2nd Order Differential Equation via Power Series

[VVS]

Homework Statement

Problem 5 on the attached Sheet here View attachment M2Uebung2.pdf

Homework Equations

We studied the Power Series Method and how to calculate a linearly independent solution if one solution is already known.
So we need to find one solution (probably) using the power series method and then using that solution we can find the 2nd linearly independent solution.

The Attempt at a Solution

See the attachment for working. View attachment Übung 5.pdf
So I have found a recursion formula.
And it can be seen that all even coefficients above n=2 are vanishing, but what about a2 and the odd coefficients?


Thanks for your help!
VVS

 

[haruspex]

It would appear that a0 and a1 can be arbitrarily assigned (unsurprising for the DE) and all further terms written in terms of them. So to get one solution you can plug in whatever you like for a0 and a1 (except both zero) and see what results. In fact, it's not hard to write out a general form for the odd terms using factorials.

 

[pasmith]

Homework Statement

Problem 5 on the attached Sheet here View attachment 62720

Homework Equations

We studied the Power Series Method and how to calculate a linearly independent solution if one solution is already known.
So we need to find one solution (probably) using the power series method and then using that solution we can find the 2nd linearly independent solution.

The Attempt at a Solution

See the attachment for working. View attachment 62721
So I have found a recursion formula.
And it can be seen that all even coefficients above n=2 are vanishing, but what about a2 and the odd coefficients?

In principle there are initial conditions on [itex]y(0) = a_0[/itex] and [itex]y'(0) = a_1[/itex].

Since you aren't given any initial conditions, you will have to choose some. Taking [itex]a_0 = 1[/itex] and [itex]a_1 = 0[/itex] seems good, since as you have observed all even terms other than [itex]a_0[/itex] and [itex]a_2[/itex] will vanish, as will all the odd terms. Then you can obtain the other solution by the suggested integral method.

 

[VVS]

Hi Haruspex and Pasmith!

Thanks for your help. Yeah, I asked my professor whether there is something wrong with the assignement.
He just wrote that I should take a closer look.
I guess Pasmith is right. Any possible solution can be regarded as a solution for the differential equation.
So we can choose a0=1 and a1=0.
I will actually try to use the recursion formula and find a general expression for the second linearly independent solution.

Thanks for your help again.
VVS

 

[VVS]

Second Linearly Independent Solution

Hey,

I got the solution for the first linearly independent solution of the Differential Equation.
But now I am facing another problem.

The integral for the second linearly independent turns out to be quite complicated.
I was able to break it down a bit by using integration by parts. See Working here View attachment Übung 5_3.pdf

But there must be simpler way.

I really appreciate your help.

thanks
VVS

 

[VVS]

Actually we don't need to calculate the integral.
Sorry for wasting posts.
So this problem is solved.