Suppose we have the following differential equation that is NOT exact, i.e. $M_y \ne N_x$: $2xy^3+y^4+(xy^3-2y)y'=0$

How would I find an integrating factor $μ(x,y)$ so that when I multiply this integrating factor by the differential equation, it become exact?

Update: Here's what I got: enter image description here

Integrating factor $\mu=\mu(\omega)$, we get from equation $$\frac{d\mu}{\mu}=\frac{M_y-N_x}{\omega_x N-\omega_y M} d\omega.$$ By replacing known values $M=2xy^3+y^4$, $M_y=6xy^2+4y^3$, $N=xy^3-2y$, $N_x=y^3$ into equation, we have $$\frac{d\mu}{\mu}=\frac{6xy^2+3y^3}{\omega_x (xy^3-2y)-\omega_y (2xy^3+y^4)} d\omega.$$ It is easy to notice identity $y(6xy^2+3y^3)=2(2xy^3+y^4)$. Because of that, we will take $\omega_y=\frac{-3}{y}$, or $$\omega=\omega(y)=-3\ln{y}.$$ By substituting this result into equation above, we finally get $$\frac{d\mu}{\mu}=\frac{-3}{y}dy,$$ $$\mu=\frac{1}{y^3}.$$ It is easy to check that $x^2+xy+\frac{2}{y}=C$ is a solution.

  • Please see my updated answer. I got -3ln(y) for the integrating factor. Did I do something wrong? I am not sure why you have a new variable ω which is -3ln(y). I thought that was the value of μ. – Blue Pony Inc. Jun 26 '13 at 1:10
  • 1
    @Blue Pony Inc. You made a calculation error before the last equation (by the way, $\mu$ is a function of $\omega$, but in this case you can take $\mu=\mu(y)$). Here is correction of your proof: $\mu(y)=e^{\int \frac{N_x-M_y}{M} dy}=e^{\int \frac{-3}{y}dy}=e^{-3\ln{y}}=e^{\ln{y^{-3}}}=y^{-3}$. Therefore, $\mu=\frac{1}{y^3}$. – alans Jun 26 '13 at 11:11

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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