multiplying coefficients into series

[SigepBrandon]

The complete problem is as follows:
Use power series to find the solution to the initial value problem y''+3xy'-6y=1, y(0)=a₀, y'(0)=0 by letting \(\displaystyle \sum_{k=0}^{\infty} a_{k}x^{k}=1\) and finding a recurrence relation for the coefficients a_k.

I found the derivatives of the series to be:

\(\displaystyle y'=\sum_{k=1}^{\infty} ka_{k}x^{k-1}\) and \(\displaystyle y''=\sum_{k=2}^{\infty} k(k-1)a_{k}x^{k-2}\)

plugging the derivatives of the series into the differential equation:

\(\displaystyle \sum_{k=2}^{\infty} k(k-1)a_{k}x^{k-2}+3x \sum_{k=1}^{\infty} ka_{k}x^{k-1} - 6 \sum_{k=0}^{\infty} a_{k}x^{k}=1\)

I know I need to multiply the coefficients (3x and 6) into their respective series and then perform the shift in order to combine. However I do not know how to get the coefficients into the same form. Any help would be greatly appreciated.

-Brandon

 

[HallsofIvy]

The complete problem is as follows:
Use power series to find the solution to the initial value problem y''+3xy'-6y=1, y(0)=a₀, y'(0)=0 by letting \(\displaystyle \sum_{k=0}^{\infty} a_{k}x^{k}=1\)

Typo? You clearly mean "= y", not "= 1".

and finding a recurrence relation for the coefficients a_k.

I found the derivatives of the series to be:

\(\displaystyle y'=\sum_{k=1}^{\infty} ka_{k}x^{k-1}\) and \(\displaystyle y''=\sum_{k=2}^{\infty} k(k-1)a_{k}x^{k-2}\)

plugging the derivatives of the series into the differential equation:

\(\displaystyle \sum_{k=2}^{\infty} k(k-1)a_{k}x^{k-2}+3x \sum_{k=1}^{\infty} ka_{k}x^{k-1} - 6 \sum_{k=0}^{\infty} a_{k}x^{k}=1\)

I know I need to multiply the coefficients (3x and 6) into their respective series and then perform the shift in order to combine. However I do not know how to get the coefficients into the same form. Any help would be greatly appreciated.

-Brandon

You can, certainly, move the "3x" and "6" into the respective series- that is simply the "distributive law".
\(\displaystyle \sum_{k=2}^\infty k(k-1)a_kx^{k-2}+ \sum_{k=1}^\infty 3ka_k x^k- \sum_{k=0}^\infty 6a_kx^k= 1\)

Now, in order to be able to combine "like terms", rearrange the sums, using new indices.
That is, in the first sum, let j= k-2 so that k= j+2. When k= 2, j= 0, k= j+2, k-1= j+1, \(\displaystyle a_k= a_{j+2}\), and \(\displaystyle x^{k-2}= x^j\) so the first term is \(\displaystyle \sum_{j=0}^\infty (j+2)(j+1)a_{j+2}x^j\).

For the last two sum, we already have \(\displaystyle x^k\) so just use \(\displaystyle j= k\). The entire sum becomes \(\displaystyle \sum_{j=0}^\infty (j+2)(j+1)a_{j+2}x^j+ \sum_{j= 1}^\infty 3ja_jx^j- \sum_{j=0}^\infty 6a_jx^j= 1\).

Now, notice that the second sum does not begin until j= 1 so for j= 0, we have
\(\displaystyle 2(1)a_2- 6a_0= 1\) which gives \(\displaystyle a_2= \frac{1}{2}+ 3a_0\).

For \(\displaystyle j\ge 1\) the coefficient of \(\displaystyle x^j\) is (j+2)(j+1)a_{j+2}+ 3ja_j+ 5a_j= 0[/tex] (remember this is the coefficient of \(\displaystyle x^j\) and there are no powers of x on the right side so this is 0, not 1).

You can solve that for \(\displaystyle a_{j+2}\) in terms of \(\displaystyle a_{j}, getting a recursive equation.
\(\displaystyle y(0)= a_0\) and \(\displaystyle y'(0)= a_1= 0\).\)

 

[SigepBrandon]

Thank you for your reply Ivy. I did incorrectly quote the problem statement and your assumptions were correct. However I believe I do see an error in the explanation provided. The second term is \(\displaystyle 3x\sum_{k=1}^{\infty} ka_{k}x^{k-1}\) not \(\displaystyle 3x\sum_{k=1}^{\infty} ka_{k}x^{k}\) I did not think the distributive law was applicable to coefficients with unlike exponent terms. In this case, would shifting the indices first then combining make more since? I've also seen examples where the coefficient is expanded and creates a new series when distributed. I'm just not certain how to approach this.