Let y = y(x)


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 :

  1. $2 e^{\pi^{2}}+5$

  2. $e^{\pi^{2}}+5$

  3. $3 e^{\pi^{2}}+5$

  4. $7 \mathrm{e}^{\pi^{2}}+5$

Correct Option: 1


$\frac{\mathrm{dy}}{\mathrm{dx}}-2 \mathrm{xy}=2(2 \sin \mathrm{x}-5) \mathrm{x}-2 \cos \mathrm{x}$

$I F=e^{-x^{2}}$

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 $x=\pi$,

$y=5+2 e^{\pi^{2}}$

Leave a comment