Solving IVP using Laplace Transforms

[daemon_dkm]

Homework Statement

"Use Laplace transforms to solve the initial value problem" from section 4.2 in Elementary Differential Equations (6th ed.) Edwards & Penny

x'' + x = cos3t; x(0) = 1 & x'(0) = 0

Homework Equations

L{coskt} = s/(s²+k²)
Apparently the answer is 1/8[9cos(t)-cos(3t)].

The Attempt at a Solution

x'' + x = cos3t
L{x''} = s²*X(s) - s*1 - 0
s²X(s) - s + X(s) = s/(s²+3²)
X(s)(s²+1) = s/(s²+3²) + s
X(s) = s/[(s²+3²)(s²+1)] + s/(s²+1)

This is where I get stuck. Do I get a common denominator then do partial fractions?

 

[vela]

There's no need to get a common denominator. Just use partial fractions to break up the first term.

 

[daemon_dkm]

What do I do with the right most part then?
Would I set it up like
s/[(s²+3²)(s²+1)] + s/(s²+1)
= (As+B)/(s²+3²) + (Cs+D)/(s²+1) + s/(s²+1)
s = (As+B)(s²+1) +(Cs+D)(s²+3) + s(s²+3²)

 

[vela]

You can ignore the last part for now, so you want to start with

\frac{s}{(s^3+3^2)(s^2+1)} = \frac{As+B}{s^2+3^2}+\frac{Cs+D}{s^2+1}

Then

X(s) = \frac{As+B}{s^2+3^2}+\frac{Cs+D}{s^2+1} + \frac{s}{s^2+1}

 

[daemon_dkm]

Thank you! That's just what I needed to know!

Because then

0s³ = A³ + C³ => A = -C
0 = B²+D²
0 = B + 3D
s = As + 9Cs
1 = -C + 9C
1 = 8C => C = 1/8 so A = -1/8
<br /> X(s) = -\frac{1}{8}\frac{s}{s^2+3^2}+\frac{1}{8}\frac{s}{s^2+1} + \frac{8}{8}\frac{s}{s^2+1}<br />

<br /> x(t) = L{X(s)} = -\frac{1}{8}\cos(3t)+\frac{9}{8}\cos(t)<br />