Question

The solution of the differential equation $\frac{dy}{dx}=\frac{y}{x}+\frac{\mathrm{\varphi }\left({\displaystyle \frac{y}{x}}\right)}{\mathrm{\varphi }\text{'}\left({\displaystyle \frac{y}{x}}\right)}$ is
(a) $\mathrm{\varphi }\left(\frac{y}{x}\right)=kx$

(b) $x\mathrm{\varphi }\left(\frac{y}{x}\right)=k$

(c) $\mathrm{\varphi }\left(\frac{y}{x}\right)=ky$

(d) $y\mathrm{\varphi }\left(\frac{y}{x}\right)=k$

Answer 1

$\left(\mathrm{a}\right)\mathrm{\varphi }\left(\frac{y}{x}\right)=kx$


$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=\frac{y}{x}+\frac{\mathrm{\varphi }\left({\displaystyle \frac{y}{x}}\right)}{\mathrm{\varphi }\text{'}\left({\displaystyle \frac{y}{x}}\right)} \\ \mathrm{Let}y=vx \\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx} \\ \therefore v+x\frac{dv}{dx}=v+\frac{\mathrm{\varphi }\left({\displaystyle v}\right)}{\mathrm{\varphi }\text{'}\left({\displaystyle v}\right)} \\ \Rightarrow x\frac{dv}{dx}=\frac{\mathrm{\varphi }\left({\displaystyle v}\right)}{\mathrm{\varphi }\text{'}\left({\displaystyle v}\right)} \\ \Rightarrow \frac{\mathrm{\varphi }\left({\displaystyle v}\right)}{\mathrm{\varphi }\text{'}\left({\displaystyle v}\right)}dv=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{\mathrm{\varphi }\text{'}\left({\displaystyle v}\right)}{\mathrm{\varphi }\left({\displaystyle v}\right)}dv=\int \frac{1}{x}dx \\ \Rightarrow \log \left|\mathrm{\varphi }\left(\mathrm{v}\right)\right|=\log \left|x\right|+\log k \\ \Rightarrow \log \left|\mathrm{\varphi }\left(\frac{y}{x}\right)\right|-\log \left|\mathrm{x}\right|=\log k \\ \Rightarrow \log \left|\frac{\mathrm{\varphi }\left(\frac{y}{x}\right)}{x}\right|=\log k \\ \Rightarrow \frac{\mathrm{\varphi }\left(\frac{y}{x}\right)}{x}=k \\ \Rightarrow \mathrm{\varphi }\left(\frac{y}{x}\right)=\mathrm{kx} \end{gathered}$$