Question

Show that the given differential equation is homogeneous and solve them
$\frac{dy}{dx}=\frac{x^2-2y^2+xy}{x^2}$

Answer 1

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

$=1-2(\frac{y}{x}{)}^2+(\frac{y}{x})$

Clearly it is a homogeneous equation

let $\frac{y}{x}=V$ or $y=vx$

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

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

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

$\int \frac{dv}{1-2v^2}-\int \frac{dx}{x}$

$\frac{1}{2}\int \frac{dv}{\frac{1}{2}-v^2}=\log x+c$

$\frac{1}{2}\int \frac{dv}{(\frac{1}{\sqrt{2}}{)}^2-v^2}=\log x+c$

Use formula

$\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log |\frac{a+x}{a-x}|+c$

$\therefore$ we got

$\frac{1}{2}\times \frac{1}{2\times \frac{1}{\sqrt{2}}}\log |\frac{\frac{1}{\sqrt{2}+v}}{\frac{1}{\sqrt{2}-v}}|=\log x+c$

$\frac{1}{2\sqrt{2}}\log \left|\frac{\frac{1}{\sqrt{2}}+\frac{y}{x}}{\frac{1}{\sqrt{2}}-\frac{y}{x}}\right|=\log x+c$

$\frac{1}{2\sqrt{2}}\log \left|\frac{x+y\sqrt{2}}{x-y\sqrt{2}}\right|=\log x+c$