Differential Equations — CBSE Class 12 · 2012 Board Question with Solution

(a) Find the particular solution of the following differential equation:

\frac{d y}{d x} = 1 + x^{2} + y^{2} + x^{2} y^{2}, given that y = 1 when x = 0.

(b) Find the particular solution of the following differential equation:

(x + 1) \frac{d y}{d x} = 2 e^{- y} - 1; y = 0 when x = 0.

x y \frac{d y}{d x} = (x + 2) (y + 2); y = - 1 when x = 1.

Answer: (a) $tan^{- 1} y = x + \frac{x ^{3}}{3} + \frac{π}{4}$ ; (b) $(x + 1) (2 - e^{y}) = 1$ ; (c) $y - 2 lo g ∣ y + 2∣ = x + 2 lo g ∣ x ∣ - 2$

Solution

(a) Factor the right side:

\frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})

\frac{d y}{1 + y ^{2}} = (1 + x^{2}) d x \Rightarrow tan^{- 1} y = x + \frac{x ^{3}}{3} + C

y = 1, x = 0

gives

C = \frac{π}{4}

, so

tan^{- 1} y = x + \frac{x ^{3}}{3} + \frac{π}{4}

.
(b)

\frac{d y}{2 e ^{- y} - 1} = \frac{d x}{x + 1} \Rightarrow \int \frac{e ^{y}}{2 - e ^{y}} d y = lo g ∣ x + 1∣ + C

. With

z = 2 - e^{y}

- lo g ∣2 - e^{y} ∣ = lo g ∣ x + 1∣ + C \Rightarrow (x + 1) (2 - e^{y}) = k

x = 0, y = 0

gives

k = 1

, so

(x + 1) (2 - e^{y}) = 1

.
(c)

\frac{y}{y + 2} d y = \frac{x + 2}{x} d x \Rightarrow \int (1 - \frac{2}{y + 2}) d y = \int (1 + \frac{2}{x}) d x

y - 2 lo g ∣ y + 2∣ = x + 2 lo g ∣ x ∣ + C

x = 1, y = - 1

gives

C = - 2

, so

y - 2 lo g ∣ y + 2∣ = x + 2 lo g ∣ x ∣ - 2

CBSE Class 12 Board · 2012 · Subjective