Question

If the differentiable equation $\frac{dy}{dx}-y=y^2(\sin x+\cos x)$ with $y(0)=1$ then $y(\pi )$ has the value equal to

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

Answer 1

Answer: (A)

$\frac{dy}{dx}-y=y^2(\sin x+\cos x)$
$\frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}=\sin x+\cos x;$ Let $-\frac{1}{y}=t\Rightarrow \frac{1}{y^2}\frac{dy}{dx}=\frac{dt}{dx}$ $\frac{dt}{dx}+t=\sin x+\cos x$
I.F. $e^x$
$\therefore t\cdot e^x=\int e^x(\sin x+\cos x)dx;-\frac{1}{y}{e}^x=e^x\sin x+C$
if $x=0,y=1\Rightarrow C=-1$
$-\frac{e^x}{y}=e^x\sin x-1$
if $x=\pi$ then $-\frac{e^x}{y}=-1\Rightarrow y=e^{\pi }$