[STUCK]Series Solution to Differential Equations

**Creebe** · January 19th 2010, 11:01 AM

the original question is:

$(x^2 - 3)y'' + 2xy' = 0$ about $x_{0} = 0$

and i got to this part:

$\sum_{n=2}^{\infty} (n-1)(n)(a_{n})(x^n) - \sum_{n=2}^{\infty} 3(n-1)(n)(a_{n})(x^{n-2}) + \sum_{n=1}^{\infty} 2(n)(a_{n})(x^n) = 0$

what do i do next?

**Jhevon** · January 19th 2010, 02:23 PM

Originally Posted by Creebe

the original question is:

$(x^2 - 3)y'' + 2xy' = 0$ about $x_{0} = 0$

and i got to this part:

$\sum_{n=2}^{\infty} (n-1)(n)(a_{n})(x^n) - \sum_{n=2}^{\infty} 3(n-1)(n)(a_{n})(x^{n-2}) + \sum_{n=1}^{\infty} 2(n)(a_{n})(x^n) = 0$

what do i do next?

you can see here. the problem is different, but when you get to the step that you're on, i explain the process.

in summary:
first, get all the powers of x to be the same, that is n. notice that in the middle term the power is n - 2. change all n's in that term to n + 2 (the index on the series will change also).

then, make sure your series all start at the same index. usually this is done by evaluating some terms of some series so that they catch up with others. in this case, you can just set all of them to begin at n = 0. this won't change what you have.

then combine all the series together, factor out the $x^n$ and set its coefficient to zero. from this you can get a recursive formula to find the coefficients of your series solution. i hope you know how to take it from there.

**HallsofIvy** · January 20th 2010, 05:33 AM

Originally Posted by Creebe

the original question is:

$(x^2 - 3)y'' + 2xy' = 0$ about $x_{0} = 0$

and i got to this part:

$\sum_{n=2}^{\infty} (n-1)(n)(a_{n})(x^n) - \sum_{n=2}^{\infty} 3(n-1)(n)(a_{n})(x^{n-2}) + \sum_{n=1}^{\infty} 2(n)(a_{n})(x^n) = 0$

what do i do next?

In the "middle sum", let k= n-2 so that n= k+ 2. Then "3(n-1)n" becomes "3(k+2-1)(k+2)= 3(k+1)(k+2). Also " $a_n$ " becomes "a_{k+2}" and " $x^{n-2}$ " becomes " $x^k$ ". Finally, when n= 2, k= 0 That is, that whole "middle sum" becomes $\sum_{k=0}^\infty 3(k+1)(k+2)a_{k+2}x^k$ .

In the other two sums, just change "n" to "k" and your whole equation becomes $\sum_{k=2}^{\infty} (k-1)(k)(a_{k})(x^k) - \sum_{k=0}^\infty 3(k+1)(k+2)a_{k+2}x^k + \sum_{k=1}^{\infty} 2(k)(a_{k})(x^k) = 0$ . Now, since those sums all have $x^k$ you can combine corresponding coefficients into a single sum. Since the only way a power series can be 0, for all x, is if the coefficients of all powers of x are 0, you can then set each coefficient equal to 0.

Be careful about the different starting values. For k= 0, you have only the middle term so the coefficient of $x^0$ , the constant term, is just $3(0+1)(0+2)a_{0+2}= 6a_2= 0$ . For k= 1, you have only the middle and third coeffient so $3(1+1)(1+2)a_{1+2}+ 2(1)(a_{1})= 18a_3+ 2a_1 = 0$ . For $n\ge 2$ , you have all three terms and can write out the general equation to get an iterative formula for $a_n$ .

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