## Solving an IVP using Laplace Transforms.

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?
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 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

Recognitions:
Homework Help
 Quote by geno678 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 $$s^2 x(s) - s x(s) - x (0) + 2 x(s) - x(0) + x(s) = \frac{5}{s+2} + \frac {1}{s^2}$$
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)##.

## Solving an IVP using Laplace Transforms.

 Quote by Mute 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)##.
Oh I see it now that's why I've been messing up. Thanks.

 Quote by Mute 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)##.
So it ends up becoming

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

Which in turn becomes

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

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