# Thread: Did I make a mistake with this ODE via Power Series?

1. ## Did I make a mistake with this ODE via Power Series?

I need to solve the following ODE using Power Series:

$xy'' + y' + xy = 0$

Now, when I do it, this is what happens:

$y = \sum_{n=0}^{\infty} a_n x^n$

$y' = \sum_{n=1}^{\infty} na_n x^{n - 1}$

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

$x \sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 2} + \sum_{n=1}^{\infty} na_n x^{n - 1} + x \sum_{n=0}^{\infty} a_n x^n = 0$

$\sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 1} + \sum_{n=1}^{\infty} na_n x^{n - 1} + \sum_{n=0}^{\infty} a_n x^{n + 1} = 0$

Respectively:

$k = n - 1$
$k = n - 1$
$k = n + 1$

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

$\sum_{k=1}^{\infty} k(k + 1) a_{k + 1} x^k + \sum_{k=1}^{\infty} (k + 1)a_{k + 1} x^k + \sum_{k=1}^{\infty} a_{k - 1} x^k + a_1 = 0$

$\sum_{k=1}^{\infty} [x^k[(k + 1)ka_{k + 1} + (k + 1)a_{k + 1} + a_{k - 1}]] + a_1 = 0$

$a_1 = 0$

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

$a_{k + 1} = \frac{-a_{k - 1}}{(k + 1)k + (k + 1)}$

$y = a_0(1 - \frac{1}{4}x^2 + \frac{1}{64}x^4 - \frac{1}{2304}x^6 + ...)$

However, Wolfram Alpha says the solution is:

$y = C_1 J_0 + C_2 Y_0$

...where $J_n$ and $Y_n$ are Bessel Functions

So, did I make a mistake? And if so, where?

2. ## Re: Did I make a mistake with this ODE via Power Series?

For solutions on the real line, it looks correct. I only checked the first two terms, but so far your solution is consistent with Wolfram Alpha for $C_1=a_0$ and $C_2=0$. However, keep in mind that your ODE is equivalent (for $x\neq 0$) to

$\left[\begin{array}{c}y_1\\y_2\end{array}\right]'=\left[\begin{array}{c}y_2\\-(y_2/x)-y_1\end{array}\right]$, which may have solutions which tend to $\pm\infty$ at zero. So you don't get to assume that every solution can be expressed in a power series about zero.

3. ## Re: Did I make a mistake with this ODE via Power Series?

Originally Posted by hatsoff
For solutions on the real line, it looks correct. I only checked the first two terms, but so far your solution is consistent with Wolfram Alpha for $C_1=a_0$ and $C_2=0$. .
Forgive me, but I'm not sure I understand. The answer from WolframAlpha had a bessel function paired with $C_2$. How does my answer of $C_2 = 0$ correspond with that?

4. ## Re: Did I make a mistake with this ODE via Power Series?

Well $C_1J_0+C_2Y_0$ is just a linear combination of Bessel functions $J_0$ and $Y_0$. Your solution is apparently the function $a_0J_0+0Y_0$.

5. ## Re: Did I make a mistake with this ODE via Power Series?

Originally Posted by hatsoff
Well $C_1J_0+C_2Y_0$ is just a linear combination of Bessel functions $J_0$ and $Y_0$. Your solution is apparently the function $a_0J_0+0Y_0$.
Okay, thanks. I guess I thought I should be getting a result that was more universal to $a_1$, rather than a specific value of $a_1$.

6. ## Re: Did I make a mistake with this ODE via Power Series?

What are you assuming a form $y = \sum_{n=0}^{\infty} x^n$ and not $y = \sum_{n=0}^{\infty} x^{n+r}$ (method of Frobenius')?

7. ## Re: Did I make a mistake with this ODE via Power Series?

Originally Posted by Lancet
I need to solve the following ODE using Power Series:

$xy'' + y' + xy = 0$

Now, when I do it, this is what happens:

$y = \sum_{n=0}^{\infty} a_n x^n$

$y' = \sum_{n=1}^{\infty} na_n x^{n - 1}$

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

$x \sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 2} + \sum_{n=1}^{\infty} na_n x^{n - 1} + x \sum_{n=0}^{\infty} a_n x^n = 0$

$\sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 1} + \sum_{n=1}^{\infty} na_n x^{n - 1} + \sum_{n=0}^{\infty} a_n x^{n + 1} = 0$

Respectively:

$k = n - 1$
$k = n - 1$
$k = n + 1$

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

$\sum_{k=1}^{\infty} k(k + 1) a_{k + 1} x^k + \sum_{k=1}^{\infty} (k + 1)a_{k + 1} x^k + \sum_{k=1}^{\infty} a_{k - 1} x^k + a_1 = 0$

$\sum_{k=1}^{\infty} [x^k[(k + 1)ka_{k + 1} + (k + 1)a_{k + 1} + a_{k - 1}]] + a_1 = 0$

$a_1 = 0$

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

$a_{k + 1} = \frac{-a_{k - 1}}{(k + 1)k + (k + 1)}$

$y = a_0(1 - \frac{1}{4}x^2 + \frac{1}{64}x^4 - \frac{1}{2304}x^6 + ...)$

However, Wolfram Alpha says the solution is:

$y = C_1 J_0 + C_2 Y_0$

...where $J_n$ and $Y_n$ are Bessel Functions

So, did I make a mistake? And if so, where?
You didn't any mistake at all!... You have hipotized a solution analytic in $x=0$ and what You have found is the series expansion of the zero order first kind Bessel function...

$y= c\ J_{0}(x)= c\ (1 - \frac{x^{2}}{2^{2}} + \frac{x^{4}}{2^{2}\ 4^{2}} - \frac{x^{6}} {2^{2}\ 4^{2}\ 6^{2}} + ...)$ (1)

The other possible solution is the zero order second kind Bessel function $Y_{0} (x)$ which in $x=0$ has a logarithmic singularity...

Kind regards

$\chi$ $\sigma$

8. ## Re: Did I make a mistake with this ODE via Power Series?

Originally Posted by Danny
What are you assuming a form $y = \sum_{n=0}^{\infty} x^n$ and not $y = \sum_{n=0}^{\infty} x^{n+r}$ (method of Frobenius')?

I'm assuming a form:

$y = \sum_{n=0}^{\infty} a_n x^n$

9. ## Re: Did I make a mistake with this ODE via Power Series?

Originally Posted by chisigma
You didn't any mistake at all!... You have hipotized a solution analytic in $x=0$ and what You have found is the series expansion of the zero order first kind Bessel function...

Thanks for the confirmation and explanation!