What is the integrating factor for $xdy + y(x+1)dx =0 $?

Question

I could do this with onlfy function forming method i.e. $M(x,y) dx + N(x,y) dy = d f(x,y)$ then the given equation is exact.

$$xdy + y(x+1)dx =0$$

My work,

$$⇒ xdy + xydx + ydx =0$$

$$⇒xdy + ydx +(xy)dy=0$$

$$⇒d(xy)+(xy)dy=0$$

$$⇒(1/xy)d(xy)+dy=0$$

integrating we get,

$$ln|xy| + y = C$$

but I could not find Integrating factor to do it by partial differential method.

and one extra related question," when both the partial derivatives are 0, being equal we can conclude that the equation is exact.( I just wanted to confirm for the 0 case)"

Answer 1

You can in fact do it with an integrating factor directly. Divide through by $xy$ (as you ultimately did) and you get $$\frac1y\,dy + \left(1+\frac1x\right)\,dx = 0,$$ which is exact.

Answer 2

It's separable

$$ xdy + y(x+1)dx =0 $$

$$ \int \frac {dy}{y}=-\int \frac {(x+1)}{x}dx=-x -ln|x|+K $$

Answer 3

First note, this is a separable differential equation $$ xy'=-y(x+1)\implies \frac{y'}{y}=-\frac{x+1}{x}\implies \ln |y|=-\int 1+\frac{1}{x}\mathrm dx\\ \implies \ln|y|=-x-\ln|x|+C\\ \implies y(x)=Axe^{-x} $$ where $A$ is a constant.

Second, if you are trying to solve this using integrating factors, the way I usually proceed is by guessing somewhat intelligently, usually some kind of exponential is involved.

Note that you have $$ \partial_x(x)=1\ne \partial_y(y(x+1))=x+1 $$ so you are off by a function of $x$, that should come down from the exponential after we use the chain rule. So guess $$ I(x,y)=e^{g(x)} $$ and solve for $g$ by enforcing $$ \partial_x(e^{g(x)}x)=xg'(x)e^{g(x)}+e^{g(x)}\\ = \partial_y(ye^{g(x)}(x+1))=e^{g(x)}x+e^{g(x)}\\ \implies g'(x)=1\implies g(x)=x+c $$