Question

If $\frac{dy}{dx}-y=y^2(\sin x+\cos x)$ with $y\left(0\right)=1$, then the value of $y\left(\pi \right)$ is

• A. $e^{\pi }$
• B. $-e^{\pi }$
• C. $e^{-\pi }$
• D. $-e^{-\pi }$

Answer 1

The correct option is A $e^{\pi }$

$\frac{dy}{dx}-y=y^2(\sin x+\cos x)$
$\Rightarrow \frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}=\sin x+\cos x$
Let $-\frac{1}{y}=t\Rightarrow \frac{1}{y^2}\frac{dy}{dx}=\frac{dt}{dx}$

Now, $\frac{dt}{dx}+t=\sin x+\cos x$
I.F. $=e^x$
$\therefore t\cdot e^x=\int e^x(\sin x+\cos x)dx$
$\Rightarrow -\frac{1}{y}{e}^x=e^x\sin x+C$
Given, if $x=0$, then $y=1$
$\Rightarrow C=-1$
Hence, equation of the curve is
$-\frac{e^x}{y}=e^x\sin x-1$
If $x=\pi$,
then $-\frac{e^{\pi }}{y}=-1$
$\Rightarrow y=e^{\pi }$