Solution of Initial Value Problems, Laplace Transform

  • Thread starter cc2203
  • Start date
  • #1
1
0
Use the Laplace transform to solve the given initial value problem.

y"+[w^(2)]y=cos2t, w^(2) does not equal 4; y(0)=1, y'(0)=0

I tried doing the problem, and I got up to Y(s)=[(s^(3)+5s]/[s^(2)+w^(2)],
which hopefully is correct. Now I'm having trouble using the Laplace
transforms to finish solving the problem.

Thanks for your help!
 

Answers and Replies

  • #2
dextercioby
Science Advisor
Homework Helper
Insights Author
13,004
554
What's the integral u have to solve to revert the Laplace transform...?

Daniel.
 
  • #3
saltydog
Science Advisor
Homework Helper
1,582
3
Taking the Laplace transform of both sides:

[tex]\mathcal{L}\{y^{''}+\omega^2 y=Cos[2t]\}[/tex]

yields:

[tex]s^2f-s+\omega^2f=\frac{s}{s^2+4}[/tex]

with:

[tex]\mathcal{L}\{y(x)\}=f(s)[/tex]

Solving for f(s) yields:

[tex]f(s)=\frac{s^3+5s}{(s^2+4)(s^2+\omega^2)}[/tex]

Now, you can reduce this using partial fractions with quadratic denominators:

[tex]\frac{s^3+5s}{(s^2+4)(s^2+\omega^2)}=\frac{As+B}{(s^2+4)}+\frac{Cs+D}{(s^2+\omega^2)}[/tex]

When solving for A,B,C,and D, you'll come up against a homogeneous system for B and D, and the constraint imposed on [itex]\omega^2[/itex] will force you to make a conclusion about what B and D can be. Try solving it to completion and report the results here. If you don't, I'll wrap it up tomorrow.
 
  • #4
saltydog
Science Advisor
Homework Helper
1,582
3
Just some wrap-up:

Solving for A and C:

[tex]A=\frac{1}{\omega^2-4}[/tex]

[tex]C=\frac{5-\omega^2}{4-\omega^2}[/tex]

The expressions for B and D are:

[tex]B+D=0[/tex]

[tex]B\omega^2+4D=0[/tex]

You know, I'm not sure about the part below deciding about B and D: Really, I think B and D HAVE to be zero in this case. Someone can correct me if my rational is not so.

Noting that a homogeneous equation has a non-zero solution iff determinant=0, we get:

[tex]4-\omega^2=0[/tex]

However, since constraint above restricts such, only the trival solution is allowed. Thus:

[tex]f(s)=A\frac{s}{s^2+4}+C\frac{s}{s^2+\omega^2}[/tex]

Finally, taking the Laplace transform of both sides:

[tex]y(t)=ACos[2t]+C(Cos[\omega t])[/tex]

A plot is attached.
 

Attachments

Related Threads on Solution of Initial Value Problems, Laplace Transform

Replies
0
Views
929
  • Last Post
2
Replies
29
Views
10K
  • Last Post
Replies
1
Views
2K
Replies
4
Views
2K
Replies
7
Views
4K
Replies
24
Views
3K
Replies
4
Views
3K
  • Last Post
Replies
1
Views
715
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
4
Views
2K
Top