If y = y(x) is the solution of the differential equation d y d x + 2 y = sin ( 2 x ) , y ( 0 ) = 3 4 then y ( 8 ) is equal to [2024]

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Solution

Q.
If 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 [2024]

1 $e^{\frac{- π}{4}}$

2 $e^{\frac{π}{4}}$

3 $e^{\frac{π}{8}}$

4 $e^{\frac{- π}{8}}$

Ans.
(1)

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

$IF = e^{\int 2 d x} = e^{2 x}$

$∴$ General solution of given differential equation is,

$y \times e^{2 x} = \int e^{2 x} \sin 2 x d x$

$\Rightarrow y e^{2 x} = \frac{e^{2 x}}{8} (2 \sin 2 x - 2 \cos 2 x) + c [∵ \int e^{a x} \sin b x d x = \frac{e^{a x}}{a^{2} + b^{2}} (a \sin b x - b \cos b x) + c]$

$\Rightarrow y e^{2 x} = \frac{e^{2 x}}{4} (\sin 2 x - \cos 2 x) + c$

Now, $y (0) = \frac{3}{4}$

$\Rightarrow \frac{3}{4} = \frac{1}{4} (- 1) + c \Rightarrow c = \frac{3}{4} + \frac{1}{4} = 1$

$∴ y (\frac{π}{8}) = \frac{1}{4} (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) + \frac{1}{e^{π / 4}} = e^{- π / 4}$

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