We are asked to find the solution to this equation:

$(x^{3}-2xy)\frac{dy}{dx} + x+2y^{2}= 0$

I tried the following:

$(1)$ $$(x^{3}-2xy)dy +(x+2y^{2})dx = 0$$

Consider a function $u(x,y) = 0$

$\Rightarrow \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy = 0$

If $u$ is continuous, it would then satisfy $(1)$. Hence we require that $\frac{\partial ^{2}u}{\partial y \partial x} = \frac{\partial ^{2}u}{\partial x \partial y}$

$(x^{3}-2xy)dy +(x+2y^{2})dx = 0$

Let $F(x,y) = x+2y^{2}$, and let $G(x,y) = x^{3}-2xy$

$\Rightarrow F = \frac{\partial u}{\partial x}, G = \frac{\partial u}{\partial y}$

Then we have that $\frac{\partial F}{\partial y} = \frac{\partial ^{2}u}{\partial y \partial x}$ and $\frac{\partial G}{\partial x} = \frac{\partial ^{2}u}{\partial x \partial y}$

We require that $\frac{\partial F}{\partial y} =\frac{\partial G}{\partial x}$ for the differential equation to be exact.

$\frac{\partial F}{\partial y} = 4y$

$\frac{\partial G}{\partial x} = 3x^{2} -2 \neq \frac{\partial F}{\partial y}$

So consider an integrating factor $\lambda (x)$:

$\frac{\partial[\lambda G]}{\partial x} = \frac{\partial[\lambda F]}{\partial y}$

$\Rightarrow \lambda\frac{\partial G}{\partial x} + G\frac{\partial \lambda}{\partial x} = \lambda \frac{\partial F}{\partial y}$

Edit: After reading the answers, I identified my error here (I missed out a $y$):

$\Rightarrow \lambda(3x^{2}-2\y) +(x^{3}-2xy)\frac{\partial \lambda}{\partial x} = \lambda 4y$

I don't see how to solve this equation, so I try an integrating factor $\lambda(y)$ instead:

$\lambda \frac{\partial F}{\partial y} + F\frac{\partial \lambda}{\partial y} = \lambda \frac{\partial G}{\partial x}$

$\Rightarrow 4y\lambda +(x+2y^{2})\frac{\partial \lambda}{\partial y} = \lambda(3x^{2} -2)$

Again I don't see how to solve this equation. However I know that I should be able to get an answer for $(1)$, so what am I missing? Thanks for any help.


Write the equation in this form $(x^{3}-2xy)\mathrm dy + (x+2y^{2})\mathrm dx= 0$

Let be $F(x,y) = x+2y^{2}$ and $G(x,y)=x^3-2xy$

$\dfrac{\partial G}{\partial x}=2x^2-2y\not=4y=\dfrac{\partial F}{\partial y}$. Let be $\lambda$ the integrating factor.

Now we have $N\mathrm dy+M\mathrm dx=0$ with $N=\lambda(x^3-2xy)$ and $M=\lambda(x+2y^{2})$

$\lambda(x^{3}-2xy)\mathrm dy + \lambda(x+2y^{2})\mathrm dx= 0$ is exact, then




We try as integrating factor a function of $x$ alone (as the third term times $x$ has the same factor as the first and then we can simplify a lot) $\lambda_y=0$

$\lambda_x(x^3-2xy)+\dfrac{3\lambda}{x}(x^3-2xy)=0\implies\dfrac{\mathrm d\lambda}{\mathrm dx}+\dfrac{3\lambda}{x}=0$

It's a separable ODE with solution $\lambda=c/x^3$. Our integrating factor is $1/x^3$.

Now, check $N_x$ and $M_y$





Now we can solve the equation. Let be $f(x,y)=C$, then

$f_y=N=1-2y/x^2\implies f=\int(1-2y/x^2)\mathrm dy=y-y^2/x^2+g(x)$

$f_x=M=1/x^2+2y^2/x^3\implies f=\int(1/x^2+2y^2/x^3)\mathrm dx=-1/x-y^2/x^2+h(y)$

$g(x)=-1/x+c_1$ and $h(y)=y+c_2$

Finally $f(x,y)=y-y^2/x^2-1/x=C$ is the solution in implicit form.

  • $\begingroup$ Thanks this answer was a lot of help. After comparing with what I had done, it turns out my mistake was a very elementary one! When I considered a integrating factor $\lambda (x)$, I had a $\lambda(3x^{2}-\2)$ instead of $\lambda(3x^{2}-2\y)$. $\endgroup$ – mrnovice Mar 13 '17 at 21:01
  • $\begingroup$ You are welcome :) $\endgroup$ – Rafa Budría Mar 13 '17 at 21:04


Considering the equation $$(x^{3}-2xy)y' + x+2y^{2}= 0$$ start defining $y=z+\frac{x^2}2$ because of the first term.

This makes the equation to be $$2 z \left(z-x z'\right)+x+\frac{x^4}{2}=0$$ Now, use $z=\sqrt u$ which makes the equation to become $$-x u'+2 u+x+\frac{x^4}{2}=0$$ which is easy to solve.

  • $\begingroup$ I'm not sure I understand what your inspiration was for the substitution $y=z+\frac{x^{2}}{2}$, but I can see it works, so +1. $\endgroup$ – mrnovice Mar 13 '17 at 21:03
  • $\begingroup$ @mrnovice. In fact, I started writing $x^3-2xy=-2xz$ but I must confess that this was not the first attempt. $\endgroup$ – Claude Leibovici Mar 14 '17 at 3:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.