Question

Let $f\left(x\right)$ be a differentiable function such that $f^{\prime }\left(x\right)+f\left(x\right)=4x{e}^{-x}\sin 2x$ and $f\left(0\right)=0$. If $\underset{n\to \infty }{lim}\sum _{k=1}^{n}f\left(k\pi \right)=\frac{-P\pi e^{\pi }}{(e^{\pi }-1{)}^2}$, then $\frac{9}{4}P$ is

Answer 1

$\frac{dy}{dx}+y=4x{e}^{-x}\sin 2x$
$P\left(x\right)=1$
$I.F.=e^{\int P\left(x\right)dx}=e^x$
$e^x.y=\int e^x\left(4x{e}^{-x}\sin 2x\right)dx$
$\Rightarrow e^x.y=4\int x\sin 2xdx$
$\Rightarrow e^x.y=\sin 2x-2x\cos 2x+c$
$\because f\left(0\right)=0\Rightarrow c=0$
$\Rightarrow f\left(x\right)=y=(\sin 2x-2x\cos 2x)e^{-x}$
$\Rightarrow f\left(k\pi \right)=(\sin 2k\pi -2k\pi \cos 2k\pi )e^{-k\pi }$
$\Rightarrow f\left(k\pi \right)=-2k\pi e^{-k\pi }=-2\pi S,[S=k{e}^{-k\pi }]$
$\underset{n\to \infty }{lim}\sum _{k=1}^{n}f\left(k\pi \right)=-2\pi \times \underset{n\to \infty }{lim}S$

$\begin{array}{cc}S=& 1.e^{-\pi }+2.e^{-2\pi }+3.e^{-3\pi }+\cdots \\ e^{-\pi }S=& e^{-2\pi }+2.e^{-3\pi }+\cdots \\ (1-e^{-\pi })S=& e^{-\pi }+e^{-2\pi }+e^{-3\pi }+\cdots \infty \end{array}$
$(1-e^{-\pi })S=\frac{e^{-\pi }}{1-e^{-\pi }}$
$\Rightarrow S=\frac{e^{-\pi }}{(1-e^{-\pi }{)}^2}=\frac{e^{\pi }}{(e^{\pi }-1{)}^2}$
$\therefore \sum _{k=1}^{\infty }f\left(k\pi \right)=-\frac{2\pi e^{\pi }}{(e^{\pi }-1{)}^2}=\frac{-P\pi e^{\pi }}{(e^{\pi }-1{)}^2}$
$\Rightarrow P=2$
$\therefore \frac{9}{4}\times P=\frac{9}{4}\times 2=4.50$