# Integration by Parts/Differential Equation

1. Sep 27, 2006

If you have $$\int e^{x^{2}}x^{2}$$ would it be reasonable to choose $$u = e^{x^{2}}, dv = x^{2}, du = 2xe^{x^{2}}, v = \frac{x^{3}}{3}$$? And then I get $$x^{3}e^{x^{2}} - 2\int x^{4}e^{x^{2}}$$. Would this be equivalent to choosing $$u = x, dv = 2xe^{x^{2}}, v = e^{x^{2}}, du = dx$$.

This was for solving a differential equation:

$$\frac{dy}{dx} + 2xy = x^{2}$$, where $$P(x) = 2x, Q(x) = x^{2}, I = e^{\int 2x} = e^{x^{2}}$$. So $$(ye^{x^{2}})' = e^{x^{2}}x^{2}$$

$$y = e^{-x^{2}} \int e^{x^{2}}x^{2} dx + C$$

Thanks

Last edited: Sep 27, 2006
2. Sep 27, 2006

### jpr0

You should note that
$$2xe^{x^2}=\frac{d}{dx}e^{x^2},$$
so that your integrand can be written as
$$e^{x^2}x^2=\frac{x}{2}\left(2xe^{x^2}\right)=\frac{x}{2}\left(\frac{d}{dx}e^{x^2}\right)$$
Then you can integrate this by parts.
By the way in your expression for $y$ you seem to have lost the homogeneous solution.
A further note is that you might find it useful to have a look at the error function when trying to evaluate your expression:

http://mathworld.wolfram.com/Erf.html

(....and you shouldnt have the constant C in your expression for $y$, it should be attached to your homogeneous solution....)

Last edited: Sep 27, 2006
3. Sep 27, 2006