Solve Laplace Transform for y'' + w^2y = cos(t), y(0) = 1, y'(0) = 0

In summary, the conversation discusses solving a differential equation with given initial conditions using Laplace transforms. The conversation also includes a discussion on expanding a partial fraction and the best approach for solving it.
  • #1
GNRtau
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Homework Statement


y'' + w^2y = cos(t)
y(0) = 1
y'(0) = 0

w^2 not equal to 4


Homework Equations


Laplace integral, transform via table/memory...
Y(s) = F(s) or whatever you like to use

The Attempt at a Solution


s^2Y(s) - sy(0) - y'(0) + w^2Y(s) = s/(s^2 + 1). Right side is L(cos(t))
Group together, you get. (s^2 + w^2)Y(s) = L(cos(t)) + s For simplicity's sake, s^2 + w^2 equals a.
Divide by left side. L(cos(t))/a + s/a = Y(s). The second term is just cos(wt), so that part is done.
I'm a little stuck on how you expand the partial fraction here. I've never really done it before Laplace transforms, so I'm having some problems doing it in situations that are a little different like this.

s/(s^2+1)(a). So I would do As+B/(s^2 + 1) + Cs+D/(a)? After that, multiply by both sides, but from there I get a mess... a(As+B) + (s^2+1)(Cs+D). I plugged in 0, got Bw^2 + D = 0. Is there a more efficient way to do this(I'm sure the people here would know a way), or do I just to need to grind through the algebra? Sorry about the notation if unfamiliar.

Thanks for all the help in advance. I appreciate it.
 
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  • #2


Assuming what you wrote for the partial fraction expansion means this$$
\frac{As+B}{s^2+\omega^2}+\frac{Cs+D}{s^2+1}$$it looks correct. Yes, you just have to grind it out. Sometimes you can shorten the work by equating powers of ##s## or picking clever values of ##s## after multiplying it out. I don't know any nice shortcut.
 

FAQ: Solve Laplace Transform for y'' + w^2y = cos(t), y(0) = 1, y'(0) = 0

What is a Laplace Transform?

A Laplace Transform is a mathematical operation that converts a function of time into a function of complex frequency. It is often used in the study of differential equations to solve for unknown functions.

How do you solve a Laplace Transform?

To solve a Laplace Transform, you must first apply the transform to both sides of the equation. Then, use tables or formulas to transform the known function into a new function in terms of the complex frequency variable. Finally, use inverse Laplace Transform techniques to find the solution for the original function.

What is the initial condition in a differential equation?

The initial condition in a differential equation refers to the values of the dependent variable and its derivatives at a specific point or time. In the given equation, y(0) = 1 and y'(0) = 0 are the initial conditions.

What is the significance of the initial conditions in the Laplace Transform?

The initial conditions in a Laplace Transform are used to determine the constants of integration in the solution. These constants are necessary to fully solve for the unknown function.

How can the Laplace Transform be applied in real-world situations?

The Laplace Transform is commonly used in engineering, physics, and other scientific fields to model and analyze systems that involve differential equations. It can also be used to solve problems in electrical circuits, heat transfer, and signal processing, among others.

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