2nd Order Differential Equation via Power Series

• VVS
In summary, the conversation discusses problem 5 on an attached sheet related to the Power Series Method for finding linearly independent solutions of a differential equation. It is mentioned that one solution can be found using the method and then the second solution can be found using that solution. The process of finding the second solution is discussed, including the use of initial conditions and integration by parts. Ultimately, it is concluded that the integral does not need to be calculated and the problem is solved.
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?

VVS

Last edited:
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.

VVS said:

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 $y(0) = a_0$ and $y'(0) = a_1$.

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

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.

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.

thanks
VVS

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

1. What is a 2nd order differential equation via power series?

A 2nd order differential equation via power series is a type of mathematical equation that involves finding a solution in the form of a power series. This type of equation is often used in physics and engineering to model systems with changing variables over time.

2. How do you solve a 2nd order differential equation via power series?

To solve a 2nd order differential equation via power series, you must first convert the equation into a series by expanding each term using the binomial theorem. Then, substitute the series into the equation and equate the coefficients of each term to zero. This will result in a recurrence relation that can be solved to find the coefficients of the series.

3. What are some real-world applications of 2nd order differential equations via power series?

Some examples of real-world applications of 2nd order differential equations via power series include modeling the motion of a pendulum, analyzing the vibrations of a guitar string, and predicting the behavior of electrical circuits. These equations are also used in fields such as economics, biology, and chemistry to model various systems.

4. What are the limitations of using power series to solve 2nd order differential equations?

One limitation of using power series to solve 2nd order differential equations is that it may not always be possible to find a solution in closed form. In some cases, the power series may converge to an infinite sum, making it difficult to find an exact solution. Additionally, power series are only valid for certain types of differential equations and may not be applicable to all systems.

5. Are there any alternative methods for solving 2nd order differential equations besides power series?

Yes, there are several alternative methods for solving 2nd order differential equations, including the method of undetermined coefficients, variation of parameters, and Laplace transforms. Each method has its own advantages and limitations, so it is important to choose the most appropriate method for a given equation.

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