Differential equations using Laplace Transform

In summary, the conversation discusses using the Second Order Differential equation to solve a problem involving the function f(t). The equation is modified to make it easier to solve and the Laplace transform is applied to both sides using the formulas L[x''] = s2X(s) - sx(0) - x'(0) and L[x'] = sX(s) - x(0). The solution involves finding X(s) = L[x(t)].
  • #1
khyvonen01
2
0

Homework Statement


((d^2)x)/(d(t^2)) + 5(dx/dt) + 6x = f(t)
f(t)=3 for (0<=t<6)
f(t)=0 for t>=6


Homework Equations



This I am not sure of, this is my question. Would I use the Second Order Differential equation to do this problem:
L{((d^2)f)/(d(t^2))} = (s^2)F(s) - sf(0") - [df/dt]_t-0

The Attempt at a Solution


Have not attempted yet as I am just trying to find the correct formula to use for this.
 
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  • #2
khyvonen01 said:

Homework Statement


((d^2)x)/(d(t^2)) + 5(dx/dt) + 6x = f(t)
f(t)=3 for (0<=t<6)
f(t)=0 for t>=6


Homework Equations



This I am not sure of, this is my question. Would I use the Second Order Differential equation to do this problem:
L{((d^2)f)/(d(t^2))} = (s^2)F(s) - sf(0") - [df/dt]_t-0

The Attempt at a Solution


Have not attempted yet as I am just trying to find the correct formula to use for this.

Make it easier on yourself by writing your differential equation as x'' + 5x' + 6x = f(t).
f(t) is the sum of two unit step functions that have been scaled by a factor of 3.
I.e., f(t) = 3u(t) - 3u(t - 6)

So your equation can be written as x'' + 5x' + 6x = 3u(t) - 3u(t - 6).
Now take the Laplace transform of both sides.

Here are a couple of the formulas you will need.
L[x''] = s2X(s) - sx(0) - x'(0)
L[x'] = sX(s) - x(0)

In these formulas, X(s) = L[x(t)].
 
  • #3
Mark44 said:
Make it easier on yourself by writing your differential equation as x'' + 5x' + 6x = f(t).
f(t) is the sum of two unit step functions that have been scaled by a factor of 3.
I.e., f(t) = 3u(t) - 3u(t - 6)

So your equation can be written as x'' + 5x' + 6x = 3u(t) - 3u(t - 6).
Now take the Laplace transform of both sides.

Here are a couple of the formulas you will need.
L[x''] = s2X(s) - sx(0) - x'(0)
L[x'] = sX(s) - x(0)

In these formulas, X(s) = L[x(t)].


Thank you! Wow, not sure why I didn't spot that. It is one of those days lol.
 

1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is used to represent physical systems and their behavior over time.

2. What is the Laplace Transform?

The Laplace Transform is a mathematical operation that converts a function of time into a function of complex frequency. It is commonly used in solving differential equations, as it simplifies the process by converting the equations into algebraic equations.

3. How is the Laplace Transform used to solve differential equations?

The Laplace Transform is used by applying it to both sides of the differential equation. This converts the equation into an algebraic equation, which can then be solved using algebraic methods. After solving for the transformed function, the inverse Laplace Transform is used to find the solution to the original differential equation.

4. What are the advantages of using the Laplace Transform to solve differential equations?

Using the Laplace Transform to solve differential equations has several advantages. It can solve both initial value and boundary value problems, and it can handle equations with discontinuous or impulsive functions. Additionally, it simplifies the process of solving differential equations and can handle a wide range of functions.

5. Are there any limitations to using the Laplace Transform to solve differential equations?

While the Laplace Transform is a powerful tool for solving differential equations, it does have some limitations. It may not work for all types of differential equations, such as those with variable coefficients or singularities. Additionally, it can be challenging to apply the inverse Laplace Transform to find the solution to the original differential equation.

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