• # question_answer The solution of the differential equation$\frac{dy}{dx}+\frac{y}{x}\log y=\frac{y}{{{x}^{2}}}{{(\log \,\,y)}^{2}}$ is A) $y=\log ({{x}^{2}}+cx)$ B) $\log \,\,y=x\left( c{{x}^{2}}+\frac{1}{2} \right)$ C) $x=\log \,\,y\left( c{{x}^{2}}+\frac{1}{2} \right)$ D) None of these

 [c] Divide the equation by $y{{(log\,y)}^{2}}$ $\frac{1}{y{{(log\,\,y)}^{2}}}\frac{dy}{dx}+\frac{1}{\log y}\cdot \frac{1}{x}=\frac{1}{{{x}^{2}}}$ Put $\frac{1}{\log \,y}=z\Rightarrow \frac{-1}{y{{(log\,\,y)}^{2}}}\frac{dy}{dx}=\frac{dz}{dx}$ Thus we get, $-\frac{dz}{dx}+\frac{1}{x}\cdot z=\frac{1}{{{x}^{2}}}$, linear in z $\Rightarrow \frac{dz}{dx}+\left( -\frac{1}{x} \right)z=-\frac{1}{{{x}^{2}}}$ I.F. $={{e}^{-\int{\frac{1}{x}dx}}}={{e}^{-\log \,\,x}}=\frac{1}{x}$ $\therefore$ The solution is,$z\left( \frac{1}{x} \right)=\int{\frac{-1}{{{x}^{2}}}\left( \frac{1}{x} \right)dx+c}$ $\Rightarrow \frac{1}{\log y}\left( \frac{1}{x} \right)=\frac{-{{x}^{-2}}}{-2}+c\Rightarrow x=\log y\left( c{{x}^{2}}+\frac{1}{2} \right)$