# Solution of Initial Value Problems, Laplace Transform

1. Apr 29, 2005

### cc2203

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.

2. Apr 30, 2005

### dextercioby

What's the integral u have to solve to revert the Laplace transform...?

Daniel.

3. May 3, 2005

### saltydog

Taking the Laplace transform of both sides:

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

yields:

$$s^2f-s+\omega^2f=\frac{s}{s^2+4}$$

with:

$$\mathcal{L}\{y(x)\}=f(s)$$

Solving for f(s) yields:

$$f(s)=\frac{s^3+5s}{(s^2+4)(s^2+\omega^2)}$$

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

$$\frac{s^3+5s}{(s^2+4)(s^2+\omega^2)}=\frac{As+B}{(s^2+4)}+\frac{Cs+D}{(s^2+\omega^2)}$$

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 $\omega^2$ 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. May 4, 2005

### saltydog

Just some wrap-up:

Solving for A and C:

$$A=\frac{1}{\omega^2-4}$$

$$C=\frac{5-\omega^2}{4-\omega^2}$$

The expressions for B and D are:

$$B+D=0$$

$$B\omega^2+4D=0$$

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:

$$4-\omega^2=0$$

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

$$f(s)=A\frac{s}{s^2+4}+C\frac{s}{s^2+\omega^2}$$

Finally, taking the Laplace transform of both sides:

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

A plot is attached.

File size:
5 KB
Views:
96