Question

Solve the following :
$(x^2-4xy-2y^2)dx+(y^2-4xy-2x^2)dy=0$

Answer 1

$\frac{dy}{dx}=\frac{-x^2+4xy+2y^2}{y^2-4xy-2x^2}=\frac{-1+\frac{4y}{x}+\frac{2y^2}{x^2}}{\frac{y^2}{x^2}-\frac{4y}{x}-2}$

Let $\frac{y}{x}=r\Rightarrow y=tx$

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

$t+x\frac{dt}{dx}=\frac{-1+4t+2t^2}{t^2-4t-2}-t\Rightarrow \frac{xdt}{dx}=-\frac{t^3+6t^x+6t-1}{t^2-4t-2}$

$\Rightarrow -\frac{t^3-1+6t(t+1)}{t^2-4t-2}=x\frac{dt}{dx}$

$\Rightarrow -\frac{(t+1)(t^2+1-t}{t^2-4t-2}=\frac{xdt}{dx}\Rightarrow \frac{t^2-4t-2}{(t-1)(t^2-t+1)}dt=-\frac{dx}{x}$

$\Rightarrow \frac{t^2-t+1-3t-3}{(t+1)(t^2-t+1)}dt=-\frac{dx}{x}\Rightarrow \int \frac{dt}{t+1}-\frac{3dt}{t^2-t+1}=\int -\frac{Dx}{x}$

$\Rightarrow ln|t+1|-3\int \frac{dt}{{(t-\frac{1}{2})}^2+\frac{3}{4}}=-ln|x|+C$

$\Rightarrow ln|t+1|-\frac{3\times 2}{\sqrt{3}}{\tan }^{-1}\left|\frac{t-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right|=-ln|x|+C$

$\Rightarrow ln|t+1|-2\sqrt{3}{\tan }^{-1}\left|\frac{2t-1}{\sqrt{3}}\right|=-ln|x|+C$

$\Rightarrow ln\left|\frac{y+x}{x}\right|-2\sqrt{3}{\tan }^{-1}\left(\frac{2y-x}{\sqrt{3}x}\right)=-ln|x|+C$