Series Solution to Second Order Linear ODE

**Kasper** · December 4th 2010, 08:18 AM

Hey I had to miss 2 lectures this week due to illness, trying to catch up. Hoping someone can help me with my snag on a problem here, not sure how to deal with 3 different coefficient terms when working with the recursion formula... Thanks! I imagine I'm going wrong somewhere here...

$y'' +ty' + (t^2 -1)y=0$ $y(0)=1, y'(0)=2$ has a power series solution about $t=0$ of the form $\displaystyle \sum_{n=0}^\infty C_n t^n<br />$

So I started by differentiating the power series and substituting into the ODE.

$\displaystyle \sum_{n=2}^\infty n(n-1)C_n t^{n-2} + t \cdot \sum_{n=1}^\infty nC_n t^{n-1} + (t^2 -1)\sum_{n=0}^\infty C_n t^n=0$

Pulling the $t$ and $(t^2 -1)$ terms into the sums.

$\displaystyle \sum_{n=2}^\infty n(n-1)C_n t^{n-2} + \sum_{n=1}^\infty nC_n t^{n} + \sum_{n=0}^\infty C_n t^{n+2} - \sum_{n=0}^\infty C_n t^n=0$

So now we need to combine the sums into a single summation so we can equate coefficients for the recursion formula. We need to shift indexes and get the same power of $t$ .

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1)C_{n+2} t^{n} + \sum_{n=0}^\infty nC_n t^{n} + \sum_{n=0}^\infty C_{n-2} t^{n} - \sum_{n=0}^\infty C_n t^n=0$

So now we can add coefficients and combine the summations.

$\displaystyle \sum_{n=0}^\infty [(n+2)(n+1)C_{n+2} +nC_n +C_{n-2} -C_n]t^n=0$

Which when we equate coefficients we obtain the following recursion formula;

$\displaystyle C_{n+2}=\frac{(n-1)C_n + C_{n-2}}{(n+2)(n+1)}$

This is where I hit a wall, I've tried plugging increasing values of n in to try to find something, but I am having trouble seeing any sort of pattern or where to go from here... I imagine i've made a mistake somewhere.

Just a push in the right direction would be great, thanks so much!
Kasper

**TheEmptySet** · December 4th 2010, 08:55 AM

Originally Posted by Kasper

$\displaystyle \sum_{n=2}^\infty n(n-1)C_n t^{n-2} + \sum_{n=1}^\infty nC_n t^{n} + \sum_{n=0}^\infty C_n t^{n+2} - \sum_{n=0}^\infty C_n t^n=0$

So now we need to combine the sums into a single summation so we can equate coefficients for the recursion formula. We need to shift indexes and get the same power of $t$ .

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1)C_{n+2} t^{n} + \sum_{n=0}^\infty nC_n t^{n} + \sum_{n=0}^\infty C_{n-2} t^{n} - \sum_{n=0}^\infty C_n t^n=0$

I am not following you reindexing.

The first and fourth sums start with a constants term. The 2nd with $t$ term and the third with a $t^2$

So we need to get all of the series starting at $t^2$

So we should get something that looks like this

$\displaystyle 2C_2+6C_3t+C_1t+C_0+C_1t+\sum_{n=4}^\infty n(n-1)C_n t^{n-2} + \sum_{n=2}^\infty nC_n t^{n} + \sum_{n=0}^\infty C_n t^{n+2} - \sum_{n=2}^\infty C_n t^n=0$
Now collecting constants gives

$\displaystyle (2C_2+C_0)+(6C_3+2C_1)t+\sum_{n=4}^\infty n(n-1)C_n t^{n-2} + \sum_{n=2}^\infty nC_n t^{n} + \sum_{n=0}^\infty C_n t^{n+2} - \sum_{n=2}^\infty C_n t^n=0$

Now we can each series to start with the $t^2$ and then combine to find the recurrence relations. Also note that That the terms we pulled out can be equated to zero to find $C_2 \text{ and } C_3$ .

I hope this gets you started.

**Kasper** · December 4th 2010, 09:57 AM

Sorry, I see how we pulled the constants out to change the index, but I'm not following on how we combine the summations now, as I thought that to add two power series $\displaystyle f(x) = \sum_{n=0}^\infty a_n(x-x_o)^n$ and $\displaystyle \sum_{n=0}^\infty b_n(x-x_o)^n$ to $\displaystyle f(x)+g(x) = \sum_{n=0}^\infty (a_n + b_n)(x-x_o)^n$

I thought we need the same index and the same power of $(x-x_o)$ .

Or am I just getting hung up on a rule that does apply to what you're doing but i'm not seeing how? Sorry! Thanks helping me understand this!

**TheEmptySet** · December 4th 2010, 10:46 AM

Originally Posted by Kasper

Sorry, I see how we pulled the constants out to change the index, but I'm not following on how we combine the summations now, as I thought that to add two power series $\displaystyle f(x) = \sum_{n=0}^\infty a_n(x-x_o)^n$ and $\displaystyle \sum_{n=0}^\infty b_n(x-x_o)^n$ to $\displaystyle f(x)+g(x) = \sum_{n=0}^\infty (a_n + b_n)(x-x_o)^n$

I thought we need the same index and the same power of $(x-x_o)$ .

Or am I just getting hung up on a rule that does apply to what you're doing but i'm not seeing how? Sorry! Thanks helping me understand this!

