Let y=y(x) be the solution of the differential equation (2x _ex)(dy)/(dx)+2y=(3)/(x) _ex,x…

Question

Let $y=y(x)$ be the solution of the differential equation $(2x{\log }_{e}x)\frac{dy}{dx}+2y=\frac{3}{x}{\log }_{e}x,x>0$ and $y({e}^{-1})=0$. Then, $y(e)$ is equal to

Accepted Answer

$\begin{gathered} (2x\log x)\frac{dy}{dx}+2y=\frac{3}{x}\log x \ \frac{dy}{dx}+\frac{y}{x\log x}=\frac{3}{2{x}^{2}} \ IF={e}^{\int \frac{1}{x\log x}}={e}^{\log (\log x)}=\log x \ y	imes \log x=\int \log x	imes \frac{3}{2{x}^{2}}dx+C \ y\log x=\frac{3}{2}[\frac{-\log x}{x}-\frac{1}{x}]+C \ y({e}^{-1})=0 \ \Rightarrow 0=\frac{3}{2}[e-e]+C \ C=0 \ y\log x=\frac{-3}{2}[\frac{\log x+1}{x}] \ y(e)	o y(e)	imes 1=\frac{-3}{2}[\frac{1+1}{e}]=-\frac{3}{e} \end{gathered}$

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