JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

\frac{d y}{d x} = 2 (y + 2 sin x - 5) x - 2 cos x

such that that

y (0) = 7

. Then

y (π)

is equal to

Answer: (B)

Solution

The equation is

\frac{d y}{d x} = 2 x (y + 2 sin x - 5) - 2 cos x

. Put

u = y + 2 sin x - 5

, so

\frac{d u}{d x} = \frac{d y}{d x} + 2 cos x = 2 xu

. Then

\frac{d u}{u} = 2 x d x

gives

u = C e^{x^{2}}

. From

y (0) = 7

u (0) = 7 - 5 = 2

, so

C = 2

and

u = 2 e^{x^{2}}

. Hence

y = 2 e^{x^{2}} - 2 sin x + 5

, and

y (π) = 2 e^{π^{2}} - 2 sin π + 5 = 2 e^{π^{2}} + 5.

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