Question

The solution of the differential equation $\frac{dy}{dx}=\frac{y}{x}+\frac{\varphi \left(\frac{y}{x}\right)}{\varphi \left(\frac{y}{x}\right)}$ is

• A. $x\varphi \left(\frac{y}{x}\right)=k$
• B. $\varphi \left(\frac{y}{x}\right)=kx$
• C. $y\varphi \left(\frac{y}{x}\right)=k$
• D. $\varphi \left(\frac{y}{x}\right)=ky$
• E. $\varphi \left(\frac{y}{x}\right)=k{e}^{y/x}$

Answer 1

Answer: (B)

The given differential equation can be written as
$\frac{dy}{dx}-\frac{y}{x}=\frac{\varphi \left(\frac{y}{x}\right)}{\varphi \left(\frac{y}{x}\right)}$
$\Rightarrow$ $x\varphi \left(\frac{y}{x}\right)\left(\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}\right)=\varphi \left(\frac{y}{x}\right)$
$\Rightarrow$ $\frac{\varphi \left(\frac{y}{x}\right)\left(\frac{x\frac{dy}{dx}-y}{x^2}\right)}{\varphi \left(\frac{y}{x}\right)}=\frac{1}{x}$
$\Rightarrow$ $\int \frac{\varphi \left(\frac{y}{x}\right)d\left(\frac{y}{x}\right)}{\varphi \left(\frac{y}{x}\right)}=\int \frac{1}{x}dx+\log k$
$\Rightarrow$ $\log \varphi \left(\frac{y}{x}\right)=\log x+\log k$
$\Rightarrow$ $\varphi \left(\frac{y}{x}\right)=kx$