Differential Equations - Power Series problem with e^t

Click For Summary
SUMMARY

The discussion centers on solving the differential equation y'' + ty' + e^ty = 0 with initial conditions y(0) = 0 and y'(0) = -1. The user struggles with the recursion relation due to the presence of the e^t term, which complicates the factoring of t^k. Key findings include that a_2 = -a_0/2, a_1 = 0, and a_3 = 0. The user is advised to utilize the Cauchy Product for multiplying series, which is essential for correctly handling the series expansion of e^ty.

PREREQUISITES
  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with power series and their convergence.
  • Knowledge of recursion relations in series solutions.
  • Basic understanding of the Cauchy Product for series multiplication.
NEXT STEPS
  • Study the application of the Cauchy Product in series multiplication.
  • Learn about the method of Frobenius for solving differential equations with series solutions.
  • Investigate the properties of exponential functions in series form, particularly e^t.
  • Explore advanced techniques for handling non-homogeneous linear differential equations.
USEFUL FOR

Students and educators in mathematics, particularly those focused on differential equations, as well as anyone seeking to deepen their understanding of power series solutions and recursion relations.

skyturnred
Messages
117
Reaction score
0

Homework Statement



The problem is to solve:

y''+ty'+e[itex]^{t}[/itex]y=0, y(0)=0 and y'(0)=-1

Homework Equations





The Attempt at a Solution



My main issue is the following: I normally find the recursion relation, and then factor out the t[itex]^{whatever}[/itex] and I know that the coefficient to this must equal 0. However, the addition of the e[itex]^{t}[/itex] means I can't factor out ALL of the t[itex]^{whatever}[/itex] so I can't exactly equate the coefficient to 0.

So specifically with the question above, I get the following:

ty'=[itex]\sum[/itex][itex]^{inf}_{k=1}[/itex]ka[itex]_{k}[/itex]t[itex]^{k}[/itex]

e[itex]^{t}[/itex]=[itex]\sum[/itex][itex]^{inf}_{k=0}[/itex]t[itex]^{k}[/itex]/k!

e[itex]^{t}[/itex]y=[itex]\sum[/itex][itex]^{inf}_{k=0}[/itex]t[itex]^{2k}[/itex]a[itex]_{k}[/itex]/k!

y''=[itex]\sum[/itex][itex]^{inf}_{k=2}[/itex]k(k-1)a[itex]_{k}[/itex]t[itex]^{k-2}[/itex]

Then I plug those into the DE, write out the first few terms so that they all start at k=2, and put them all under the same sum to get as follows:

a[itex]_{0}[/itex]+a[itex]_{1}[/itex]t+a[itex]_{1}[/itex]t[itex]^{2}[/itex]+2a[itex]_{2}[/itex]+6a[itex]_{3}[/itex]t+[itex]\sum[/itex][itex]^{inf}_{k=2}[/itex][(k+2)(k+1)a[itex]_{k+2}[/itex]+ka[itex]_{k}[/itex]+a[itex]_{k}[/itex]t[itex]^{k}[/itex]]t[itex]^{k}[/itex]=0

From this you find that a[itex]_{2}[/itex]=-a[itex]_{0}[/itex]/2

and

a[itex]_{1}[/itex]=0

and

a[itex]_{3}[/itex]=0


Like I said, normally at this point I find the recursion relation and factor out a t[itex]^{k}[/itex] so that I know that the entire coefficient is 0. But in this case, after factoring out the t[itex]^{k}[/itex] one of the terms in the coefficient still have a t[itex]^{k}[/itex], so I don't know if I can't just equate this whole thing to 0.

Can anyone help?

Thank-you
 
Physics news on Phys.org
skyturnred said:

Homework Statement



The problem is to solve:

y''+ty'+e[itex]^{t}[/itex]y=0, y(0)=0 and y'(0)=-1

Homework Equations





The Attempt at a Solution



My main issue is the following: I normally find the recursion relation, and then factor out the t[itex]^{whatever}[/itex] and I know that the coefficient to this must equal 0. However, the addition of the e[itex]^{t}[/itex] means I can't factor out ALL of the t[itex]^{whatever}[/itex] so I can't exactly equate the coefficient to 0.

So specifically with the question above, I get the following:

[itex]ty'=\sum^{\infty}_{k=1}ka_{k}t^{k}[/itex]

You don't need to put the itex tags around every item. Just put them around the whole equation. Right click on the above expression and see how I changed it for you.

e[itex]^{t}[/itex]=[itex]\sum[/itex][itex]^{inf}_{k=0}[/itex]t[itex]^{k}[/itex]/k!

e[itex]^{t}[/itex]y=[itex]\sum[/itex][itex]^{inf}_{k=0}[/itex]t[itex]^{2k}[/itex]a[itex]_{k}[/itex]/k!

You don't multiply two series like that. You need the Cauchy Product. See:
http://planetmath.org/cauchyproduct
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
8
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K