Solving xy'(x) - y(x) = x^2 Exp[x] using the power series method

1. Jan 19, 2007

Repetit

I've tried solving the equation xy'(x) - y(x) = x^2 Exp[x] using the power series method. I assume that y has the form:

$$y = \sum_{n=0}^{\infty} a_n x^n$$

Inserting this in the diff. eq. gives:

$$\sum_{n=0}^{\infty} n a_n x^n - \sum_{n=0}^{\infty} a_n x^n = x^2 e^x$$

Now, in the other types of diff. eqations I've solved using this method I found a recursive formula for the coefficients, but I don't see how I can do that here? How would I got about solving this problem?

Thanks

2. Jan 19, 2007

cristo

Staff Emeritus
You could try using the series for ex on the RHS. It's just a suggestion, so I don't know whether it'll work, but give it a go!

3. Jan 19, 2007

HallsofIvy

Yes, if you are required to use the series method, expand ex in Taylor's series and compare corresponding powers of x.

Of course, that's an "Euler type" equation so it should be easier to solve it directly, perhaps by converting to an equation with constant coefficients.

4. Jan 20, 2007

Repetit

My problem is that I don't see how to obtain a recusion formula for the coefficients. Even when trying to solve just the corresponding homogeneous equation $$x y'(x) - y(x) = 0$$ I end up with something like $$n a_n - a_n = 0$$ and I cant see what to do.

5. Jan 20, 2007

HallsofIvy

Obviously, $na_n- a_n= 0$ is true as long as $a_n= 0$ for any n other than 1. For n= 1, the equation becomes $a_1= a_1$ which is also true- $a_1$ becomes the undetermined constant in your general solution. What that tells you is that the general solution to xy'(x)- y(x)= 0 is y(x)= Cx.
Since $e^x= \Sigma_{n=0}^\infty\frac{1}{n!}x^n$, $x^2e^x= \Sigma_{n=0}^\infty\frac{1}{n!}x^{n+2}$. If you let i= n+2, then you get $x^2e^x= \Sigma_{i=2}^\infty\frac{1}{(i-2)!}x^i$ so you can compare "like powers".
You get $(0)a_0- a_0= 0$ so a_0= 0. $(1)a_1- a_1= 0$ which is always true- $a_1$ will be an "unknown constant".

For n> 1, $na_n- a_n= a_n(n-1)= \frac{1}{(n-2)!}$ so
[tex]a_n= \frac{1}{(n-2)!(n-1)}[/itex].

6. Jan 21, 2007

Repetit

Hmm, of course. Why did I not see that. I guess I was to focused on obtaining a recursion formula for the coefficients, because I thought I always had to do that. I see now that I don't. Thanks alot for the help!