# Prove the following

Question:

If $y \frac{d y}{d x}=x\left[\frac{y^{2}}{x^{2}}+\frac{\phi\left(\frac{y^{2}}{x^{2}}\right)}{\phi^{\prime}\left(\frac{y^{2}}{x^{2}}\right)}\right], x>0, \phi>0$, and $y(1)=-1$,

then $\phi\left(\frac{y^{2}}{4}\right)$ is equal to:

1. $4 \phi(2)$

2. $4 \phi(1)$

3. $2 \phi$ (1)

4. $\phi$ (1)

Correct Option: , 2

Solution:

Let, $\mathrm{y}=\mathrm{tx}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{t}+\mathrm{x} \frac{\mathrm{dt}}{\mathrm{dx}}$

$\therefore \operatorname{tx}\left(\mathrm{t}+\mathrm{x} \frac{\mathrm{dt}}{\mathrm{dx}}\right)=\mathrm{x}\left(\mathrm{t}^{2}+\frac{\varphi\left(\mathrm{t}^{2}\right)}{\varphi^{\prime}\left(\mathrm{t}^{2}\right)}\right)$

$\mathrm{t}^{2}+\mathrm{xt} \frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{t}^{2}+\frac{\varphi\left(\mathrm{t}^{2}\right)}{\varphi^{\prime}\left(\mathrm{t}^{2}\right)}$

$\int \frac{t \varphi^{\prime}\left(t^{2}\right)}{\varphi\left(t^{2}\right)} d t=\int \frac{d x}{x}$

Let $\varphi\left(\mathrm{t}^{2}\right)=\mathrm{p}$

$\therefore \varphi^{\prime}\left(\mathrm{t}^{2}\right) 2 \mathrm{tdt}=\mathrm{dp}$

$\Rightarrow \int \frac{\mathrm{dy}}{2 \mathrm{p}}=\int \frac{\mathrm{dx}}{\mathrm{x}}$

$\frac{1}{2} \ell \mathrm{n} \varphi\left(\mathrm{t}^{2}\right)=\ell \mathrm{n} \mathrm{x}+\ell \mathrm{nc}$

$\varphi\left(\mathrm{t}^{2}\right)=\mathrm{x}^{2} \mathrm{k}$

$\therefore \varphi^{\prime}\left(\mathrm{t}^{2}\right) 2 \mathrm{tdt}=\mathrm{dp}$

$\Rightarrow \int \frac{d y}{2 p}=\int \frac{d x}{x}$

$\frac{1}{2} \ln \varphi\left(\mathrm{t}^{2}\right)=\ln \mathrm{x}+\ell \mathrm{nc}$

$\varphi\left(\mathrm{t}^{2}\right)=\mathrm{x}^{2} \mathrm{k}$

$\varphi\left(\frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}\right)=\mathrm{kx}^{2}, \varphi(1)=\mathrm{k}$

$\varphi\left(\frac{\mathrm{y}^{2}}{4}\right)=4 \varphi(1)$