Answer to Question #143651 in Differential Equations for Nikhil Singh

Question

Solve the differential equation:
(x-y^2)dx+2xy.dy=0

Answer 1

First of all we can remark that $2ydy = d(y^2)$ and therefore our equation can be written as follows :

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

With the substitution $u=y^2$ we can write it as :

$(x-u)dx+xdu =0$

By "dividing" (strictly speaking, it is not a division) we find :

$x-u+xu'=0$

$u-xu'=x$

This is a linear non-homogeneous differential equation of 1st order, so we will first find the general solution of the corresponding homogeneous equation (it is an equation with separable variables, so it is pretty straightforward) :

$u-xu'=0$

$u=Cx, C\in\mathbb{R}$

Now we need to find the particular solution of the non-homogeneous equation. We can find it with the method of variation of a constant (we treat C as a function of x) :

$Cx - x (Cx)'=x$

$Cx - x(C'x+C)=x$

$-x^2C'=x$

$C'=-\frac{1}{x}$

$C=-ln(x)+A, A\in\mathbb{R}$

So our general solution is :

$u=-x\ln(x)+Ax$

Now by going from our substituion we find :

$y=\pm\sqrt{Ax-x\ln(x)},A\in\mathbb{R}, x\in(0; e^A]$ (the whole solution can be either only $\sqrt{Ax-x\ln(x)},$ or $-\sqrt{Ax-x\ln(x)}$ for continuity reasons, as $\sqrt{Ax-x\ln(x)}=0$ only for $x=e^A$ )