Solve: $ (3xy-2ay^2)dx + (x^2-2axy)dy = 0 $

Question

Solve:

$ (3xy-2ay^2)dx + (x^2-2axy)dy = 0 $

This is a homogeneous differential equation. So why can't I solve it in the following manner?

Answer 1

The given differential equation is

$$(3xy−2ay^2)dx+(x^2−2axy)dy=0\tag{1}$$

where

$$M=3xy−2ay^2,~N=x^2−2axy$$

therefore

$$\frac{\partial M}{\partial y}=3x−4ay,~\frac{\partial N}{\partial x}=2x−2ay$$

as $\frac{\partial M}{\partial y}\neq \frac{\partial N}{\partial x}$, the differential equation is not exact. However, we can proceed by

$$\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N}=\frac{3x−4ay-2x+2ay}{x^2−2axy}=\frac{x-2ay}{x(x-2ay)}=\frac{1}{x}=f(x)$$

so that our integrating factor is

$$\text{I.F.}=\text{exp}\left(\int f(x)dx\right)=\text{exp}\left(\int \frac{1}{x} dx\right)=x$$

if we then multiply $(1)$ by our integrating factor of $x$

$$(3x^2y−2axy^2)dx+(x^3−2ax^2y)dy=0\tag{2}$$

we now see that $(2)$ is exact since

$$M=3x^2y−2axy^2,~N=x^3−2ax^2y$$

hence

$$\frac{\partial M}{\partial y}=3x^2−4axy,~\frac{\partial N}{\partial x}=3x^2−4axy$$

and $\frac{\partial M}{\partial y}= \frac{\partial N}{\partial x}$, which can be solved as an exact differential equation. Your approach doesn't work since

$$\text{I.F.}=\frac{1}{M_x+N_y}$$

provided $M_x+N_y$ consists of only one term. When you simplify, you have more than one term.