Solution to $y''+(1-x)y'+y=0$ using power series, where $y=\sum_{n=0}^{\infty}a_{n}x^{n}$

Question

I have, $$y=\sum_{n=0}^{\infty}a_{n}x^{n},\quad y'=\sum_{n=0}^{\infty}na_{n}x^{n-1}, \quad y''=\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2},$$ subbing these back into the original equation, $$y''+(1-x)y'+y=0,$$ I obtain, $$\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2} + \sum_{n=0}^{\infty}na_{n}x^{n-1} - \sum_{n=0}^{\infty}na_{n}x^{n} + \sum_{n=0}^{\infty}a_{n}x^{n}=0,$$ Then I sub in $k=n-2$ and $k=n-1$ appropriately to get, $$\sum_{k=0}^{\infty}(k+1)(k+2)a_{k+2}x^{k} + \sum_{k=0}^{\infty}(k+1)a_{k+1}x^{k} - \sum_{n=1}^{\infty}na_{n}x^{n} + \sum_{n=0}^{\infty}a_{n}x^{n}=0.$$ It is after this point in my calculation I get stuck, if anyone could show me where to go from where, or where I have gone wrong that would be appreciated. Thank you in advance.

Answer 1

You are on the correct path. First let me confirm you that.

Keep in mind that $$\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2}=\sum_{n=2}^{\infty}n(n-1)a_{n}x^{n-2}$$ Since the terms of the sum for $n=0,1$ are all $0$.

Hence, we can now sub $k=n-2$ and we get, $$\sum_{n=2}^{\infty}n(n-1)a_{n}x^{n-2}=\sum_{k=0}^{\infty}(k+2)(k+1)a_{k+2}x^{k}$$

Similarly we have $$\sum_{n=0}^{\infty}na_{n}x^{n-1}=\sum_{k=0}^{\infty}(k+1)a_{k+1}x^{k}$$

Hence, the expression $$\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2} + \sum_{n=0}^{\infty}na_{n}x^{n-1} - \sum_{n=0}^{\infty}na_{n}x^{n} + \sum_{n=0}^{\infty}a_{n}x^{n}=0,$$

finally becomes $$\sum_{k=0}^{\infty}(k+2)(k+1)a_{k+2}x^{k}+\sum_{k=0}^{\infty}(k+1)a_{k+1}x^{k}-\sum_{k=0}^{\infty}ka_{k}x^{k} + \sum_{k=0}^{\infty}a_{k}x^{k}=0$$

That is, $$\sum_{k=0}^{\infty}\left[(k+2)(k+1)a_{k+2}+(k+1)a_{k+1}-ka_{k}+a_{k}\right]x^k=0$$

Then you solve the recurrence equation by equating the coefficients of $x^k$ on both sides.

$$(k+2)(k+1)a_{k+2}+(k+1)a_{k+1}+(1-k)a_{k}=0$$

This is a 3-term recurrence relation. So you need to have $2$ initial guesses, i.e. for $a_0$ and $a_1$.

And you have to proceed to find values for $a_k$ to find your solution.

Hope this helps.