# Solving the non exact differential equation

Solve the following differential equation:

$$a(x\frac{dy}{dx}+2y)=xy\frac{dy}{dx}$$. --Edited: see edit notes

I am having trouble solving this equation, problems that I run into are outlined below.

First, this is a non-exact differential equation. I will not put the work here, but it can be seen if you put the equation in the form $$\frac{dM}{dy}=\frac{dN}{dx}$$, and it comes out that $$\frac{dM}{dy}\neq\frac{dN}{dx}$$. So, integrating factors must be used in order to move forward solving.

But this becomes very difficult. If you choose to multiply your ODE by some function $$\mu(x)$$ or a $$\mu(y)$$ or $$\mu(x,y)$$, attempting to find the integrating factor, nothing cancels and you are left integrating something that cannot be integrated.

I also tried to multiply the ODE by $$x^\alpha y^\beta$$ (because I originally thought that $$N$$ and $$M$$ were sums of products of powers of $$x$$ and $$y$$, but this also proved inadequate, as the $$a$$s in the ODE did not cancel, and you are left with two sides of an equation that you cannot get to equal one another.

Leaving me back where I started- ground zero. Does anyone have any ideas on how to find this integrating factor?

The equation is separable,

$$2a^2y=x(y-a)\frac{dy}{dx},$$

$$2a^2\frac{dx}{x}=\frac{y-a}{y}dy,$$

$$2a^2\log x+C=y-a\log y.$$

• Apologies, I miswrote the ODE, I will nonetheless check and see if it will be able to be separated the way that you did – Pascal Sep 30 '18 at 21:42

It's separable $$a(x\frac{dy}{dx}+2ay)=xy\frac{dy}{dx}$$ $$a(xy'+2ay)=xyy'$$ $$2a^2y=xy'(y-a)$$ $$2a^2\int \frac {dx}x=\int \frac {(y-a)}{y}dy$$

$$a(x\frac{dy}{dx}+2y)=xy\frac{dy}{dx}$$ $$a(xy'+2y)=xyy'$$ $$y'x(a-y)=-2ay$$ This last equation is separable $$\int \frac {a-y}{y}dy=-2a\int \frac {dx}x$$

• Apologies, I miswrote the ODE, I will nonetheless check and see if it will be able to be separated the way that you did – Pascal Sep 30 '18 at 21:43
• @Pascal no problem edit your question – Isham Sep 30 '18 at 21:44
• Great, I think I was thinking too hard and getting ahead of myself with this one. Thank you for your clarity and work. – Pascal Sep 30 '18 at 21:47
• I added some lines @Pascal – Isham Sep 30 '18 at 21:49

Starting from Isham's answer (extracting $$y'$$ from the equation $$\int \frac {a-y}{y}dy=-2a\int \frac {dx}x$$ that is to say $$a \log(y)-y=-2a\log(x)+c$$ from which $$y=-a\, W\left(-\frac{e^{\frac{c}{a}}}{a x^2}\right)$$ where appears Lambert function.

$$a\left(x\frac{dy}{dx}+2y\right)=xy\frac{dy}{dx}$$ we can see a common factor of $$x\frac{dy}{dx}$$ so we can obtain: $$x\frac{dy}{dx}\left(y-a\right)=2ya$$ so: $$\int\frac{y-a}{2ya}dy=\int\frac1xdx$$ so: $$2\ln|x|=\int\left(\frac{1}{a}-\frac{1}{y}\right)dy$$ $$2\ln|x|=\frac{y}{a}-\ln|y|+C$$ $$x^2y=e^{\frac{y}{a}+C}$$ $$x=\sqrt{y^{-1}e^{\frac{y+C_1}{a}}}$$ and this can then be used solving the Lambert W function: https://en.wikipedia.org/wiki/Lambert_W_function