- #1
Kaguro
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- TL;DR 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.
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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