x}\)) x
∴ x3 + y3 = 3x3 + 3xy2 \(\frac{d y}{d x}\)
∴ 2x3 + 3xy2 \(\frac{d y}{d x}\) – y3 = 0
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

(ii) Ax2 + By2 = 1
Solution:
Ax2 + By2 = 1
Differentiating both sides w.r.t. x, we get
A × 2x + B × 2y \(\frac{d y}{d x}\) = 0
∴ Ax + By \(\frac{d y}{d x}\) = 0 ……..(1)
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (ii)
Substituting the value of A in (1), we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (ii).1
This is the required D.E.

Alternative Method:
Ax2 + By2 = 1 ……..(1)
Differentiating both sides w.r.t. x, we get
A × 2x + B × 2y \(\frac{d y}{d x}\) = 0
∴ Ax + By \(\frac{d y}{d x}\) = 0 ……….(2)
Differentiating again w.r.t. x, we get,
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (ii).2
The equations (1), (2) and (3) are consistent in A and B.
∴ determinant of their consistency is zero.
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (ii).3
This is the required D.E.

(iii) y = A cos(log x) + B sin(log x)
Solution:
y = A cos(log x) + B sin (log x) ……. (1)
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (iii)

(iv) y2 = (x + c)3
Solution:
y2 = (x + c)3
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (iv)
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

(v) y = Ae5x + Be-5x
Solution:
y = Ae5x + Be-5x ……….(1)
Differentiating twice w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (v)
This is the required D.E.

(vi) (y – a)2 = 4(x – b)
Solution:
(y – a)2 = 4(x – b)
Differentiating both sides w.r.t. x, we get
2(y – a) . \(\frac{d}{d x}\)(y – a) = 4 \(\frac{d}{d x}\)(x – b)
∴ 2(y – a) . (\(\frac{d y}{d x}\) – 0) = 4(1 – 0)
∴ 2(y – a) \(\frac{d y}{d x}\) = 4
∴ (y – a) \(\frac{d y}{d x}\) = 2 ……..(1)
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (vi)
This is the required D.E.

(vii) y = a + \(\frac{a}{x}\)
Solution:
y = a + \(\frac{a}{x}\)
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (vii)
Substituting the value of a in (1), we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (vii).1
This is the required D.E.

(viii) y = c1e2x + c2e5x
Solution:
y = c1e2x + c2e5x ………(1)
Differentiating twice w.r.t. x, we get
\(\frac{d y}{d x}\) = c1e2x × 2 + c2e5x × 5
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (viii)
The equations (1), (2) and (3) are consistent in c1e2x and c2e5x
∴ determinant of their consistency is zero.
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (viii).1
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

Alternative Method:
y = c1e2x + c2e5x
Dividing both sides by e5x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (viii).2
This is the required D.E.

(ix) c1x3 + c2y2 = 5.
Solution:
c1x3 + c2y2 = 5 ……….(1)
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (ix)
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (ix).1
The equations (1), (2) and (3) in c1, c2 are consistent.
∴ determinant of their consistency is zero.
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (ix).2
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

(x) y = e-2x(A cos x + B sin x)
Solution:
y = e-2x(A cos x + B sin x)
∴ e2x . y = A cos x + B sin x ………(1)
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (x)
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (x).1
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (x).2
This is the required D.E.

Question 2.
Form the differential equation of family of lines having intercepts a and b on the coordinate axes respectively.
Solution:
The equation of the line having intercepts a and b on the coordinate axes respectively, is
\(\frac{x}{a}+\frac{y}{b}=1\) ……….(1)
where a and b are arbitrary constants.
[For different values of a and b, we get, different lines. Hence (1) is the equation of family of lines.]
Differentiating (1) w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q2
Differentiating again w.r.t. x, we get \(\frac{d^{2} y}{d x^{2}}=0\)
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

Question 3.
Find the differential equation all parabolas having length of latus rectum 4a and axis is parallel to the X-axis.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q3
Let A(h, k) be the vertex of the parabola whose length of latus rectum is 4a.
Then the equation of the parabola is (y – k)2 = 4a (x – h), where h and k are arbitrary constants.
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q3.1
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q3.2
This is the required D.E.

Question 4.
Find the differential equation of the ellipse whose major axis is twice its minor axis.
Solution:
Let 2a and 2b be lengths of major axis and minor axis of the ellipse.
Then 2a = 2(2b)
∴ a = 2b
∴ equation of the ellipse is
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
i.e., \(\frac{x^{2}}{(2 b)^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ \(\frac{x^{2}}{4 b^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ x2 + 4y2 = 4b2
Differentiating w.r.t. x, we get
2x + 4 × 2y \(\frac{d y}{d x}\) = 0
∴ x + 4y \(\frac{d y}{d x}\) = 0
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

Question 5.
Form the differential equation of family of lines parallel to the line 2x + 3y + 4 = 0.
Solution:
The equation of the line parallel to the line 2x + 3y + 4 = 0 is 2x + 3y + c = 0, where c is an arbitrary constant.
Differentiating w.r.t. x, we get
2 × 1 + 3 \(\frac{d y}{d x}\) + 0 = 0
∴ 3 \(\frac{d y}{d x}\) + 2 = 0
This is the required D.E.

