Variation of parameters for a second order ODE

Click For Summary

Homework Help Overview

The problem involves solving the second-order ordinary differential equation (ODE) given by ##y''+2y'+2y=e^{-t}\sin t## using the method of variation of parameters. The original poster expresses difficulty in applying this method effectively, despite its theoretical applicability.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to solve the homogeneous ODE and derive the complementary solution, but struggles with the particular solution. Some participants question the correctness of the original poster's expression for the particular solution and suggest checking for missing exponents and collecting real and imaginary terms. Others propose an alternative form for the solution that may simplify the process.

Discussion Status

The discussion is active, with participants providing guidance on rewriting the complementary solution and suggesting alternative approaches. There is no explicit consensus, but some productive directions have been offered to help clarify the original poster's understanding of the problem.

Contextual Notes

The original poster references using external resources, such as Wolfram Alpha, to assist with integral evaluations, indicating a reliance on computational tools for verification. There may be assumptions about the method's applicability that are being questioned by participants.

fluidistic
Gold Member
Messages
3,932
Reaction score
283

Homework Statement


I must solve ##y''+2y'+2y=e^{-t}\sin t##.
I know variation of parameters might not be the fastest/better way to solve this problem but I wanted to practice it as I never, ever, could solve a DE with it. (Still can't with this one). Though the method is supposed to work.

Homework Equations


Variation of parameters.


The Attempt at a Solution


I quickly solved the homogeneous ODE, the complementary solution is ##y_c (t)=c_1\underbrace{e^{(-1-i)t}}_{y_1(t)}+c_2\underbrace{e^{(-1+i)t}}_{y_2(t)}##, I even manually derivated it twice and plugged it into the ODE, it does satisfy the homogeneous ODE.
I've calculated the Wronskian of ##y_1(t)## and ##y_2(t)##, it's worth ##2ie^{-2t}##.
Now according to all sources of information I found, the particular solution is ##y_p =-y_1(t) \int \frac{y_2(t)e^{-t}\sin t}{W(y_1,y_2)(t)}dt+y_2(t) \int \frac{y_1 e^{-t}\sin t}{W(y_1,y_2)(t)}dt##.
Using a "cheat" taken from Wolfram Alpha, namely that ##\int e^{it}\sin t dt= \frac{it}{2}-\frac{1}{4}e^{2it}##, I reach a total nonsense : ##y_p(t)=e^{(-1-i)t}\left ( \frac{i}{8}- \frac{t}{4} \right ) - e^{(-1+i)t} \left ( \frac{i}{8}+ \frac{t}{4} \right )##. The answer should be ##y_p(t)=-\frac{e^t}{2}t \cos t##.
I'm totally clueless on where is/are my error(s). It's a simple integral's evaluation and I sought help from wolfram alpha ( http://www.wolframalpha.com/input/?i=integral+e^(i*x)sin+x+dx to be exact). There's no reason the method should fail, yet it "seems that it does fail".
 
Physics news on Phys.org
Check your yp. Is not some exponents missing? Write e±it=cos(t)± isin(t). Collect the real and imaginary terms in yp and see what you get.

ehild
 
Last edited:
In terms of real variables, the general solution to the associated homogeneous equation, y''+ 2y'+ 2y= 0 is y= e^{-t}(Acos(t)+ Bsin(t)).

You will find it simpler to use y= u(t)e^{-t}sin(t)+ v(t)e^{-t}cos(t) as your solution.
 
Thanks guys, the rewriting of the complementary solution did the job. I'll keep this in mind for next time(s).
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K