Using Laplace Transforms to solve IVP's

Click For Summary
To solve the initial value problem (IVP) y'' + 2y' + 2y = 0 with conditions y(0) = 1 and y'(0) = -3 using Laplace transforms, the transformation leads to Y(s) = (s - 1) / (s^2 + 2s + 2). The denominator can be simplified by completing the square, resulting in Y(s) = (s - 1) / [(s - 1)^2 + 1]. This form allows the application of the inverse Laplace transform, specifically using the formula for the inverse of a shifted cosine function. The final step involves applying the inverse transform to find the solution in the time domain.
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

Y(s)=\frac{s-1}{(s-1)^2+1}

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

f(t)=L^{-1}\left\{\frac{s-a}{(s-a)^2+b^2} \right\} = e^{at}\cos{(bt)}
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

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