How Do You Find a Recurrence Relation for a Power Series Solution of an ODE?

[james weaver]

Homework Statement

Solve the DE in terms of a power series

Relevant Equations

power series

I believe I am doing everything right up until the point where I have to try and find a recurrence relation. I honestly have no idea what to do from there. I've listed my work in getting the powers of n and the indicies to all match. Any help appreciated.

Here is the original DE:
$y^{'}-(x-1)y=0$

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

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

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

$2a_0+\sum_{n=1}^\infty {(n+1)}a_{n+1} x^{n}-\sum_{n=1}^\infty a_{n-1} x^{n}+\sum_{n=1}^\infty a_n x^{n}=0$

$2a_0+\sum_{n=1}^\infty \left[{(n+1)}a_{n+1} -a_{n-1}+a_n\right]x^n=0$$a_0=0$
$a_1=-2a_2$
$a_2=-a_3$
$a_3=-2a_4$
$a_4=-\frac 5 3 a_5$
$a_5=-\frac 9 4 a_6$

And this is where I get stuck.

 

[FactChecker]

Write down the general equation of the coefficient for a power of x. Then solve for the coefficient with the highest index in terms of the coefficients with lower indices. That gives the recurrence relation that you will need.

 

[Delta2]

As it is deduced from the last line of his derivation, in order for that polynomial to be zero for every $x$, the coefficient of every power $x^n$ must be zero so $$(n+1)a_{n+1}-a_{n-1}+a_n=0$$ is the recurrence relation for $a_n$. (and $a_0=0$).

 

[Orodruin]

Your constant term looks wrong to me, it should be $a_0 + a_1$.

It is also unclear to me if your problem is finding the recursion relation itself (as answered above) -since you have quoted several coefficients- or solving the recursion relation.

 

[FactChecker]

Your constant term looks wrong to me, it should be $a_0 + a_1$.

Or is it that better to say that $a_0$ can be any initial value and then $a_0+a_1=0 \implies a_1= - a_0$. As is typical of these problems, the first few coefficients have to be done carefully as special cases taking the initial conditions into account. Then the general equation of the n'th coefficient can be routinely applied for the coefficients of the higher powers. EDIT: Hopefully, you can see the pattern and put all of the n'th coefficients in terms of $a_0$.

It is also unclear to me if your problem is finding the recursion relation itself (as answered above) -since you have quoted several coefficients- or solving the recursion relation.

Assuming that the title has been carefully stated, I interpret the problem as just defining the Taylor series.

 

[Orodruin]

Or is it that better to say that a0 can be any initial value and then a0+a1=0⟹a1=−a0. As is typical of these problems, the first few coefficients have to be done carefully as special cases taking the initial conditions into account. Then the general equation of the nth coefficient can be routinely applied for the coefficients of the higher powers.

This is always the case. However, in this case we have a first order ODE leading to a recursion relation that contains three different coefficients. The first requires a single initial value, which translates to $a_0$ whereas the second requires two. The second ($a_1$) is determined by the differential equation through the constant term. The recursion relation can then be solved by z-transform.

Assuming that the title has been carefully stated, I interpret the problem as just defining the Taylor series.

The title just asks to solve the ODE using a power series. I agree that the actual wording of the question makes it seem like the recursion relation is what the OP has trouble with, but the work shown at least partially contradicts this seeing that the OP quotes several coefficients that are actually found. The step from the OP’s expression to the recursion relation is almost infinitesimal.

 

[FactChecker]

I agree that the actual wording of the question makes it seem like the recursion relation is what the OP has trouble with, but the work shown at least partially contradicts this seeing that the OP quotes several coefficients that are actually found.

Yes. It is not clear if the OP was going in the correct direction or if it went in the wrong direction and got stuck.

The step from the OP’s expression to the recursion relation is almost infinitesimal.

It's not clear if the title should be taken literally. If so, then the Taylor series is the goal.

 

[pasmith]

The question doesn't actually require that the power series be centered at zero, so it is more natural here to set y(x) = \sum_{n=0}^\infty b_n(x - 1)^n. Also the ODE is easily solved by separation, so you can compare your answer to the power series of that solution.

 

[FactChecker]

The question doesn't actually require that the power series be centered at zero, so it is more natural here to set y(x) = \sum_{n=0}^\infty b_n(x - 1)^n.

Good point! Although it is not mentioned in the OP, I wonder if the x-location of initial conditions would make that x value a better place to expand at.