We do but we need to "take out" a few terms so that the powers are the same. Now that we have done that we can reindex the series Since we are starting at the with the $t^2$ term the summations should now start at $n=2$ so both the 2nd and 4th series are where they are supposed to be Now we can reindex the 1st series with $j = n-2$ this gives

$\displaystyle \sum_{n=4}^\infty n(n-1)C_n t^{n-2} =\sum_{n=2}^\infty (n+2)(n+1)C_{n+2} t^{n}$

and the 3rd with $j=n+2$

$\displaystyle \sum_{n=0}^\infty C_n t^{n+2} = \sum_{n=2}^\infty C_{n-2} t^{n}$

Now this gives

$\displaystyle (2C_2+C_0)+(6C_3+2C_1)t+\sum_{n=2}^\infty (n+2)(n+1)C_{n+2} t^{n} + \sum_{n=2}^\infty nC_n t^{n} + \sum_{n=2}^\infty C_{n-2} t^{n} - \sum_{n=2}^\infty C_n t^n=0$

Now all of the series start at the same power $t^2$ and start at the same index $n=2$ . Now we can combine the sums to get

$\displaystyle (2C_2+C_0)+(6C_3+2C_1)t+\sum_{n=2}^\infty \bigg[(n+2)(n+1)C_{n+2} +nC_n + C_{n-2} - C_n \bigg]t^n=0$

Note: it is not always possible to write solutions in a nice closed form, but as many terms as you need can be calculated.

**Kasper** · December 4th 2010, 12:47 PM

Oh ok that makes good sense, I see now what we are doing with the sums.

One final question sorry!

The question asks me to find the first 6 coefficients of the power series solution $C_o, C_1, C_2, ... , C_6$ , so we can find $C_2$ in terms of $C_o$ and $C_3$ in terms of $C_1$ , along with the recursion formula;

$C_2 = -\frac{1}{2}C_o$

$C_3 = -\frac{1}{3}C_1$

$\displaystyle C_n = -\frac{(n+2)(n+1)c_{n+2} + C_{n-2}}{n-1}$

But I'm not sure how to start finding my coefficients, namely $C_o$ , because we get a $C_{-2}$ term in the recursion formula. Does $C_{-2}$ exist?

Sorry, and thanks again for your patience, slowly making progress!

[EDIT] Scratch that, forgot about initial conditions, I am writing out the solution power series as a degree 6 polynomial and tinkering around with ICs to start back tracking the coefficients.

**TheEmptySet** · December 4th 2010, 01:01 PM

Originally Posted by Kasper

Oh ok that makes good sense, I see now what we are doing with the sums.

One final question sorry!

The question asks me to find the first 6 coefficients of the power series solution $C_o, C_1, C_2, ... , C_6$ , so we can find $C_2$ in terms of $C_o$ and $C_3$ in terms of $C_1$ , along with the recursion formula;

$\displaystyle C_n = -\frac{(n+2)(n+1)c_{n+2} + C_{n-2}}{n-1}$

But I'm not sure how to start finding my coefficients, namely $C_o$ , because we get a $C_{-2}$ term in the recursion formula. Does $C_{-2}$ exist?

Sorry, and thanks again for your patience, slowly making progress!

No You are doing fine, but you need to solve for the highest order coefficient. That is $C_{n+2}$ So we have

$(n+2)(n+1)C_{n+2} +nC_n + C_{n-2} - C_n = 0$

$(n+2)(n+1)C_{n+2} =(1-n)C_n - C_{n-2}$

$\displaystyle C_{n+2} =\frac{(1-n)C_n - C_{n-2}}{(n+2)(n+1)}$

and remember that $n$ is starting at $n=2$ and we can solve for $c_2 \text{ and } c_3$ for what was pulled out of the equation.

**Kasper** · December 4th 2010, 01:08 PM

Using the initial conditions, I got

$y(0)=1=C_o$
$y'(0)=2=C_1$

Which are both correct, but I'm not sure where I went wrong with $C_2$ and $C_3$ , because they should look like;

$C_2=-\frac{1}{2}C_o$
and
$C_3=-\frac{1}{3}C_1$

Right?

But the answers $C_2 = -\frac{1}{2}$ and $C_3 = -\frac{2}{3}$ are incorrect.

But once I find those, then

$C_4 = \frac{-C_2 + C_o}{4\cdot 3}$
and
$C_5 = \frac{-2C_3 + C_1}{5 \cdot 4}$

**TheEmptySet** · December 4th 2010, 01:28 PM

It took me a bit to find it but back in post 2 when I pulled out the coeffeints

$\displaystyle 2C_2+6C_3t+C_1t+C_0+C_1t+\sum_{n=4}^\infty n(n-1)C_n t^{n-2} + \sum_{n=2}^\infty nC_n t^{n} + \sum_{n=0}^\infty C_n t^{n+2} - \sum_{n=2}^\infty C_n t^n=0$

I made a sign error. The very last sum has a negative sign so it should look like this

$\displaystyle 2C_2+6C_3t+C_1t-C_0-C_1t+\sum_{n=4}^\infty n(n-1)C_n t^{n-2} + \sum_{n=2}^\infty nC_n t^{n} + \sum_{n=0}^\infty C_n t^{n+2} - \sum_{n=2}^\infty C_n t^n=0$

This gives

$(2c_2-c_0)+6c_3t$ This will be nice because now $c_3$ must be zero.

**Kasper** · December 4th 2010, 01:45 PM

Beauty, got all the answers correct now and i'm feeling better about this topic! Time to crush more problems like this.

Thanks so much for your help man, it's really appreciated!

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