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Series Solution to Second Order DE

  • #1

Homework Statement


Consider a power series solution about x0 = 0 for the differential equation y'' + xy' + 2y = 0.
a) Find the recurrence relations satisfied by the coefficients an of the power series solution.
b) Find the terms a2, a3, a4, a5, a6, a7, a8 of this power series in terms of the first two terms a0, a1.
c) Deduce the general form of the coefficients an and write down the general solution as a linear combination of 2 linearly independent power series solutions.
d) What is the radius of convergence of each one of the two linearly independent power series solutions.
f) Find the power series solution of the initial value problem y'' + xy' + 2y = 0. y(0) = 3, y' (0) = −1. g) Give the polynomial approximations of degrees 4 and 5 for the solution of the initial value problem in (f).

Homework Equations



y=Σanxn

The Attempt at a Solution


[/B]
For a) I let y=∑anxn

and through taking the derivative and substituting I got

∑xn[(n+2)(n+1)an+ 2 + nan +2an ]= 0
Which showed that the recurrence relation is

(n+2)(n+1)an+2 +nan+2an=0
which simplifies to an+2 = -an/(n+1)

b) I found that

a2 = -a0
a4 = -a0/(3)
a6 = -a0/(3*5)
a8 = -a0/(3*5*7)

Similarly,

a3 = -a1/2
a5 = -a1/(2*4)
a7 = -a1/(2*4*6)

c) I know I have to try to find a solution where I can write a0 ∑ + a1
but I don't know any series that has the pattern that I observed so I am lost.
 
Last edited:

Answers and Replies

  • #2
LCKurtz
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You don't need to know the pattern hoping it comes out something familiar like and exponential or trig function. Just write down the two series that have ##a_0## and ##a_1## in them. It's best if you can figure out a formula for the nth term of each so you can test them for convergence. And, of course, ##a_0## and ##a_1## are your two arbitrary constants.
 
  • #3
Would I be correct in writing it as


y=a0 [-1 + x3/3 - x5 /(3*5) + x7 /(3*5*7) + ...+ ] + a1 [ -x2/2 + x4/(2*4) - x6/(2*4*6)]
 
  • #4
LCKurtz
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Assuming your arithmetic is correct, yes but not complete. You need to figure out the general term of each to test for convergence. For example, look at your ##a_0## series. You have odd powers of ##x##. So the general exponent on ##x## could be written ##x^{2n-1}##. Your calculus book probably has examples of series with denominators like that. You could write that denominator like ##3\cdot 5\cdot 7\cdots (?)## where you need the formula for the last term in terms of ##n##. Similarly for the even terms.
 
  • #5
I think I would get a general solution of

y= a0 ∑(-1)n x2n-1/(2n-1)] for my first term however I am still working on my second term
 
  • #6
LCKurtz
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I think I would get a general solution of

y= a0 ∑(-1)n x2n-1/(2n-1)] for my first term however I am still working on my second term
That isn't correct. Notice that the denominators start with 3 and have more factors as they get larger. Also you will likely need to check if the first term or two agree with the general formula. And put the correct index on the sum. For the ##a_1## series, a hint is to take a ##2## out of each factor.
 

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