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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.

Please help.

(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.

Please help.

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