Non-linear first order differential equations in product form I am trying to solve the following equation $$y'^3-yy'^2-x^2y'+x^2y=0,\;y(0)=1 $$ $$y'^2(y' - y) -x^2(y' - y) =0$$ $$ (y'^2 - x^2)(y' -y)=0\implies (y' + x)(y' - x)(y' - y)=0$$
My question is that the product of the solution of these three equations $$ y=(1-x^4/4)e^x $$ can be taken as the general solution.
 A: No. You can check that the function doesn't satisfy your equation. The general solutions are the union of solutions of each equation.
If you want to join them in one equation, you can do it as follows.
If $y_i = f_i(x)$ are solutions of each of the equation, then those solutions satisfy $(y - f_1(x))(y - f_2(x))(y - f_3(x)) = 0$
A: Because of the continuity of the function and derivative, you can divide a solution $y$ into segments so that one of the factors is zero on each segment. The factors that are "active" can only change if two of them are zero at the same time. Thus you can change from $(y'-x)=0$ to $(y'+x)=0$ only at $x=0$, and from either of them to $(y'-y)=0$ only at points where $\pm x=y(x)=y'(x)$.
So if $x=0$ is inside such a segment, then the solution is locally equal to $y(x)=1\pm x^2/2$ or $y(x)=e^x$. If one changes between the two first solutions at $x=0$, one gets locally $y(x)=1\pm x|x|/2$. There are points where $-x=1-x^2/2$, so that changes from this solution variant to an exponential segment are possible.
