Register to reply

Solving ODE using power series

by jessawells
Tags: power, series, solving
Share this thread:
jessawells
#1
Mar31-08, 11:18 PM
P: 20
1. The problem statement, all variables and given/known data

solve the initial value problem:

[tex] x(2-x)y'' - 6(x-1)y' - 4y = 0[/tex]
[tex] y(1)=1 [/tex]
[tex]y'(1) = 0 [/tex]

hint: since the initial condition is given at [tex]x_0 = 1 [/tex], it is best to write the solution as a series centered at [tex]x_0 = 1 [/tex].

2. Relevant equations
I have attempted the question, but I got stuck in one of the steps. Here's what I have so far:

assume the solution can be written as a power series centered about x = 1. Then the solution and its derivatives are:


[tex] y(t) = \displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} [/tex]

[tex] y'(t) = \displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^{n-1}} [/tex]

[tex] y''(t) = \displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}} [/tex]

where the [tex] a_n [/tex] are constant coefficients.



3. The attempt at a solution

Substitute y, y', and y'' into the ODE in order to determine the [tex] a_n [/tex]:

[tex]x(2-x)\displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}}-6(x-1)\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^{n-1}}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0[/tex]

[tex](1-(x-1)^2)\displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}}-6(x-1)\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^{n-1}}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0[/tex]

[tex]\displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n ((x-1)^{n-2}- (x-1)^n)}-6\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^n}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0[/tex]

[tex]\displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}}- \displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^n} -6\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^n}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0[/tex]

now I need to shift indices to make all the exponents of (x-1) equal to n, and also to try to make all the summations start at n=0:

[tex]\displaystyle\sum_{n=0}^{\infty}{(n+2)(n+1) a_{n+2} (x-1)^n}- \displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^n} -6\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^n}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0[/tex]


I don't know how to proceed from here. I have 4 summations. Two of them start at n=0 and has the [tex](x-1)^n[/tex] factor. The other two summations start at n=2 or n=1. I can't get them to start at n=0 because if I try to do that, then I get:

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

and

[tex]6\displaystyle\sum_{n=0}^{\infty}{(n+1) a_{n+1} (x-1)^{n+1}}[/tex]

As you can see, these have the factors [tex] (x-1)^{n+2} [/tex] and [tex] (x-1)^{n+1} [/tex] instead of [tex] (x-1)^n [/tex]. So I don't know how to combine all this, since the summations do not start at the same number.

--------------------------------------------------------------
I'm told that in the end, the answer to the coefficients should be:

[tex]a_{2n} = (n+1)a_0 [/tex] and [tex]a_{2n+1} = \frac{2n+3}{3}a_1 [/tex]

So I just don't know how to get from the point I'm stuck to this answer. I'm also quite confused as to how to solve the initial value problem, after obtaining the expected answer. The initial conditions are: [tex] y(1)=1 [/tex], [tex]y'(1) = 0 [/tex]. Using the initial condition y(1) = 1, I get:

[tex] y(1)= \displaystyle\sum_{n=0}^{\infty}{a_n ((1)-1)^n} [/tex]
[tex] 1 = \displaystyle\sum_{n=0}^{\infty}{a_n (0) [/tex]
[tex] 1 = 0 [/tex]

This does not make sense...what am I doing wrong?
Phys.Org News Partner Science news on Phys.org
Sapphire talk enlivens guesswork over iPhone 6
Geneticists offer clues to better rice, tomato crops
UConn makes 3-D copies of antique instrument parts

Register to reply

Related Discussions
Solving an ODE using a power series Calculus & Beyond Homework 1
Solving for work...Power equations Introductory Physics Homework 2
Power series & Taylor series Calculus & Beyond Homework 4
Solving xy'(x) - y(x) = x^2 Exp[x] using the power series method Calculus & Beyond Homework 5
Solving ODE's by Power Series Introductory Physics Homework 2