(a) Find the particular solution of the differential equation $$\dfrac{dy}{dx}-2xy=3x^{2}e^{x^{2}};\quad y(0)=5$$ OR (b) Solve the following differential equation: $$x^{2}\,dy+y(x+y)\,dx=0$$

Differential Equations — CBSE Class 12 · 2024 Board Question with Solution

(a) Find the particular solution of the differential equation

\frac{d y}{d x} - 2 x y = 3 x^{2} e^{x^{2}}; y (0) = 5

OR
(b) Solve the following differential equation:

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

Answer: (a) $y = x^{3} e^{x^{2}} + 5 e^{x^{2}}$ ; (b) $x^{2} y = C (y + 2 x)$

(a) Linear in

y

with

P = - 2 x

Q = 3 x^{2} e^{x^{2}}

I.F. = e^{\int - 2 x d x} = e^{- x^{2}}

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

y = (x^{3} + C) e^{x^{2}}

. Using

y (0) = 5

5 = C

. Hence

y = x^{3} e^{x^{2}} + 5 e^{x^{2}}

(b)

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

is homogeneous. Put

y = v x

\frac{d y}{d x} = v + x \frac{d v}{d x}

v + x \frac{d v}{d x} = - v (1 + v) \Rightarrow x \frac{d v}{d x} = - (v^{2} + 2 v)

\int \frac{d v}{v ( v + 2 )} = - \int \frac{d x}{x} \Rightarrow \frac{1}{2} lo g \frac{v}{v + 2} = - lo g x + lo g C

Back-substitute

v = \frac{y}{x}

x^{2} y = C (y + 2 x)

CBSE Class 12 Board · 2024 · Subjective