How can the given differential equation be solved using power series?

In summary, the conversation discusses solving a given differential equation using a power series and finding the recurrence relation and two linearly independent solutions. The main difficulty is that the terms in the recurrence relation are expressed in terms of both a0 and a1, and it is unclear how to obtain two linearly independent solutions.
  • #1
cepheid
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I'm having some difficulty with this question:

Solve the given differential equation by means of a power series about the given point x0. Find the recurrence relation. Also find the first four terms in each of two lin. indep. solutions. If possible, find the general term in each solution:

[tex] y'' - xy' - y = 0 , \ \ x_0 = 1 [/tex]

First of all, like in all the preceding problems, I noted that the coefficient of the y'' term, P(x), was equal to 1, so every point, including x0, is an ordinary point. I therefore assumed that the power series was of the form:

[tex] \sum_{n=0}^{\infty}{a_n(x - x_0)^n} [/tex]

and converged over some interval [itex] |x - x_0| < \rho [/itex], which I guess I would normally confirm later, if the question actually required it. So after differentiating the power series representation of y to give me y' and y'', and doing some appropriate manipulations, I was able to express the DE as follows:

[tex] 2a_2 - a_1 - a_0 \ \ + \ \ \sum_{n=1}^{\infty}{(n+2)(n+1)a_{n+2}(x - 1)^n} \ \ - \ \ \sum_{n=1}^{\infty}{(n+1)a_{n+1}(x - 1)^n} \ \ - \ \ \sum_{n=1}^{\infty}{na_{n}(x - 1)^n} \ \ - \ \ \sum_{n=1}^{\infty}{a_{n}(x - 1)^n} = 0 [/tex]

This gave the recurrence relation:

[tex] (n+2)(n+1)a_{n+2} - a_{n+1} - a_n = 0 [/tex]

This recurrence relation is correct (I checked the answers). But the problem I'm having is that each term is expressed in terms of both a1 and a0:

[tex] a_2 = \frac{a_1 + a_0}{2} [/tex]

[tex] a_3 = \frac{3a_1 + a_0}{6} [/tex]

Normally, the even terms are expressed in terms of a0, and the odd terms in terms of a1. So, both sets of terms satisfy the recurrence relation, and you can split them up to obtain two lin. indept. solutions with arbitrary constants a0 and a1. However, in this case, you get (x-1)^3 (ie odd terms, for example) with coefficients that include both a1 and a0. What they seem to have done in the answers is split up the (x-1)^n term into two parts, one with a0 and the other with a1. So for example, one solution would get the (a1 / 2) *(x-1)^3 term, and the other solution would get the (a0 / 6) *(x-1)^3 term. But neither of these terms sastifies the recurrence relation on its own, only when they are together! So if you split them up and put them in two separate series, then neither series is a soln' to the DE! How DO you get two linearly independent solutions from this recurrence relation?!
 
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  • #2
Just a little bump, as this thread seems to have gotten lost in the woodwork. I know it's long and convoluted, but the question is fairly straightforward.
 
  • #3
cepheid said:
Just a little bump, as this thread seems to have gotten lost in the woodwork. I know it's long and convoluted, but the question is fairly straightforward.

You have two linearly independent solutions by taking {a0=1 and a1=0} and
{a0=0 and a1=1}, for example.

ehild
 
Last edited:

1. What is the purpose of solving ODE's by power series?

The purpose of solving ODE's (ordinary differential equations) by power series is to find an approximate solution to a given differential equation. This method can be useful when it is difficult or impossible to find an exact solution using other methods.

2. How does the power series method work?

The power series method involves representing the unknown function in the differential equation as a power series, which is an infinite sum of terms with increasing powers of the variable. By substituting this series into the differential equation and comparing coefficients, we can solve for the unknown coefficients and find an approximate solution.

3. What types of differential equations can be solved using this method?

The power series method can be used to solve both linear and nonlinear differential equations. However, it is most effective for linear equations with constant coefficients. It can also be used for some special types of nonlinear equations, such as Riccati equations.

4. What are the advantages of using the power series method?

One advantage of the power series method is that it can be used to find approximate solutions to a wide range of differential equations. It also allows for a systematic and straightforward approach to solving these equations, making it easier to understand and apply. Additionally, it can be used to find solutions that may not be easily obtained using other methods.

5. Are there any limitations to using the power series method?

While the power series method can be a powerful tool for solving differential equations, it does have some limitations. It may not always yield an exact solution, and the convergence of the series may be limited to a specific region. It also requires some knowledge of power series and Taylor series expansions, which may be challenging for some individuals.

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