Solving DE using power series (w/initial conditions)

In summary, the conversation discussed solving the initial value problem of y'' = y' + y, where y(0) = 0 and y(1) = 1. The method of deriving the power series solution of y(x) = \sum_{n=1}^{\infty}{(F_{n}x^n)/n!} was also mentioned, where {Fn} is the sequence of Fibonacci numbers. There was confusion about the second boundary condition, but it was resolved by realizing it was a misprint and the correct solution was obtained by using y'(0) = 1.
  • #1
Fuzedmind
9
0

Homework Statement



Solve the initial value problem y'' = y' + y where y(0) = 0 and y(1) = 1

derive the power series solution y(x) = [tex]
\ \ \sum_{n=1}^{\infty}{(F_{n}x^n)/n!} \ \ [/tex] where {Fn} is the sequence 0,1,1,2,3,5,8,13... of Fibonacci numbers defined by F0 = 0 and F1 = 1

Homework Equations


The Attempt at a Solution



I plugged in the series and got this equation:

[tex]
\ \ \sum_{n=2}^{\infty}{(n)(n-1)c_{n}x^(n-2)} \ \ + \ \ \sum_{n=1}^{\infty}{c_{n}x^(n-1)} \ \ + \ \ \sum_{n=0}^{\infty}{c_{n}x^n} \ \ = x
[/tex]

from which i got the recurrence equation:

Fn = Fn-1/n + Fn-2/n(n-1)

Problem is I don't know how to get F0 and F1 from the initial conditions. My teacher never taught us how to solve initial condition power series ODE's nor are there any examples in the book. Can anyone explain it to me?
 
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  • #2
You are assuming y = c0 + c1x + c2x2+...

y(0) is just c0. And y'(0) would be c1 , which would be the usual conditions for an IVP. Are you sure you have the second boundary condition written correctly? My guess it is supposed to be y'(0) = 1.
 
  • #3
no y(0) = 0 and y(1) = 1, I am looking at it right now
 
  • #4
I don't have time to work through the solution right now, but what you have then is a boundary value problem, not an initial value problem as claimed. I'm suspecting a misprint. You might just try using y'(0)=1 and working it through to see if you get the correct answer.
 
  • #5
I had a little time later in the day. Your text indeed has a misprint. If you take your equation with the initial conditions y(0) = 0 and y'(0) = 1, which give you the values of c0 and c1 and work out the series solution, you get exactly what the problem claims you will get.
 
  • #6
Ok thanks a bunch man. I did that and it worked out for me too.
 

FAQ: Solving DE using power series (w/initial conditions)

How do you solve a differential equation using power series?

To solve a differential equation using power series, you need to first rewrite the equation into a series form. This involves substituting the power series expansion of the unknown function into the differential equation. Then, you can solve for the coefficients of the series by comparing coefficients of like terms on both sides of the equation. Finally, use the initial conditions to determine the value of the coefficients and find the solution to the differential equation.

What are the advantages of using power series to solve differential equations?

Power series provide a systematic and efficient method for solving differential equations. They can be used to find exact solutions and can also approximate solutions to a desired degree of accuracy. Additionally, power series can be used to solve a wide range of differential equations, including those that cannot be solved using traditional methods.

Can power series solutions be used for any type of initial conditions?

Yes, power series solutions can be used for any type of initial conditions. The initial conditions are used to determine the values of the coefficients in the power series expansion, which in turn determines the solution to the differential equation. As long as the initial conditions are well-defined and the series converges, power series can be used to find a solution.

How do you know if a power series solution is valid?

A power series solution is valid if the series converges to a finite value for all values of the independent variable. This can be determined by checking the radius of convergence of the series. If the radius of convergence is greater than the value of the independent variable, the solution is valid. It is also important to check for any singularities or points where the series may not converge.

Are there any special cases where power series may not be the best method for solving a differential equation?

While power series are a powerful tool for solving differential equations, there may be cases where they are not the most efficient method. For example, if the differential equation has a simple analytical solution using traditional methods, it may be easier to use those methods instead of finding a power series solution. Additionally, if the initial conditions are not well-defined or the series does not converge, power series may not be a suitable method for solving the differential equation.

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