Question:
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 $y(0)=7$. Then $y(\pi)$ is equal to :
Correct Option: 1
Solution:
$\frac{\mathrm{dy}}{\mathrm{dx}}-2 \mathrm{xy}=2(2 \sin \mathrm{x}-5) \mathrm{x}-2 \cos \mathrm{x}$
$I F=e^{-x^{2}}$
so, $y \cdot e^{-x^{2}}=\int e^{-x^{2}}(2 x(2 \sin x-5)-2 \cos x) d x$
$\Rightarrow y \cdot e^{-x^{2}}=e^{-x^{2}}(5-2 \sin x)+c$
$\Rightarrow y=5-2 \sin x+c \cdot e^{x^{2}}$
Given at $x=0, y=7$
$\Rightarrow 7=5+\mathrm{c} \Rightarrow \mathrm{c}=2$
So, $y=5-2 \sin x+2 e^{x^{2}}$
Now at $x=\pi$,
$y=5+2 e^{\pi^{2}}$
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