# Solving an ODE with power series

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## Summary:

In power series method, after putting in the summations in the DE and simplifying, I found three powers of x.. how do I solve it?

## Main Question or Discussion Point

I have an ODE:
(x-1)y'' + (3x-1)y' + y = 0
I need to find the solution about x=0. Since this is an ordinary point, I can use the regular power series solution.

Let y = $\sum_{r=0}^\infty a_r x^r$
after finding the derivatives and putting in the ODE, I have:
$\sum_{r=0}^\infty a_r (r)(r-1) x^{r-1} - \sum_{r=0}^\infty a_r (r)(r-1) x^{r-2} + 3 \sum_{r=0}^\infty a_r (r) x^{r} - \sum_{r=0}^\infty a_r (r) x^{r-1} + \sum_{r=0}^\infty a_r x^{r}=0$

Now I have three powers of x, so If I transform the index I'll still end up with a recurrence relation relating three coefficients.
I don't know how to proceed in that case.

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FactChecker
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When you put the power series into the ODE, what happened to the " = 0" part? It is necessary to use that to solve for the coefficients.

FactChecker
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2018 Award
For each power of x, the total on the left must equal that power of x on the right, which is zero. Work on that and see how far it gets you.

HallsofIvy
Homework Helper
Summary:: In power series method, after putting in the summations in the DE and simplifying, I found three powers of x.. how do I solve it?

I have an ODE:
(x-1)y'' + (3x-1)y' + y = 0
I need to find the solution about x=0. Since this is an ordinary point, I can use the regular power series solution.

Let y = $\sum_{r=0}^\infty a_r x^r$
after finding the derivatives and putting in the ODE, I have:
$\sum_{r=0}^\infty a_r (r)(r-1) x^{r-1} - \sum_{r=0}^\infty a_r (r)(r-1) x^{r-2} + 3 \sum_{r=0}^\infty a_r (r) x^{r} - \sum_{r=0}^\infty a_r (r) x^{r-1} + \sum_{r=0}^\infty a_r x^{r}=0$

Now I have three powers of x, so If I transform the index I'll still end up with a recurrence relation relating three coefficients.
I don't know how to proceed in that case.
Actually you have a many powers of x hidden in each sum. Also, though all your sums start at r= 0, in the first sum you have coefficients r and r- 1 so the first two terms are 0, It would be better to write it as $\sum_{r= 2}^\infty a_r r(r-1)x^{r- 2}$. In order to get $x^i$, let i= r- 2 so that r= i+ 2 and the sum becomes $\sum_{i= 0}^\infty a_{i+2}(i+2)(i+1)x^{i}$. Do the same with each of the other sums so that you have $x^i$ in each sum and can combine coefficients of "like powers".