Solving an Initial value problem using Laplace transform

In summary, we use the Laplace transform to solve the initial value problem \ddot{y} +2y = 0, y0 = C1, \dot{y} = C2 by converting it into the form s2Y + 2Y = sy(0) + y'(0). By comparing the transformed equation to the form s/(s2+k2), we can determine that the solution to the problem is Y(s) = C_1 \frac{s}{s^2+\omega^2} + (C_2/\omega)\frac{\omega}{s^2+\omega^2}, where \omega^2 = 2. By referencing the Laplace transform table for cosine and sine
  • #1
Rubik
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0

Homework Statement


Solve the Initial value problem using Laplace transform
[itex]\ddot{y}[/itex] +2y = 0, y0 = C1, [itex]\dot{y}[/itex] = 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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  • #2
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
Oh so I use cos[itex]\omega[/itex]t
 
  • #4
Actually I am confused does that mean for the first bit my answer is cos([itex]\sqrt{2}[/itex] t) what happens to the C1?
 
  • #5
Rubik said:
Oh so I use cos[itex]\omega[/itex]t

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([itex]\sqrt{2}[/itex] 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.
 
  • #6
so is my answer just C1cos([itex]\sqrt{2}[/itex] t) otherwise I am really confused..
 
  • #7
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 + [itex]\omega[/itex]2) and not a numerator of [itex]\omega[/itex]
 
  • #8
Rubik said:
so is my answer just C1cos([itex]\sqrt{2}[/itex] 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 + [itex]\omega[/itex]2) and not a numerator of [itex]\omega[/itex]

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
so [itex]\omega[/itex]2 = 2 which implies [itex]\omega[/itex] = [itex]\sqrt{2}[/itex]?
 
  • #10
Good. So you have [tex]Y(s) = C_1 \frac{s}{s^2+\omega^2} + (C_2/\omega)\frac{\omega}{s^2+\omega^2}[/tex]where [itex]\omega^2 = 2[/itex]. 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 [itex]C_1 \cos \omega t[/itex], 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/[itex]\sqrt{2}[/itex]) sin([itex]\sqrt{2}[/itex]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.
 

FAQ: Solving an Initial value problem using Laplace transform

1. What is an initial value problem?

An initial value problem is a type of differential equation that involves finding a function that satisfies a given differential equation and a set of initial conditions. These initial conditions typically specify the value of the function at a particular point.

2. What is the Laplace transform?

The Laplace transform is a mathematical tool used to solve differential equations, particularly initial value problems. It transforms a function of time into a function of complex frequency, making it easier to solve certain types of differential equations.

3. How do you use the Laplace transform to solve an initial value problem?

To use the Laplace transform to solve an initial value problem, you first apply the transform to both sides of the differential equation. This converts the differential equation into an algebraic equation. You can then solve for the transformed function and use the inverse Laplace transform to find the solution to the original differential equation.

4. What are the advantages of using the Laplace transform to solve initial value problems?

Using the Laplace transform can make solving initial value problems easier and more efficient. It allows you to solve certain types of differential equations that may be difficult to solve using other methods. Additionally, the Laplace transform can handle a wider range of initial conditions compared to other techniques.

5. Are there any limitations to using the Laplace transform for initial value problems?

While the Laplace transform is a powerful tool for solving differential equations, it may not be suitable for all types of initial value problems. Some equations may not have a Laplace transform or the transform may be difficult to obtain. Additionally, the Laplace transform may not be the most efficient method for solving certain types of initial value problems.

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