Differential equation

Question

$$(x^2+y^2)dx+(x^2-xy)dy=0$$

Solving this differential equation. I think it can be solved by using integration factor but I couldn't find its reasonable integration factor. How should I solve this?

Answer 1

You want to solve $$x^2+y(x)^2+\left(x^2-xy(x)\right)y'(x)=0.$$

Now let $y(x)=x w(x)$ i.e. $y'(x)=w(x)+xw'(x)$ and plugging this in we get

\begin{align} x^2+x^2 w(x)^2+\left(x^2-x^2w(x)\right)\left( w(x)+xw'(x) \right)&=0 \\ x^2 \left(xw'(x)+w(x)-xw'(x)w(x)+1 \right)&=0 \\ w'(x) \frac{w-1}{w+1}&=\frac{1}{x} \\ \int w'(x) \frac{w-1}{w+1}&=\int \frac{1}{x} \\ -2\ln(w+1)+w&=\ln(x)+C\end{align}

where we assumed $x \neq 0$; For this case you just get $y=0$. Now solving for $w$ (with the Lambert-W-function) and plugging this into $y=xw$ yields your solution.

Answer 2

rewrite your equation in the form $$x^2\left(1+\left(\frac{y}{x}\right)^2\right)+x^2\left(1+\frac{y}{x}\right)\frac{dy}{dx}=0$$ and set $$\frac{y}{x}=u$$

Answer 3

This is a nonlinear ODE, but of homogeneous nonlinearity. Using the tranformation $$ w=\frac{y}{x}, $$ such equations become Separable Equations, i.e., of the form $u'=f(x)g(u)$.