Differential equations using Laplace Transform

[khyvonen01]

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.

 

[Mark44]

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)].

 

[khyvonen01]

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.