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CBSE Class 12 · 2016 · Differential EquationsLong Answer

Find the particular solution of the differential equation

\frac{d y}{d x} = - \frac{x + y cos x}{1 + sin x}

, given that

y = 1

when

x = 0

.

Answer: $y = \frac{2 - x ^{2}}{2 ( 1 + sin x )}$

Solution

Rewrite in linear form:

\frac{d y}{d x} + \frac{cos x}{1 + sin x} y = - \frac{x}{1 + sin x}

I.F. = e^{\int \frac{c o s x}{1 + s i n x} d x} = e^{l o g (1 + s i n x)} = 1 + sin x

y (1 + sin x) = \int - x d x = - \frac{x ^{2}}{2} + C

Apply

y = 1, x = 0

:

1 = C

. Hence

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

:

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

CBSE Class 12 Board · 2016 · Subjective

Practise more: all CBSE board PYQs on Differential Equations →

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