Integration by Parts/Differential Equation

[courtrigrad]

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

 

[jpr0]

You should note that
<br /> 2xe^{x^2}=\frac{d}{dx}e^{x^2},<br />
so that your integrand can be written as
<br /> e^{x^2}x^2=\frac{x}{2}\left(2xe^{x^2}\right)=\frac{x}{2}\left(\frac{d}{dx}e^{x^2}\right)<br />
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 shouldn't have the constant C in your expression for y, it should be attached to your homogeneous solution...)

 

[courtrigrad]

thank you for your help