# Cant get to the next step.

#### Sonal7

##### New member
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?

#### Attachments

• 26.7 KB Views: 2
• 792.9 KB Views: 3

#### HallsofIvy

##### Elite Member
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 $$\displaystyle y= 1- \frac{1}{x}+ \frac{C}{x}e^{-x}$$.

#### Subhotosh Khan

##### Super Moderator
Staff member
A constraint on the solution above is:

$$\displaystyle x \ne 0$$

#### HallsofIvy

##### Elite Member
Yes, of course. Thank you.