# Doubt regarding a differential equation.

There's a problem in a book regarding exact differential equation :
Solve : $$x\,dx+y\,dy = \frac{a^2(x\,dy-y\,dx)}{x^2+y^2}$$ The author further proceeds to rearrange above in the form $$M\,dx+N\,dy=0$$ where $$M=x+\frac{a^2y}{x^2+y^2} ; N= y-\frac{a^2x}{x^2+y^2}$$ Which further implies $$M_y=N_x=\frac{a^2(x^2-y^2)}{(x^2+y^2)^2}$$ This is the required condition for the given equation to be exact and the solution is obtained using standard formula.

But, what I did is as follows : $$x\,dx+y\,dy = \frac{a^2(x\,dy-y\,dx)}{x^2+y^2}$$ $$\to (x^2+y^2)x\,dx + (x^2+y^2)y\,dy=a^2(x\,dy -y\,dx)$$ Therefore $$(x^3+xy^2+a^2y)\,dx + (y^3+yx^2-a^2x)\,dy =0$$ Comparing it with the equation $$M\,dx+N\,dy=0$$ we get, $$M=x^3+xy^2+a^2y; N=y^3+yx^2-a^2x$$ But $$M_y=2xy+a^2; N_x=2xy-a^2$$ Evidently, $$M_y\neq N_x$$.
So what am I doing wrong here?
At first I thought I shouldn't/couldn't just simply multiply both sides of the given equation with the denominator ($$x^2+y^2$$) of right hand side if it's zero, but since as it's already in denominator isn't it understood that it isn't/can't be zero ??
Any help is really appreciated.

$$x\,dx+y\,dy = \frac{a^2(x\,dy-y\,dx)}{x^2+y^2}\tag 1$$ As you correctly found : Equation $$(1)$$ is an exact ODE.

You transformed Eq.$$(1)$$ into Eq.$$(2)$$: $$(x^3+xy^2+a^2y)\,dx + (y^3+yx^2-a^2x)\,dy =0\tag 2$$

Again you correctly found : Equation $$(2)$$ is not an exact ODE.

This is not contradictory because Eqs.$$(1)$$ and $$(2)$$ are not the same ODE.

In order to transform Eq.$$(2)$$ into an exact ODE one have to multiply it with an "integration factor", say $$\mu(x,y)$$.

One find that $$\mu(x,y)=\frac{1}{x^2+y^2}$$. $$\mu(x,y)\Big( (x^3+xy^2+a^2y)\,dx + (y^3+yx^2-a^2x)\,dy\Big) =0\tag 3$$ $$\frac{x^3+xy^2+a^2y}{x^2+y^2}\,dx + \frac{y^3+yx^2-a^2x}{x^2+y^2}\,dy =0\quad\text{is exact.}$$ In fact Eq.$$(3)$$ is equivalent to Eq.$$(1)$$ thanks to the correct $$\mu(x,y)$$.

You have checked your equation $$(x^3+xy^2+a^2y)dx+(y^3+yx^2-a^2x)dy=0$$ is not exact.

That's why we need integrating factor. Here the integrating factor is exactly $$\frac{1}{x^2+y^2}$$.

By the integrating factor, we make your equation an exact equation deliberately, then we can solve the ODE by exact equations.

If $$x^2+y^2=0$$, then this ODE is not $$\textbf{well-defined}$$.

p.s. It's my first time to answer a problem. Perhaps I misunderstand your problem, please tell me and I would try to correct my answer.