# Solving an IVP using Laplace Transforms.

1. Sep 3, 2012

### geno678

Solving an IVP using Laplace Transforms. HELP!

Ok I'm supposed to Solve this problem using Laplace Transforms.

$$\frac{d^2 x}{dt^2}+2\frac{dx}{dt}+x = 5e^{-2t} + t$$

Initial Conditions

x (0) = 2 ; $$\frac{dx}{dt} (0) = -3$$

so I transformed the the IVP and it looks like this

$$s^2 x(s) - s x(s) - x (0) + 2 x(s) - x(0) + x(s) = \frac{5}{s+2} + \frac {1}{s^2}$$

Then I plugged in my initial conditions

$$s^2 x(s) + 3s - 2 + 2 x(s) - 2 + x(s) = \frac{5}{s+2} + \frac {1}{s^2}$$

Then I factored the x(s) on the left side, and found a common denominator on the right side

$$s^2 x(s) + 2 x(s) + x(s) + 3s - 4 = \frac{5}{s+2} + \frac {1}{s^2}$$

$$s^2 x(s) + 2 x(s) + x(s) + 3s - 4 = \frac{5}{s+2}*\frac {s^2}{s^2} + \frac {1}{s^2} *\frac {s+2}{s+2}$$

$$s^2 x(s) + 2 x(s) + x(s) + 3s - 4 = \frac{5s^2}{(s+2)s^2} + \frac {s+2}{(s^2)(s+2)}$$

$$s^2 x(s) + 2 x(s) + x(s) + 3s - 4 = \frac{5s^2 + s + 2}{(s+2)s^2}$$

$$(s^2 + 2s + 1) x(s) = \frac{5s^2 + s + 2}{(s+2)s^2} - 3s +4$$

Then I found x(s)

$$(s+1)(s+1) x(s) = \frac{5s^2 + s + 2}{(s+2)s^2} (- 3s +4)*\frac{(s+2)s^2}{(s+2)s^2}$$

$$(-3s +4)*(s^3 + 2s^2) = -3s^4 - 6s^3 + 4s^3 + 8s^2$$

$$x(s) = \frac{5s^2 + s + 2 -3s^4 - 6s^3 + 4s^3 + 8s^2 }{(s+2)(s^2)(s+1)^2}$$

$$x(s) = \frac{13s^2 + s + 2 -3s^4 -2s^3 }{(s+2)(s^2)(s+1)^2}$$

Now I used the heaviside theorem to find the residues.

$$x(s) = \frac{13s^2 + s + 2 -3s^4 -2s^3 }{(s+2)(s^2)(s+1)^2} = \frac{R1}{s} + \frac{R2}{s^2} + \frac{R3}{s+2} + \frac{R4}{(s+1)^2} + \frac{R5}{s+1}$$

Ok so here is where I'm stuck. Say I try to solve for R3.

$$R3 = (s+2) x(s) |s = -2 , \frac{13s^2 + s + 2 -3s^4 -2s^3 }{(s^2)(s+1)^2} | s = -2 ] = 5$$

Ok so I got the correct residue for R3, but now I'm stuck what do I do after this?

Last edited: Sep 3, 2012
2. Sep 3, 2012

### JJacquelin

Hi !

OK. R3=5
Why not doing the same for the other coefficients ?
multiply by (s+1)² to compute R4
multiply by s² to obtain R2

3. Sep 3, 2012

### Mute

Re: Solving an IVP using Laplace Transforms. HELP!

Before you get too deep into your calculation, you might want to double check this result. The Laplace transform of the second derivative is

$$\mathcal L[\ddot{x}(t)](s) = s^2 X(s) - s x(t=0) - \dot{x}(t=0),$$
whereas you seem to have written

$$s^2 X(s) - sX(s) - x(0).$$
(I'm writing $X(s)$ for the transform to make it easier to distinguish from x(t)).

When you transform the first derivative term, it also looks like you didn't factor through the 2 : you wrote $2X(s)-x(t=0)$ instead of $2X(s) - 2x(t=0)$.

4. Sep 3, 2012

### geno678

Re: Solving an IVP using Laplace Transforms. HELP!

Oh I see it now that's why I've been messing up. Thanks.

Last edited: Sep 3, 2012
5. Sep 3, 2012

### geno678

Re: Solving an IVP using Laplace Transforms. HELP!

So it ends up becoming

$$s^2 X(s) - s X(0) - \dot{x}(0)$$

Which in turn becomes

$$s^2 X(s) - 2s +3$$