Question

Let $y=y\left(x\right)$ be the solution of the differential equation $\frac{dy}{dx}=2(y+2\sin x-5)x-2\cos x$ such that $y\left(0\right)=7$. Then $y\left(\pi \right)$ is equal to

• A. $e^{{\pi }^2}+5$
• B. $2e^{{\pi }^2}+5$
• C. $7e^{{\pi }^2}+5$
• D. $3e^{{\pi }^2}+5$

Answer 1

The correct option is B $2e^{{\pi }^2}+5$

Given: $\frac{dy}{dx}=2(y+2\sin x-5)x-2\cos x$
$\Rightarrow \frac{dy}{dx}-2xy=4x\sin x-2\cos x-10x$
$I.F.=e^{\int -2xdx}=e^{-x^2}$
Now, the general solution:
$y\cdot e^{-x^2}=\int e^{-x^2}(4x\sin x-2\cos x-10x)dx+C$
$\Rightarrow y{e}^{-x^2}=4\int e^{-x^2}\left(x\right)\left(\sin x\right)dx-2\int \left(\cos x\right)e^{-x^2}dx+5\int (-2x{e}^{-x^2})dx+C$
$\Rightarrow y{e}^{-x^2}=-2\sin x\cdot e^{-x^2}+5e^{-x^2}+C$
Put $x=0\Rightarrow 7=0+5+C$
$\Rightarrow C=2$
$\therefore y{e}^{-x^2}=-2\sin x\cdot e^{-x^2}+5e^{-x^2}+2$
$\Rightarrow y=-2\sin x+5+2e^{x^2}$
Put $x=\pi \Rightarrow y=5+2e^{{\pi }^2}$