Question

Solve the following :
$\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}$

Answer 1

We have,

$\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}$

This is a homogeneous differential equation then,

Put,

$y=vx$

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Now,

$\frac{dy}{dx}+\frac{vx}{x}=\frac{v^2{x}^2}{x^2}$

$v+x\frac{dv}{dx}+\frac{vx}{x}=\frac{v^2{x}^2}{x^2}$

$v+x\frac{dv}{dx}+v=v^2$

$x\frac{dv}{dx}+2v=v^2$

$xdv+2vdx=v^{2}dx$

$xdv=v^{2}dx-2vdx$

$xdv=(v^2-2v)dx$

$\frac{dv}{(v^2-2v)}=\frac{dx}{x}$

$\frac{dv}{(v^2-2v+1-1)}=\frac{dx}{x}$

$\frac{dv}{{(v-1)}^2-1^2}=\frac{dx}{x}$

$\text{On}\text{intigrating}\text{and}\text{we}\text{get,}$

$\int \frac{dv}{{(v-1)}^2-1^2}=\int \frac{dx}{x}\therefore \int \frac{1}{x^2-a^2}dx=\frac{1}{2}\ln \left(\frac{x-a}{x+a}\right)$

$\frac{1}{2}\ln \left(\frac{v-1-1}{v-1+1}\right)=\ln x+C$

$\frac{1}{2}\ln \left(\frac{\frac{y}{x}-2}{\frac{y}{x}}\right)=\ln x+C\therefore v=\frac{y}{x}$

$\frac{1}{2}\ln \left(\frac{\frac{y-2x}{x}}{\frac{y}{x}}\right)=\ln x+C$

$\frac{1}{2}\ln \left(\frac{y-2x}{y}\right)=\ln x+C$

$\ln {\left(\frac{y-2x}{y}\right)}^{\frac{1}{2}}=\ln x+C$

$\ln \sqrt{y-2x}-\ln y=\ln x+C$

Hence, this is the answer.