# Series Solution to Second Order Linear ODE

• Dec 4th 2010, 08:18 AM
Kasper
Series Solution to Second Order Linear ODE
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...

Quote:

$\displaystyle y'' +ty' + (t^2 -1)y=0$ $\displaystyle y(0)=1, y'(0)=2$ has a power series solution about $\displaystyle t=0$ of the form $\displaystyle \displaystyle \sum_{n=0}^\infty C_n t^n$
So I started by differentiating the power series and substituting into the ODE.

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

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

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

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 $\displaystyle t$.

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

So now we can add coefficients and combine the summations.

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

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

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

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
• Dec 4th 2010, 08:55 AM
TheEmptySet
Quote:

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

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 $\displaystyle t$.

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

I am not following you reindexing.

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

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

So we should get something that looks like this

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

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

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

I hope this gets you started.
• Dec 4th 2010, 09:57 AM
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 $\displaystyle \displaystyle f(x) = \sum_{n=0}^\infty a_n(x-x_o)^n$ and $\displaystyle \displaystyle \sum_{n=0}^\infty b_n(x-x_o)^n$ to $\displaystyle \displaystyle f(x)+g(x) = \sum_{n=0}^\infty (a_n + b_n)(x-x_o)^n$

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

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!
• Dec 4th 2010, 10:46 AM
TheEmptySet
Quote:

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 $\displaystyle \displaystyle f(x) = \sum_{n=0}^\infty a_n(x-x_o)^n$ and $\displaystyle \displaystyle \sum_{n=0}^\infty b_n(x-x_o)^n$ to $\displaystyle \displaystyle f(x)+g(x) = \sum_{n=0}^\infty (a_n + b_n)(x-x_o)^n$

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

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 $\displaystyle t^2$ term the summations should now start at $\displaystyle n=2$ so both the 2nd and 4th series are where they are supposed to be Now we can reindex the 1st series with $\displaystyle j = n-2$ this gives

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

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

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

Now this gives

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

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

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

Note: it is not always possible to write solutions in a nice closed form, but as many terms as you need can be calculated.
• Dec 4th 2010, 12:47 PM
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 $\displaystyle C_o, C_1, C_2, ... , C_6$, so we can find $\displaystyle C_2$ in terms of $\displaystyle C_o$ and $\displaystyle C_3$ in terms of $\displaystyle C_1$, along with the recursion formula;

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

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

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

But I'm not sure how to start finding my coefficients, namely $\displaystyle C_o$, because we get a $\displaystyle C_{-2}$ term in the recursion formula. Does $\displaystyle C_{-2}$ 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.
• Dec 4th 2010, 01:01 PM
TheEmptySet
Quote:

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 $\displaystyle C_o, C_1, C_2, ... , C_6$, so we can find $\displaystyle C_2$ in terms of $\displaystyle C_o$ and $\displaystyle C_3$ in terms of $\displaystyle C_1$, along with the recursion formula;

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

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

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

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

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

and remember that $\displaystyle n$ is starting at $\displaystyle n=2$ and we can solve for $\displaystyle c_2 \text{ and } c_3$ for what was pulled out of the equation.
• Dec 4th 2010, 01:08 PM
Kasper
Using the initial conditions, I got

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

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

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

Right?

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

But once I find those, then

$\displaystyle C_4 = \frac{-C_2 + C_o}{4\cdot 3}$
and
$\displaystyle C_5 = \frac{-2C_3 + C_1}{5 \cdot 4}$
• Dec 4th 2010, 01:28 PM
TheEmptySet
It took me a bit to find it but back in post 2 when I pulled out the coeffeints

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

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

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

This gives

$\displaystyle (2c_2-c_0)+6c_3t$ This will be nice because now $\displaystyle c_3$ must be zero.
• Dec 4th 2010, 01:45 PM
Kasper
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!