# power series and differentual equations guidance

• Apr 24th 2013, 01:21 PM
LinearOperator
power series and differentual equations guidance
I am working on a post exam worksheet and I want to master before my exam tomorrow. I am a little sketchy so if you can please help i really appreciate!! :)

Q. Find an interval around x = 0 for which the initial value problem: (x-2)y'' +3y = x, y(0) = 0, y'(0)=0 = 1 has a unique solution.

I understand that I need to use power series. this is non cauchy-euler equation. My main concern is the power series and the shifts. Because sometimes, I think i read by ferbini's theorem, you can guess: y = sum Cnx^n or y = sum c_n(x-x0)^n
how do you know which guess to make??
• Apr 24th 2013, 01:25 PM
LinearOperator
Re: power series and differentual equations guidance
would it be ok to guess y = sum c_n(x-2)^n ?
• Apr 24th 2013, 03:36 PM
HallsofIvy
Re: power series and differentual equations guidance
If you do that, you are going to have trouble with the initial conditions which are at x= 0. I would recommend using $y= \sum a_nx^n$ and writing the equation as xy''- 2y''+ 3y= x. Now, $y'= \sum na_nx^{n-1}$, [tex]y''= \sum n(n-1)a_nx^{n-2} $. Putting those into the equation, we get
img.top {vertical-align:15%;} $sum_{n= 2}^\infty n(n- 1)a_n x^{n-1}- \sum_{n= 2}^infty 2n(n-1)a_n x^{n- 2}+ \sum_{j= 0}^\infty a_nx^n= x$. In the first sum let j= n-1 so that n= j+1, n-1= j, and tex]a_n= a_{j+1}" alt=". Putting those into the equation, we get
$sum_{n= 2}^\infty n(n- 1)a_n x^{n-1}- \sum_{n= 2}^infty 2n(n-1)a_n x^{n- 2}+ \sum_{j= 0}^\infty a_nx^n= x$. In the first sum let j= n-1 so that n= j+1, n-1= j, and tex]a_n= a_{j+1}" /> and the first sum becomes $\sum_{j=0}^\infty j(j+1)a_{j+1}x^j$. In the second sum, let j= n- 2 so that n= j+ 2, n- 1= j+ 1, and [tex]a_n a_{j+ 2} and the sum becomes $\sum_{j= 1}^\infty (j+1)(j+ 2)a_{j+2}x^j$. In the third sum, let j= n so the sum becomes $\sum_{j= 0}^\infty a_j x^j$.

That sets the equation as $\sum_{j= 0}^\infty j(j+1)a_{j+1}x^j-\sum_{j= 1}^\infty (j+ 2)(j+1)a_{j+2}x^j+ \sum_{j= 0}^\infty a_j x^j$. Now that those all have x to the "j" power, you can combine them (being careful about the j= 0 term) to get recursion relations for the $a_j$. The initial conditions tell you that $a_0= 0$, $a_1= 1$.
• Apr 24th 2013, 03:40 PM
LinearOperator
Re: power series and differentual equations guidance
Oh! Ok! I get it, thank you so much!!

I am just having trouble seeing when I factor out terms and when to use! do you know of a thread post that has an explanation of that?