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.


2 Answers 2


$$x\,dx+y\,dy = \frac{a^2(x\,dy-y\,dx)}{x^2+y^2}\tag 1$$ 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 $\mu(x,y)$.


In my mind, your question is about integrating factor.

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.

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