Iffy diffy q

1. Jun 22, 2004

Math Is Hard

Staff Emeritus
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. :grumpy:
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?

2. Jun 22, 2004

HallsofIvy

Staff Emeritus
You are doing integration by parts THE WRONG WAY!

let dv= exdx so that v= ex 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.

3. Jun 22, 2004

Gokul43201

Staff Emeritus
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.

4. Jun 22, 2004

Math Is Hard

Staff Emeritus
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.

5. Jun 22, 2004

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

6. Jun 22, 2004

Math Is Hard

Staff Emeritus
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.

7. Jun 22, 2004

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

8. Jun 23, 2004

Math Is Hard

Staff Emeritus

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

9. Jun 23, 2004

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.

10. Jun 23, 2004

Math Is Hard

Staff Emeritus