Answer to Question #222892 in Differential Equations for Hizaam

Question

solve differential equation dy/dx=(2x^2+y^2)/(-2xy+3y^2)

Answer 1

Let us solve the differential equation $\frac{dy}{dx}=\frac{2x^2+y^2}{-2xy+3y^2},$ which is equivalent to $\frac{dy}{dx}=\frac{2+(\frac{y}{x})^2}{-2\frac{y}{x}+3(\frac{y}{x})^2}.$ Let us use the transformation $y=ux.$ Then $\frac{dy}{dx}=\frac{du}{dx}x+u.$ We get the equation $\frac{du}{dx}x+u=\frac{2+u^2}{-2u+3u^2},$ which is equivalent to $\frac{du}{dx}x=\frac{2+u^2}{-2u+3u^2}-u,$ and hence to $\frac{du}{dx}x=\frac{2+u^2+2u^2-3u^3}{-2u+3u^2}.$ Then we have the equation $\frac{-2u+3u^2}{2+3u^2-3u^3}du=\frac{dx}{x}.$ It follows that

$\int\frac{-2u+3u^2}{2+3u^2-3u^3}du=\int\frac{dx}{x}$

$-\frac{1}{3}\int\frac{6u-9u^2}{2+3u^2-3u^3}du=\int\frac{dx}{x}$

$-\frac{1}{3}\int\frac{d(2+3u^2-3u^3)}{2+3u^2-3u^3}=\int\frac{dx}{x}$

$\ln|x|=-\frac{1}{3}\ln|2+3u^2-3u^3|+\ln|C|$

$\ln|x|+\frac{1}{3}\ln|2+3u^2-3u^3|=\ln|C|$

$\ln|x(2+3u^2-3u^3)^{\frac{1}{3}}|=\ln|C|$

$x\sqrt[3]{2+3u^2-3u^3}=C$

$x\sqrt[3]{2+3(\frac{y}{x})^2-3(\frac{y}{x})^3}=C$

$\sqrt[3]{2x^3+3y^2x-3y^3}=C$

It follows that the general solution of the differential equation $\frac{dy}{dx}=\frac{2x^2+y^2}{-2xy+3y^2}$ is

$2x^3+3y^2x-3y^3=C.$