I want to solve the following DE using integration factors
$(x + 2y) dx − x dy = 0$.
My attempt:
Letting $M(x,y) = x+2y, N(x,y)=-x$,then $\frac {M_{y}-N_{x}}{N} = \frac {2-(-1)}{-x}=\frac {-3}{x}$ is a function of $x$ say $f(x)$ and so the integrating factor according to my notes should be $e^{\int f(x)}=e^{\int -3/xdx}=e^{ln|1/x^3|}=1/x^3$
When I multiply $M,N$ by this function it does not seem to make an exact equation. Any tips for what I have done wrong much appreciated.
On
$$(x + 2y) dx − x dy = 0$$ $$Pdx+Qdy=0$$ For $\mu(x)=\frac 1 {x^3}$ the differential is exact $$ \frac {(x + 2y)}{x^3} dx − \frac 1 {x^2} dy = 0$$ $$P'dx+Q'dy=0$$ $$\partial_y P'=\partial_y \left(\frac {x+2y}{x^3}\right)=\frac 2 {x^3}$$ $$\partial_x Q'=\partial_x(\frac {-1}{x^2})=\frac 2 {x^3}$$ We have $$\partial_y P'=\partial_x Q'$$
If you multiply by $1/x^3$, it is exact. You get $M = 1/x^2 + 2y/x^3$ and $N = - 1/x^2.$ Then $$M_y = 2/x^3 = N_x.$$ Note that you may need to rewrite $-1/x^2$ as $-x^{-2}$ when you take the derivative, maybe you are making a computation error.