Question

The solution of the differential equation
$y \frac{dy}{dx}=x\left[\frac{y^{2}}{x^{2}}+\frac{\varphi\left(\frac{y^{2}}{x^{2}}\right)}{\varphi'\left(\frac{y^{2}}{x^{2}}\right)}\right]$
is (where $c$ is a constant)

• A. $\varphi \left(\frac{y^{2}}{x^{2}}\right)=cx$
• B. $x\varphi \left(\frac{y^{2}}{x^{2}}\right)=c$
• C. $\varphi \left(\frac{y^{2}}{x^{2}}\right)=cx^{2}$
• D. $x^{2}\varphi \left(\frac{y^{2}}{x^{2}}\right)=c$

Answer 1

Answer: (C)

Given differential equation can be rewritten as
$\frac{d y}{d x}=\frac{y}{x}+\frac{x \phi\left(\frac{y^{2}}{x^{2}}\right)}{y \phi^{\prime}\left(\frac{y^{2}}{x^{2}}\right)}$
Put $y=v x$
$ \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore $ Given equation becomes,
$v+x \frac{d v}{d x}=\frac{v x}{x}+\frac{x \phi\left(\frac{v^{2} x^{2}}{x^{2}}\right)}{v x \phi^{'}\left(\frac{v^{2} x^{2}}{x^{2}}\right)}$
$\Rightarrow x \frac{d v}{d x}=\frac{\phi\left(v^{2}\right)}{v \phi^{'}\left(v^{2}\right)}$
$\Rightarrow \frac{v \phi^{'}\left(v^{2}\right)}{\phi\left(v^{2}\right)} d v=\frac{d x}{x}$
On integrating both sides, we get
$\frac{1}{2} \log \phi\left(v^{2}\right)=\log x+\log c_{1}$
$\Rightarrow \log \phi\left(v^{2}\right)=2 \log x c_{1}$
$\Rightarrow \phi\left(v^{2}\right)=\left(x c_{1}\right)^{2} $
$\Rightarrow \phi\left(\frac{y^{2}}{x^{2}}\right)=x^{2} c_{1}^{2}$
$\Rightarrow \phi\left(\frac{y^{2}}{x^{2}}\right)=x^{2} c$
[put $\left.c_{1}^{2}=c\right]$