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:
1 Answer
$\begingroup$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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