Solving a Linear Equation with e^(x^2) Integrating Factor

[Math Is Hard]

I am trying to solve a linear equation and getting stuck.

y' + 2xy = x^2

I am using e^{x^2} as my integrating factor and multiplying that to both sides.

Afterwards, I am able to wrap up the LHS as [y e^{x^2}]'

and I have [y e^{x^2}]' = x^2 e^{x^2}

Now all I need to do is integrate both sides and I and home free, but I haven't found a way to integrate x^2 e^{x^2}.
Using integration by parts just makes things more and more complicated.
I am letting u = e^{x^2} and dV = x^2 dx
du = e^{x^2} 2x dx and V = (x^3)/3
I don't think I have any other choice for this.

Am I missing something really obvious or have I made a mistake along the way? Or is there another technique I can apply?

Thanks in advance for your responses.

 

[HallsofIvy]

You are doing integration by parts THE WRONG WAY!

let dv= e^xdx so that v= e^x and
u= x² so that du= 2x dx. Any time you have x to a power as a factor of an integrand, consider using that as the "u" since differentiating reduces the power.
Repeating that will eventually reduce it to a constant.

 

[Gokul43201]

Here's a useful aid for Integration by parts : Remember ILATE, a mnemonic for "inverse trig - log - algebraic - trig - exp." When you have 2 functions, often the right way to pick u and dv is in the order they appear in ILATE.

In your case, you have an Algebraic term, and an Exponential term. In the mnemonic, A comes before E. So, you pick u and dv as the A term and the E term, respectively, as HallsofIvy has shown.

 

[Math Is Hard]

Thank you for your responses. But I am unclear on why dV should be
e^{x} dx instead of e^{x^2} dx

And thanks, for the tip, Gokul! I had not heard of this mnemonic. That's very cool.

 

[cookiemonster]

It should be e^{x^2}. The integral doesn't turn out nicely.

cookiemonster

 

[Math Is Hard]

Thank you for looking at this, Cookie. Here's the solution for y from my textbook by the way:

y = x/2 + Ce^{-x^2} - e^{-x^2} {\int \ (1/2) e^{x^2}\, dx}

sorry about the less than perfect latex. Still learning.

 

[cookiemonster]

To get that form, use u = x and dv = xe^{x^2}dx.

cookiemonster

 

[Math Is Hard]

To get that form, use u = x and dv = xe^{x^2}dx.

cookiemonster

whoa! I hadn't even thought about that! interesting split!

oh, well. Thank goodness that one didn't show up on the final tonight.

 

[cookiemonster]

To figure it out, I just noticed that letting u = e^(x^2) didn't accomplish anything, and I couldn't integrate it when u = x^2 because then dv = e^(x^2)dx, which I notice can't integrate without an x in front. But I just so happened to have an x in the u term, so I moved it over and voila.

cookiemonster

 

[Math Is Hard]

You are one smart cookie!

Here - have a bag of Oreos. It's on the house.

 

[HallsofIvy]

Oops, I missed that it was e^(x^2) rather than e^x!