JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Differential Equations

y = y (x)

is the solution of the differential equation

\frac{d y}{d x} + 2 y = sin (2 x), y (0) = \frac{3}{4}

, then

y (\frac{π}{8})

is equal to:

Answer: (C)

Solution

Linear ODE

\frac{d y}{d x} + 2 y = sin 2 x

, I.F.

= e^{2 x}

y e^{2 x} = \int e^{2 x} sin 2 x d x = \frac{e ^{2 x} ( 2 sin 2 x - 2 cos 2 x )}{8} + C .

x = 0, y = \frac{3}{4}

\frac{3}{4} = \frac{2 ( 0 ) - 2 ( 1 )}{8} + C \Rightarrow C = 1

. So

y = \frac{sin 2 x - cos 2 x}{4} + e^{- 2 x}

.
At

x = \frac{π}{8}

sin \frac{π}{4} = cos \frac{π}{4}

, so the bracket is 0 and

y = e^{- π /4} .

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