# Thread: solve this Differential equation by indicial equation

1. ## solve this Differential equation by indicial equation

2. Originally Posted by szpengchao
Given the form of the DE, I would assume that the first part requires you to find a solution of the form $Ax^n$.

Where are you stuck in the second part?

3. ## problem

i have problem with the first part

4. Originally Posted by szpengchao
Hint:
$z=\ln(x)$

Use the chain rule

$\frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=\frac{1}{ x}\frac{dy}{dz}$

Now we need the 2nd derivative using the product and chain rule

$\frac{d^2y}{dx^2}=-\frac{1}{x^2}\frac{dy}{dz}+\frac{1}{x}\left( \frac{d^2y}{dz^2} \cdot \frac{1}{x}\right)$

from here sub into the equation

Good luck

5. ## no

no...i mean the 1st question...not the second

6. Originally Posted by szpengchao
no...i mean the 1st question...not the second
Are you talking about using the method of frobenius to obtain series solutions to the ODE?

7. ## yeah

yeah.... i have problem with choosing point for this particular question...
would you like to post ur steps?

8. ## choosing ordinary point

well, x=0 is a regular singular point, right?
then , i set y= sigma( A_n * x^(n+b)) into the equation. then i got something like:

sigma( A_n * (n^2+1) = 0) well, this means A_n = 0 for all n...then how can i continue?

9. Originally Posted by szpengchao
well, x=0 is a regular singular point, right?
then , i set y= sigma( A_n * x^(n+b)) into the equation. then i got something like:

sigma( A_n * (n^2+1) = 0) well, this means A_n = 0 for all n...then how can i continue?
$y=\sum_{n=0}^{\infty}c_nx^{n+r}$

$y'=\sum_{n=0}^{\infty}c_n(n+r)x^{n+r-1}$

$y''=\sum_{n=2}^{\infty}c_n(n+r)(n+r-1)x^{n+r-2}$

$x^2y''+xy'+y=0$

$x^2\sum_{n=0}^{\infty}c_n(n+r)(n+r-1)x^{n+r-2}+x\sum_{n=0}^{\infty}c_n(n+r)x^{n+r-1}+ \sum_{n=0}^{\infty}c_nx^{n+r}=0$

$\sum_{n=0}^{\infty}c_n(n+r)(n+r-1)x^{n+r}+\sum_{n=0}^{\infty}c_n(n+r)x^{n+r}+ \sum_{n=0}^{\infty}c_nx^{n+r}=0$

$c_nx^{n+r}\left(\sum_{n=0}^{\infty}(n+r)(n+r-1)+(n+r)+ 1 \right)=0$

$c_nx^{n+r}\left(\sum_{n=0}^{\infty}(n+r)[(n+r-1)+1]+ 1 \right)=0$

$c_nx^{n+r}\left(\sum_{n=0}^{\infty}(n+r)^2+ 1 \right)=0$

I think this is where you are stuck.

10. ## yup

yup...then? how can that be zero?

11. Originally Posted by szpengchao
yup...then? how can that be zero?
I don't know besides the trivial solution. Maybe someone can find my mistake.

12. Originally Posted by TheEmptySet
I don't know besides the trivial solution. Maybe someone can find my mistake.

Using the substitution given in the second part, I get $y = A \sin (\ln x) + B \cos (\ln x)$ ......

13. Originally Posted by mr fantastic
Given the form of the DE, I would assume that the first part requires you to find a solution of the form $Ax^n$.

Where are you stuck in the second part?
Do this and you get $n^2 + 1 = 0 \Rightarrow n = \pm i$.

Therefore

$y = A_1 x^i + A_2 x^{-i} = A_1 e^{i \ln x} + A_2 e^{-i \ln x} = B_1 \cos (\ln x) + B_2 \sin (\ln x)$

by the magic of Euler's equation and the usual identity.

14. Thanks Mr. Fantastic,

I forgot that $c_0 \ne 0$ in this method so

$c_nx^{n+r}\left(\sum_{n=0}^{\infty}(n+r)^2+ 1 \right)=0 \iff c_0x^{r}(r^2+1)+c_nx^{n+r}\left(\sum_{n=1}^{\infty }(n+r)^2+ 1 \right)=0$

so $c_n=0, \forall n \ge 1$

$r^2=1 \iff r=\pm i$

Then the rest follows.

Thanks again!