One method for doing an ODE is to try to make it an exact ODE by multiplying by an integrating factor. Why don't you have to remove the integrating factor afterwards?

For instance, in this example, the problem was not an exact DiffEQ:

$\displaystyle \left( 3xy+y^2 \right) + \left( x^2+xy \right) y' = 0$

$\displaystyle M(x,y)=3xy+y^2 \hspace{20 mm} N(x,y)=x^2+xy$

$\displaystyle M_y=3x+2y \hspace{30 mm} N_x=2x+y$

$\displaystyle M_y \ne N_x$, so NOT an exact DiffEQ.

Next is to look for an integrating factor, which we call $\displaystyle \mu(x)$:

$\displaystyle \mu(x) \left( 3xy+y \right) + \mu(x) \left( x^2+xy \right) y' = 0$

$\displaystyle M(x,y)=\mu(x)\left( 3xy+y^2 \right) \hspace{20 mm} N(x,y)=\mu(x)\left( x^2+xy \right)$

$\displaystyle M_y=\mu(x)(3x+2y) \hspace{30 mm}= N_x=\mu'(x)(x^2+xy)+\mu(x)(2x+y)$

$\displaystyle \mu(x)(x+y)=\mu'(x)(x^2+xy)$

$\displaystyle \mu(x)=\mu'(x)x=\fracd{d\mu}{dx}x$

$\displaystyle \frac{dx}{x}=\frac{d\mu}{\mu} \implies \mu(x)=x$

So, back to the top, multiply the whole equation by x:

$\displaystyle \left( 3x^2y+xy^2 \right) + \left( x^3+x^2y \right) y' = 0$

$\displaystyle \int(3x^2y+xy^2)dx = x^3y+\frac12x^2y^2+f(y)$

$\displaystyle \int(x^3+x^2y)dy = x^3y+\frac12x^2y^2+g(x)$

$\displaystyle f(x)=g(x)=C$(some constant)

$\displaystyle \psi(x,y)=x^3y+\frac12x^2y^2=C$

Why don't we have to compensate for $\displaystyle \mu(x)=x$ after this last step?Notice the $\displaystyle \psi_x$ and $\displaystyle \psi_y$ do not equal the starting $\displaystyle M$ and $\displaystyle N$:

$\displaystyle \psi_x=3x^2y+xy^2 \hspace{30 mm} \psi_y=x^3+x^2y$

And dividing psi by x also doesn't work (although dividing the partial derivatives of psi by x works). What gives here?

Thanks (sorry for the too long post),