# Solving an Initial value problem using Laplace transform

1. Aug 22, 2011

### Rubik

1. The problem statement, all variables and given/known data
Solve the Initial value problem using Laplace transform
$\ddot{y}$ +2y = 0, y0 = C1, $\dot{y}$ = C2

2. Relevant equations

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

3. 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?

2. Aug 22, 2011

### rock.freak667

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)

3. Aug 22, 2011

### Rubik

Oh so I use cos$\omega$t

4. Aug 22, 2011

### Rubik

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

5. Aug 22, 2011

### rock.freak667

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

The C1 should be there, even if your book says it isn't there, it should be there still.

6. Aug 22, 2011

### Rubik

so is my answer just C1cos($\sqrt{2}$ t) otherwise I am really confused..

7. Aug 22, 2011

### Rubik

8. Aug 22, 2011

### rock.freak667

No, remember you have two functions.

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.

9. Aug 22, 2011

### Rubik

so $\omega$2 = 2 which implies $\omega$ = $\sqrt{2}$?

10. Aug 23, 2011

### vela

Staff Emeritus
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. Aug 24, 2011

### Rubik

Is it right to do the second term as C2(1/(s2 + 2) which corresponds to (1/$\sqrt{2}$) sin($\sqrt{2}$t)?

12. Aug 24, 2011

### vela

Staff Emeritus
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. Aug 24, 2011

### Rubik

Thank you so much for all your help :D I am finally beginning to understand!!

14. Aug 25, 2011

### rude man

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.