1st Order Differential Equation - Power Series Method

Click For Summary
SUMMARY

The discussion focuses on solving a first-order differential equation using the power series method. The participant identifies the general solution format as ##x = x_{HOM} + x_{Inhom}##, concluding that the homogeneous solution ##x_{HOM}## results in ##c_0 = -t \cdot \sin(t)##, with all coefficients ##c_n = 0## for ##n \geq 1##. The participant expresses uncertainty regarding the validity of this solution and suggests rewriting ##\sin(t)## as a power series to explore further simplifications and solutions that yield a sum of zero.

PREREQUISITES
  • Understanding of first-order differential equations
  • Familiarity with power series expansions
  • Knowledge of homogeneous and inhomogeneous solutions
  • Basic skills in manipulating infinite series
NEXT STEPS
  • Explore the method of power series for solving differential equations
  • Learn about the convergence of power series and their applications
  • Study the properties of sine functions in relation to Taylor series
  • Investigate the implications of homogeneous versus inhomogeneous solutions in differential equations
USEFUL FOR

Students and educators in mathematics, particularly those studying differential equations, as well as researchers and practitioners interested in analytical methods for solving such equations.

NicolaiTheDane
Messages
100
Reaction score
10

Homework Statement


upload_2018-10-26_11-27-21.png

The Attempt at a Solution


upload_2018-10-26_11-28-2.png


I have deliberately made several obvious steps, because I keep ending up here. However I have no idea what to do from here. I thought about the fact, that differential equations have the solution ##x = x_{HOM} + x_{Inhom}##, but the ##x_{HOM}## ends up the same, except equal 0, which suggests that the only solution is ##c_0 = -t \cdot sin(t)##. as all ##c_n=0, n \geq 1, t \in (0,\infty)##. But that can't be right, because that actually constitute a solution?
 

Attachments

  • upload_2018-10-26_11-27-21.png
    upload_2018-10-26_11-27-21.png
    4.4 KB · Views: 638
  • upload_2018-10-26_11-28-2.png
    upload_2018-10-26_11-28-2.png
    3.5 KB · Views: 670
Physics news on Phys.org
try writing ##sin(t)## as ##\sum_{0}^{\infty}\frac{(-1)^n t^{2n+1}}{(2n+1)!}##, simplify the expression and then try to find solutions for which the sum is always going to equal to zero
 

Similar threads

Replies
6
Views
3K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
7
Views
2K
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K