Solving an Initial value problem using Laplace transform

Rubik
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Homework Statement


Solve the Initial value problem using Laplace transform
\ddot{y} +2y = 0, y0 = C1, \dot{y} = C2


Homework Equations



[s2 - sy(0) - y'(0)] + a[sY - y(0)] + bY

The Attempt at a Solution


s2Y - sy(0) - y'(0) + 2y = 0
s2Y + 2Y = sy(0) + y'(0)
(s2 + 2)Y = s(C1) + (C2)
Y = (s(C1))/[s2 + 2] + (C2)/[s2 + 2]

And that is as far as I can get as I am unsure what to do now?
 
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Well the laplace transform of what function gives the form s/(s2+k2)?

Similarly what function's transform give k/(s2+k2)?



(Hint:Think trig functions)
 
Oh so I use cos\omegat
 
Actually I am confused does that mean for the first bit my answer is cos(\sqrt{2} t) what happens to the C1?
 
Rubik said:
Oh so I use cos\omegat

Right so cosωt would give ω/(s22), so comparing this to C1* s/(s2+2) what is ω ?

Rubik said:
Actually I am confused does that mean for the first bit my answer is cos(\sqrt{2} t) what happens to the C1?

The C1 should be there, even if your book says it isn't there, it should be there still.
 
so is my answer just C1cos(\sqrt{2} t) otherwise I am really confused..
 
rock.freak667 said:
Right so cosωt would give ω/(s22), so comparing this to C1* s/(s2+2) what is ω ?



My book says it is s/(s2 + \omega2) and not a numerator of \omega
 
Rubik said:
so is my answer just C1cos(\sqrt{2} t) otherwise I am really confused..

No, remember you have two functions.

Rubik said:
rock.freak667 said:
Right so cosωt would give ω/(s22), so comparing this to C1* s/(s2+2) what is ω ?


My book says it is s/(s2 + \omega2) and not a numerator of \omega

Sorry, I wrote the other one.

s/(s22) compared to s/(s2+2) gives ω as?


If you are confused as to what I am trying to get you to see, look up the laplace transforms for cosine and sine.
 
so \omega2 = 2 which implies \omega = \sqrt{2}?
 
  • #10
Good. So you have Y(s) = C_1 \frac{s}{s^2+\omega^2} + (C_2/\omega)\frac{\omega}{s^2+\omega^2}where \omega^2 = 2. Note I multiplied and divided the second term by ω to get the Laplace transform to look like one in the table. The first term corresponds to C_1 \cos \omega t, as you noted earlier. What do you get for the second term?
 
  • #11
Is it right to do the second term as C2(1/(s2 + 2) which corresponds to (1/\sqrt{2}) sin(\sqrt{2}t)?
 
  • #12
Sure. I just wrote it with ω on top because that's how it most likely appears in the table, rather than 1/(s22).
 
  • #13
Thank you so much for all your help :D I am finally beginning to understand!
 
  • #14
Y = (s(C1))/[s2 + 2] + (C2)/[s2 + 2]

And that is as far as I can get as I am unsure what to do now?


Just use tables! Of course you have to jiggle your constants around a bit in order to accommodate the particular form your tables happen to be in.
 
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