Initial Value Problem. Laplace Transform

In summary, the question is asking to solve for y using Laplace transforms, with the given conditions of y(0) = 0 and y'(0) = 0. The solution is y = t/2 - 1/4 + 1/4*e^(-2t). The solution involves using the laplace transform table and partial fractions to find the inverse laplace transform.
  • #1
bob29
18
0

Homework Statement


The question is solve using laplace transforms: y" - 4y = 1-2t where y(0) =0 and y'(0) = 0

Ans is: y = t/2 - 1/4 + 1/4*e^(-2t)

Homework Equations


http://www.stanford.edu/~boyd/ee102/laplace-table.pdf"
partial fractions

The Attempt at a Solution


(s^2)Y - sy(0) - y'(0) - 4Y = L{1-2t}
(s^2)Y - 4Y = L{1-2t}
Y(s-4) = (1/s) - (2/s^2)
Y = [ (1/(s(s-4)) ] - [ 2/(s^2(s-4)) ]
L^-1 { [ s/(s^2(s-4)) ] - [ 2/(s^2(s-4)) ] = { (s - 2) / (s^2(s-4)) }

partial fractions to get
A/s + B/(s^2) + C/(s-4) = -1/(8s) + 1/(2s^2) + 1/8((s-4))

inverse laplace transform to get
y = -1/8 + t/2 + 1/8e^(4t)
 
Last edited by a moderator:
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  • #2
bob29 said:

Homework Statement


The question is solve using laplace transforms: y" - 4y = 1-2t where y(0) =0 and y'(0) = 0

Ans is: y = t/2 - 1/4 + 1/4*e^(-2t)

Homework Equations


http://www.stanford.edu/~boyd/ee102/laplace-table.pdf"
partial fractions


The Attempt at a Solution


(s^2)Y - sy(0) - y'(0) - 4Y = L{1-2t}
(s^2)Y - 4Y = L{1-2t}
Y(s-4) = (1/s) - (2/s^2)
The above should be Y(s2 - 4) = 1/s - 2/s2
bob29 said:
Y = [ (1/(s(s-4)) ] - [ 2/(s^2(s-4)) ]
L^-1 { [ s/(s^2(s-4)) ] - [ 2/(s^2(s-4)) ] = { (s - 2) / (s^2(s-4)) }

partial fractions to get
A/s + B/(s^2) + C/(s-4) = -1/(8s) + 1/(2s^2) + 1/8((s-4))

inverse laplace transform to get
y = -1/8 + t/2 + 1/8e^(4t)
 
Last edited by a moderator:

1. What is an initial value problem?

An initial value problem is a mathematical equation that involves an unknown function and its derivatives, along with an initial condition. The initial condition specifies the value of the unknown function at a particular point in the domain. The goal of solving an initial value problem is to find the function that satisfies the equation and the initial condition.

2. What is the Laplace transform?

The Laplace transform is a mathematical tool used to solve initial value problems in differential equations. It transforms a function of time into a function of a complex variable, making it easier to solve differential equations by converting them into algebraic equations. It is frequently used in engineering and physics applications to model systems and analyze their behavior.

3. How is the Laplace transform used to solve initial value problems?

The Laplace transform is used to solve initial value problems by first applying the transform to both sides of the equation, which converts the differential equation into an algebraic equation. The initial condition is then used to determine the constants in the resulting equation, yielding the solution to the initial value problem.

4. What types of initial value problems can be solved using the Laplace transform?

The Laplace transform can be used to solve a wide range of initial value problems, including linear and nonlinear problems, as well as problems with constant or variable coefficients. It is also commonly used to solve systems of differential equations and boundary value problems.

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

While the Laplace transform can be a powerful tool for solving initial value problems, there are some limitations to its use. It may not be applicable to problems with discontinuous functions or problems with non-constant coefficients. Additionally, the inverse Laplace transform may not always exist, making it difficult to obtain the original solution in some cases.

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