1. ## Power Series Solution

Question: The differential equation $\displaystyle y'' + ty' + (t^2-2)y = 0, y(0) = -2, y'(0) = -2$ has a power series solution about t = 0 of the form $\displaystyle$\displaystyle\sum\limits_{n=0}^\infty c_{n}t^{n}. The first six coefficients are: $\displaystyle C_{0}, C_{1}, C_{2}, C_{3}, C_{4}, C_{5} = ?$

I applied the shift when trying to make the all the summations having $\displaystyle t^{n}$, but i get stuck at this step:

$\displaystyle$\displaystyle\sum\limits_{n=0}^\infty c_{n+2}(n+2)(n+1)t^{n}$+$\displaystyle\sum\limits_{n=0}^\infty c_{n+1}(n+1)t^{n}$+$\displaystyle\sum\limits_{n=0}^\infty c_{n}t^{n+2}$+$\displaystyle\sum\limits_{n=0}^\infty 2c_{n}t^{n}

Like how do i get rid of the $\displaystyle t^{n+2}$ in the third summation there?

2. Hmm. Using the ansatz

$\displaystyle \displaystyle y=\sum_{n=0}^{\infty}c_{n}t^{n}$ yields

$\displaystyle \displaystyle y'=\sum_{n=1}^{\infty}n c_{n}t^{n-1}$ and

$\displaystyle \displaystyle y''=\sum_{n=2}^{\infty}n(n-1)c_{n}t^{n-2}.$

Plugging these into the DE yields

$\displaystyle \displaystyle \sum_{n=2}^{\infty}n(n-1)c_{n}t^{n-2} +t\sum_{n=1}^{\infty}n c_{n}t^{n-1} +(t^{2}-2)\sum_{n=0}^{\infty}c_{n}t^{n}=0,$ or

$\displaystyle \displaystyle\sum_{n=2}^{\infty}n(n-1)c_{n}t^{n-2}+\sum_{n=1}^{\infty}nc_{n}t^{n}+\sum_{n=0}^{\inf ty}c_{n}t^{n+2}-\sum_{n=0}^{\infty}2c_{n}t^{n}=0.$

Now, at this point, what I would normally do is break off any pieces I need in order for each sum to start at the same power of $\displaystyle t$ thus:

$\displaystyle \displaystyle 2c_{2}+6c_{3}t+\sum_{n=4}^{\infty}n(n-1)c_{n}t^{n-2}+c_{1}t+\sum_{n=2}^{\infty}nc_{n}t^{n}+\sum_{n=0 }^{\infty}c_{n}t^{n+2}-2c_{0}-2c_{1}t-\sum_{n=2}^{\infty}2c_{n}t^{n}=0.$

Now I would do the index shifting. Does this help?

3. How did you get the first 2 terms and the others? I dont get the break off part

4. What you do is you compare the smallest number that the dummy variable can be in each summation with the power of t in each summation. So, in the expression

$\displaystyle \displaystyle\sum_{n=2}^{\infty}n(n-1)c_{n}t^{n-2}+\sum_{n=1}^{\infty}nc_{n}t^{n}+\sum_{n=0}^{\inf ty}c_{n}t^{n+2}-\sum_{n=0}^{\infty}2c_{n}t^{n}=0,$

the first summation has t to the zeroth power for the first term, the second summation has t to the first power for the first term, the third summation has t to the second power as the first term, and the fourth summation has t to the zeroth power as the first term. Make sense?

You have to have all the summations start at the same power of t. So you break off enough terms from the sum in order to get that to happen.

5. Originally Posted by Belowzero78
Question: The differential equation $\displaystyle y'' + ty' + (t^2-2)y = 0, y(0) = -2, y'(0) = -2$ has a power series solution about t = 0 of the form $\displaystyle$\displaystyle\sum\limits_{n=0}^\infty c_{n}t^{n}. The first six coefficients are: $\displaystyle C_{0}, C_{1}, C_{2}, C_{3}, C_{4}, C_{5} = ?$

I applied the shift when trying to make the all the summations having $\displaystyle t^{n}$, but i get stuck at this step:

$\displaystyle$\displaystyle\sum\limits_{n=0}^\infty c_{n+2}(n+2)(n+1)t^{n}$+$\displaystyle\sum\limits_{n=0}^\infty c_{n+1}(n+1)t^{n}$+$\displaystyle\sum\limits_{n=0}^\infty c_{n}t^{n+2}$+$\displaystyle\sum\limits_{n=0}^\infty 2c_{n}t^{n}

Like how do i get rid of the $\displaystyle t^{n+2}$ in the third summation there?
An alternative approach that doesn't require great efforts is the following: first You suppose that $\displaystyle y(t)$ is analytic in $\displaystyle t=0$ so that is...

$\displaystyle \displaystyle y(t)= \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ t^{n}$ (1)

Then You can use the 'initial conditions' to find $\displaystyle c_{0}$ and $\displaystyle c_{1}$...

$\displaystyle \displaystyle y(0)= -2 \implies c_{0}= -2$

$\displaystyle \displaystyle y^{'}(0)= -2 \implies c_{1}= -2$

For $\displaystyle c_{2}$ You can write the DE as...

$\displaystyle \displaystyle y^{''}= -t\ y^{'} + (2-t^{2})\ y$ (2)

... and from (2) You obtain...

$\displaystyle \displaystyle y^{''}(0)= -4 \implies c_{2} = -2$

For $\displaystyle c_{3}$ You can derive (2)...

$\displaystyle \displaystyle y^{(3)}= -t\ y^{''} + (2-t-t^{2})\ y^{'} -2 t y$ (3)

... and from (3) You obtain...

$\displaystyle \displaystyle y^{(3)} (0) = -4 \implies c_{3} = -\frac{2}{3}$

... and so one... the computation of $\displaystyle c_{4}$ and $\displaystyle c_{5}$ is left to You as exercize...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$