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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

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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 $y(6xy^2+3y^3)=3(2xy^3+y^4)$. 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.

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  • $\begingroup$ 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. $\endgroup$ Commented Jun 26, 2013 at 1:10
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    $\begingroup$ @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, $\mu=\frac{1}{y^3}$. $\endgroup$ Commented Jun 26, 2013 at 11:11
  • $\begingroup$ What is $ \omega$ $\endgroup$ Commented Sep 26, 2020 at 18:31
  • $\begingroup$ @Buraian $\omega$ is function of $x,y$. $\endgroup$ Commented Sep 29, 2020 at 7:31

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