Using Laplace Transforms to solve IVP's

Click For Summary
SUMMARY

The discussion focuses on solving the initial value problem (IVP) defined by the differential equation y'' + 2y' + 2y = 0 with initial conditions y(0) = 1 and y'(0) = -3 using Laplace transforms. The solution involves transforming the equation into the Laplace domain, yielding Y(s) = (s - 1) / (s^2 + 2s + 2). The key step is recognizing that the denominator can be simplified to (s - 1)^2 + 1 through completing the square, allowing for the application of the inverse Laplace transform formula f(t) = L^{-1}{(s - a) / ((s - a)^2 + b^2)} = e^{at}cos(bt).

PREREQUISITES
  • Understanding of Laplace transforms and their properties
  • Familiarity with solving ordinary differential equations (ODEs)
  • Knowledge of initial value problems (IVPs)
  • Ability to complete the square in algebraic expressions
NEXT STEPS
  • Study the properties of Laplace transforms in detail
  • Learn how to apply inverse Laplace transforms for different functions
  • Practice solving more complex initial value problems using Laplace transforms
  • Explore the method of partial fractions for Laplace transform solutions
USEFUL FOR

Students studying differential equations, mathematicians interested in Laplace transforms, and educators teaching ODEs and IVPs.

dm59
Messages
5
Reaction score
0

Homework Statement


solve the ivp using laplace tranforms

y''+2y'+2y=0 y(0)=1 y'(0)=-3


Homework Equations





The Attempt at a Solution



get to Y(s)[s^2+2s+2]=s-1

Y=(s-1)/[s^2+2s+2]

^^^
don't know how to simplify the denominator to solve using Laplace transforms. If I had to guess I would say maybe partial fractions but keep getting the wrong answer when I try to use them.
 
Physics news on Phys.org
You've pretty much finished it, all you need to recognize that

[tex]Y(s)=\frac{s-1}{(s-1)^2+1}[/tex]

through completing the square in the denominator. Now can you get that to work with

[tex]f(t)=L^{-1}\left\{\frac{s-a}{(s-a)^2+b^2} \right\} = e^{at}\cos{(bt)}[/tex]
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
9
Views
3K