1. ## [SOLVED] Integrating Factor

I want to find out the integrating factor of $\displaystyle (y^4 + 2y)dx + (xy^3 + 2y^4 - 4x)dy = 0$

As $\displaystyle M$ is $\displaystyle y^4 + 2y$and $\displaystyle N$ is $\displaystyle xy^3 + 2y^4 - 4x$ I took the partial derivative of $\displaystyle N$ with respect to $\displaystyle x$ and that of $\displaystyle M$ with respect to $\displaystyle y$ and then dividing by $\displaystyle M$ I got :-

$\displaystyle \frac {(y^3 - 4) - (4y^3 + 2)}{y^4 +2y}$

Whereas the answer at the back is simply $\displaystyle \frac{1}{y^3}$.

2. Originally Posted by Altair
I want to find out the integrating factor of $\displaystyle (y^4 + 2y)dx + (xy^3 + 2y^4 - 4x)dy = 0$

As $\displaystyle M$ is $\displaystyle y^4 + 2y$and $\displaystyle N$ is $\displaystyle xy^3 + 2y^4 - 4x$ I took the partial derivative of $\displaystyle N$ with respect to $\displaystyle x$ and that of $\displaystyle M$ with respect to $\displaystyle y$ and then dividing by $\displaystyle M$ I got :-

$\displaystyle \frac {(y^3 - 4) - (4y^3 + 2)}{y^4 +2y}$

Whereas the answer at the back is simply $\displaystyle \frac{1}{y^3}$.

$\displaystyle \frac {(y^3 - 4) - (4y^3 + 2)}{y^4 +2y} \iff \frac{-3(y^3+2)}{y(y^3+2)}=\frac{-3}{y^3}$

so

$\displaystyle I(y)=\int \frac{-3}{y}dy=\frac{1}{y^3}$

Good luck.

3. Or if there is a different method to solve this then please do tell about that.