Cant get to the next step.

[Sonal7]

I am trying to get to the general solution for this DE and the Integrating factor is correct but dont seem to be able to think of the next step. I feel that there needs to be a 2 in front one of the Xs to make it a differential of x^2 e^x
Am i right?

 

[HallsofIvy]

An "integrating factor" for $\displaystyle x\frac{dy}{dx}+ (x+ 1)y$ is a function, u(x), such that $\displaystyle \frac{d(u(x)xy}{dx}= ux\frac{dy}{dx}+ (x\frac{du}{dx}+ u)y= u(x\frac{dy}{dx}+ (x+ 1)y)= ux\frac{dy}{dx}+ u(x+ 1)y$. That is, we must have $\displaystyle \frac{du}{dx}+ xu= xu+ u$. $\displaystyle \frac{du}{dx}= u$ so $\displaystyle u(x)= e^x$.

Checking, $\displaystyle \frac{xe^xy}{dx}= e^xy+ xe^xy+ xe^x\frac{dy}{dx}= xe^x\frac{dy}{dx}+ e^x(x+1)y= e^x(x\frac{dy}{dx}+ (x+ 1)y)$.

Once you have that, $\displaystyle e^x\left(\frac{dy}{dx}+ (x+ 1)y\right)= xe^x$, $\displaystyle \frac{d(xe^xy)}{dy}= xe^x$. Integrating the left side gives, of course, $\displaystyle xe^xy$. Integrating the right side, using "integration by parts", let u= x and $\displaystyle dv= e^xdx$ so that $\displaystyle du= dx$ and $\displaystyle v= e^x$ so $\displaystyle \int u dv= uv- \int vdu= xe^x- \int e^xdx= xe^x- e^x$. Integrating both sides gives $\displaystyle xe^xy= xe^x- e^x+ C$. Divide both side by $\displaystyle xe^x$ to get $\displaystyle y= 1- \frac{1}{x}+ \frac{C}{x}e^{-x}$.

 

[Deleted member 4993]

A constraint on the solution above is:

$\displaystyle x \ne 0$

 

[HallsofIvy]

Yes, of course. Thank you.