If $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4{\mathrm{y}}^2+4\mathrm{xy}+{\mathrm{x}}^2}{4{\mathrm{x}}^2}$ and y(1)=0 then $\mathrm{y}\left({\mathrm{e}}^{\frac{\mathrm{\pi }}{2}}\right)=$

$\begin{array}{l}\frac{dy}{dx}=\frac{4y^2+4xy+x^2}{4x^2}\\ \frac{dy}{dx}={\left(\frac{y}{x}\right)}^2+\left(\frac{y}{x}\right)+\frac{1}{4}\\ Puty=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\\ \Rightarrow v+x\frac{dv}{dx}=v^2+v+\frac{1}{4}\\ \Rightarrow x\frac{dv}{dx}=v^2+\frac{1}{4}\\ \Rightarrow \int \frac{1}{v^2+{\left(\frac{1}{2}\right)}^2}dv=\int \frac{1}{x}dx\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{\tan }^{-1}\left(\frac{v}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)=\log x+c\\ \Rightarrow 2{\tan }^{-1}\left(\frac{2y}{x}\right)=\log x+c\\ Putx=1\\ \Rightarrow 2{\tan }^{-1}\left(2y\left(1\right)\right)=\log x+c\\ \Rightarrow 2{\tan }^{-1}\left(0\right)=0+c\\ \Rightarrow c=0\\ \Rightarrow 2{\tan }^{-1}\left(\frac{2y}{x}\right)=\log x\end{array}$

$\begin{array}{l}\Rightarrow \frac{2y}{x}=\tan \left(\frac{\log x}{2}\right)\\ \Rightarrow 2y=x\tan \left(\frac{\log x}{2}\right)\\ Putx=e^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ \Rightarrow 2y\left(e^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)=e^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\tan \left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)\\ \Rightarrow y\left(e^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)=\frac{1}{2}{e}^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}$