Solving an IVP using Laplace Transforms.

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Solving an IVP using Laplace Transforms. HELP!

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

[tex] \frac{d^2 x}{dt^2}+2\frac{dx}{dt}+x = 5e^{-2t} + t [/tex]

Initial Conditions

x (0) = 2 ; [tex] \frac{dx}{dt} (0) = -3 [/tex]

so I transformed the the IVP and it looks like this

[tex] s^2 x(s) - s x(s) - x (0) + 2 x(s) - x(0) + x(s) = \frac{5}{s+2} + \frac {1}{s^2} [/tex]

Then I plugged in my initial conditions


[tex] s^2 x(s) + 3s - 2 + 2 x(s) - 2 + x(s) = \frac{5}{s+2} + \frac {1}{s^2} [/tex]


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

[tex] s^2 x(s) + 2 x(s) + x(s) + 3s - 4 = \frac{5}{s+2} + \frac {1}{s^2} [/tex]


[tex] 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} [/tex]

[tex] 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)}[/tex]

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

[tex] (s^2 + 2s + 1) x(s) = \frac{5s^2 + s + 2}{(s+2)s^2} - 3s +4 [/tex]

Then I found x(s)

[tex] (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} [/tex]

[tex](-3s +4)*(s^3 + 2s^2) = -3s^4 - 6s^3 + 4s^3 + 8s^2 [/tex]

[tex] x(s) = \frac{5s^2 + s + 2 -3s^4 - 6s^3 + 4s^3 + 8s^2 }{(s+2)(s^2)(s+1)^2} [/tex]

[tex] x(s) = \frac{13s^2 + s + 2 -3s^4 -2s^3 }{(s+2)(s^2)(s+1)^2} [/tex]

Now I used the heaviside theorem to find the residues.

[tex] 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} [/tex]

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

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

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

Answers and Replies

  • #2
798
34
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
Mute
Homework Helper
1,388
10


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

[tex] \frac{d^2 x}{dt^2}+2\frac{dx}{dt}+x = 5e^{-2t} + t [/tex]

Initial Conditions

x (0) = 2 ; [tex] \frac{dx}{dt} (0) = -3 [/tex]

so I transformed the the IVP and it looks like this

[tex] s^2 x(s) - s x(s) - x (0) + 2 x(s) - x(0) + x(s) = \frac{5}{s+2} + \frac {1}{s^2} [/tex]
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
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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.
 
Last edited:
  • #5
22
0


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

[tex] s^2 X(s) - s X(0) - \dot{x}(0) [/tex]

Which in turn becomes

[tex] s^2 X(s) - 2s +3 [/tex]
 

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