Power Series Solution

**Belowzero78** · November 30th 2010, 06:47 PM

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

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

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

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

**Ackbeet** · November 30th 2010, 07:12 PM

Hmm. Using the ansatz

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

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

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

Plugging these into the DE yields

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

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

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 $t$ thus:

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

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

**Belowzero78** · November 30th 2010, 07:26 PM

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

**Ackbeet** · November 30th 2010, 07:31 PM

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

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

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.

**chisigma** · November 30th 2010, 10:36 PM

Originally Posted by Belowzero78

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

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

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

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

An alternative approach that doesn't require great efforts is the following: first You suppose that $y(t)$ is analytic in $t=0$ so that is...

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

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

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

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

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

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

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

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

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

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

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

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

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

Kind regards

$\chi$ $\sigma$

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