Mathematics - Differential Equation Question with Solution | TestHub

Given $\frac{dy}{dx}=\frac{y}{x}+\varphi \left(\frac{y}{x}\right)$ Let $\left(\frac{y}{x}\right)=v$ so that $y=xv$ or $\frac{dy}{dx}=x\frac{dv}{dx}+v$ from (1) \& (2), $x\frac{dv}{dx}+v=v+\varphi \left(\frac{1}{v}\right)$ or, $\frac{dv}{\varphi \left(\frac{1}{v}\right)}=\frac{dx}{x}$ Integrating both sides, we get $\int \frac{dx}{x}=\int \frac{dv}{\varphi \left(\frac{1}{v}\right)}\Rightarrow \ln x+c=\int \frac{dv}{\varphi \left(\frac{1}{v}\right)}$ (where $\mathrm{c}$ being constant of integration) But, given $\mathrm{y}=\frac{x}{\ln |cx|}$ is the general solution so that $\frac{x}{y}=\frac{1}{v}=\ln |cx|=\int \frac{dv}{\varphi \left(\frac{1}{v}\right)}$ Differentiating w.r.t $v$ both sides, we get $\varphi \left(\frac{1}{v}\right)=\frac{-1}{v^2}\Rightarrow \varphi \left(\frac{x}{y}\right)=-\frac{y^2}{x^2}$ when $\frac{x}{y}=2$ i.e. $\varphi (2)$ $$=-{\left(\frac{y}{x}\right)}^2=-{\left(\frac{1}{2}\right)}^2=\left(\frac{-1}{4}\right)$$