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CBSE Class 12 · 2020 · Differential EquationsLong Answer

(a) Find the general solution of the differential equation

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

OR
(b) Find the particular solution of the differential equation

x \frac{d y}{d x} = y - x tan (\frac{y}{x})

, given that

y = \frac{π}{4}

at

x = 1

.

Solution

(a) Divide by

y e^{y}

to get a linear equation in

x

:

\frac{d x}{d y} - \frac{2}{y} x = y^{2} e^{- y}

I.F. = e^{- 2 \int d y / y} = \frac{1}{y ^{2}}

\frac{x}{y ^{2}} = \int y^{2} e^{- y} \cdot \frac{1}{y ^{2}} d y = \int e^{- y} d y = - e^{- y} + C

x = - y^{2} e^{- y} + C y^{2}

(b) Homogeneous: with

y = v x

,

cot v d v = - \frac{d x}{x}

gives

x sin \frac{y}{x} = C

. Apply

y = \frac{π}{4}, x = 1

:

sin \frac{π}{4} = C \Rightarrow C = \frac{1}{2}

:

x sin (\frac{y}{x}) = \frac{1}{2}

CBSE Class 12 Board · 2020 · Subjective

Practise more: all CBSE board PYQs on Differential Equations →

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