Question 6.
Find the differential equation of all circles having radius 9 and centre at point (h, k).
Solution:
Equation of the circle having radius 9 and centre at point (h, k) is
(x – h)2 + (y – k)2 = 81 …… (1)
where h and k are arbitrary constant.
Differentiating (1) w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q6
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q6.1
From (2), x – h = -(y – k) \(\frac{d y}{d x}\)
Substituting the value of (x – h) in (1), we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q6.2
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q6.3
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

Question 7.
Form the differential equation of all parabolas whose axis is the X-axis.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q7
The equation of the parbola whose axis is the X-axis is
y2 = 4a(x – h) …… (1)
where a and h are arbitrary constants.
Differentiating (1) w.r.t. x, we get
2y \(\frac{d y}{d x}\) = 4a(1 – 0)
∴ y \(\frac{d y}{d x}\) = 2a
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q7.1
This is the required D.E.

Class 12 Maharashtra State Board Maths Solution 

Differential Equations Class 12 Maths 2 Exercise 6.1 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.1 Questions and Answers.

12th Maths Part 2 Differential Equations Exercise 6.1 Questions And Answers Maharashtra Board

1. Determine the order and degree of each of the following differential equations:

Question (i).
\(\frac{d y}{d x^{2}}+X\left(\frac{d y}{d x}\right)+y=2 \sin x\)
Solution:
The given D.E. is \(\frac{d y}{d x^{2}}+X\left(\frac{d y}{d x}\right)+y=2 \sin x\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. is of order 2 and degree 1.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1

Question (ii).
\(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
Solution:
The given D.E. is \(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
On cubing both sides, we get
\(1+\left(\frac{d y}{d x}\right)^{2}=\left(\frac{d^{2} y}{d x^{2}}\right)^{3}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 3.
∴ the given D.E. is of order 2 and degree 3.

Question (iii).
\(\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}\)
Solution:
The given D.E. is \(\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}\)
∴ \(\left(\frac{d y}{d x}\right)^{2}\) = 2 sin x + 3
This D.E. has highest order derivative \(\frac{d y}{d x}\) with power 2.
∴ the given D.E. is of order 1 and degree 2.

Question (iv).
\(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x=\sqrt{1+\frac{d^{3} y}{d x^{3}}}\)
Solution:
The given D.E. is \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x=\sqrt{1+\frac{d^{3} y}{d x^{3}}}\)
On squaring both sides, we get
\(\left(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x\right)^{2}=1+\frac{d^{3} y}{d x^{3}}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 1.
∴ the given D.E. has order 3 and degree 1.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1

Question (v).
\(\frac{d^{2} y}{d t^{2}}+\left(\frac{d y}{d t}\right)^{2}+7 x+5=0\)
Solution:
The given D.E. is \(\frac{d^{2} y}{d t^{2}}+\left(\frac{d y}{d t}\right)^{2}+7 x+5=0\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. has order 2 and degree 1.

Question (vi).
(y”‘)2 + 3y” + 3xy’ + 5y = 0
Solution:
The given D.E. is (y”‘)2 + 3y” + 3xy’ + 5y = 0
This can be written as:
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}+3 \frac{d^{2} y}{d x^{2}}+3 x \frac{d y}{d x}+5 y=0\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 2.
∴ The given D.E. has order 3 and degree 2.

Question (vii).
\(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0\)
Solution:
The given D.E. is \(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\)
∴ order = 2
Since this D.E. cannot be expressed as a polynomial in differential coefficients, the degree is not defined.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1

Question (viii).
\(\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}=8 \frac{d^{2} y}{d x^{2}}\)
Solution:
The given D.E. is \(\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}=8 \frac{d^{2} y}{d x^{2}}\)
On squaring both sides, we get
\(\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}=8^{2} \cdot\left(\frac{d^{2} y}{d x^{2}}\right)^{2}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 2.
∴ the given D.E. has order 2 and degree 2.

Question (ix).
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{2}} \cdot\left(\frac{d y}{d x}\right)^{\frac{1}{3}}=20\)
Solution:
The given D.E. is \(\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{2}} \cdot\left(\frac{d y}{d x}\right)^{\frac{1}{3}}=20\)
∴ \(\left(\frac{d^{3} y}{d x^{3}}\right)^{3} \cdot\left(\frac{d y}{d x}\right)^{2}=20^{6}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 3.
∴ the given D.E. has order 3 and degree 3.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1

Question (x).
\(x+\frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}\)
Solution:
The given D.E. is \(x+\frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}\)
On squaring both sides, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1 (x)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. has order 2 and degree 1.

Class 12 Maharashtra State Board Maths Solution