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

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

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

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^{-1}x)^{2} + c; (1 – x^{2}) \(\frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=2\)

Solution:

y = (sin^{-1} x)^{2} + c …….(1)

Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

(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

∴ \(e^{x} \frac{d^{2} y}{d x^{2}}=1\)

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 = x^{m}; \(x^{2} \frac{d^{2} y}{d x^{2}}-m x \frac{d y}{d x}+m y=0\)

Solution:

y = x^{m}

Differentiating twice w.r.t. x, we get

This shows that y = x^{m} 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

Differentiating again w.r.t. x, we get

Hence, y = a + \(\frac{b}{x}\) is a solution of the D.E.

\(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)

(vi) y = e^{ax}; x \(\frac{d y}{d x}\) = y log y

Solution:

y = e^{ax}

log y = log e^{ax} = 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 = e^{ax} 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:

(ii) log(\(\frac{d y}{d x}\)) = 2x + 3y

Solution:

(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.

(iv) sec^{2}x . tan y dx + sec^{2}y . tan x dy = 0

Solution:

sec^{2}x . tan y dx + sec^{2}y . 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 c_{1} = 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 = c_{1}

∴ log|sin y| – [-log|cos x|] = log c, where c_{1} = 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.

(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 cos^{2}y dy + x cos^{2}x 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 = c_{1} ……..(1)

Using integration by parts

Multiplying throughout by 4, this becomes

2x^{2} + 2y^{2} + 2x sin 2x + cos 2x + 2y sin 2y + cos 2y = 4c_{1}

∴ 2(x^{2} + y^{2}) + 2(x sin 2x + y sin 2y) + cos 2y + cos 2x + c = 0, where c = -4c_{1}

This is the general solution.

(viii) \(y^{3}-\frac{d y}{d x}=x^{2} \frac{d y}{d x}\)

Solution:

(ix) 2e^{x+2y} dx – 3 dy = 0

Solution:

(x) \(\frac{d y}{d x}\) = e^{x+y} + x^{2} e^{y}

Solution:

∴ 3e^{x} + 3e^{-y} + x^{3} = -3c_{1}

∴ 3e^{x} + 3e^{-y} + x^{3} = c, where c = -3c_{1}

This is the general solution.

Question 3.

For each of the following differential equations, find the particular solution satisfying the given condition:

(i) 3e^{x} tan y dx + (1 + e^{x}) sec^{2}y dy = 0, when x = 0, y = π

Solution:

3e^{x} tan y dx + (1 + e^{x}) sec^{2}y dy = 0

(ii) (x – y^{2}x) dx – (y + x^{2}y) dy = 0, when x = 2, y = 0

Solution:

(x – y^{2}x) dx – (y + x^{2}y) dy = 0

∴ x(1 – y^{2}) dx – y(1 + x^{2}) dy = 0

When x = 2, y = 0, we have

(1 + 4)(1 – 0) = c

∴ c = 5

∴ the particular solution is (1 + x^{2})(1 – y^{2}) = 5.

(iii) y(1 + log x) \(\frac{d x}{d y}\) – x log x = 0, y = e^{2}, when x = e

Solution:

y(1 + log x) \(\frac{d x}{d y}\) – x log x = 0

(iv) (e^{y} + 1) cos x + e^{y} sin x \(\frac{d y}{d x}\) = 0, when x = \(\frac{\pi}{6}\), y = 0

Solution:

(e^{y} + 1) cos x + e^{y} sin x \(\frac{d y}{d x}\) = 0

\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log|f(x)| + c

∴ from (1), the general solution is

log|sin x| + log|e^{y} + 1| = log c, where c_{1} = log c

∴ log|sin x . (e^{y} + 1)| = log c

∴ sin x . (e^{y} + 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 . (e^{y} + 1) = √2

(v) (x + 1) \(\frac{d y}{d x}\) – 1 = 2e^{-y}, y = 0, when x = 1

Solution:

This is the general solution.

Now, y = 0, when x = 1

∴ 2 + e^{0} = c(1 + 1)

∴ 3 = 2c

∴ c = \(\frac{3}{2}\)

∴ the particular solution is 2 + e^{y} = \(\frac{3}{2}\) (x + 1)

∴ 2(2 + e^{y}) = 3(x + 1).

(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^{-1}a

∴ 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:

(ii) (x – y)^{2} \(\frac{d y}{d x}\) = a^{2}

Solution:

(iii) x + y \(\frac{d y}{d x}\) = sec(x^{2} + y^{2})

Solution:

Integrating both sides, we get

∫cos u du = 2 ∫dx

∴ sin u = 2x + c

∴ sin(x^{2} + y^{2}) = 2x + c

This is the general solution.

(iv) cos^{2}(x – 2y) = 1 – 2 \(\frac{d y}{d x}\)

Solution:

Integrating both sides, we get

∫dx = ∫sec^{2}u du

∴ x = tan u + c

∴ x = tan(x – 2y) + c

This is the general solution.

(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 \(1-\frac{d y}{d x}=\frac{d u}{d x}\)

∴ 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