The discussion focuses on solving the second-order differential equation x'' + 5x' + 6x = f(t) using the Laplace Transform. The function f(t) is defined as f(t) = 3u(t) - 3u(t - 6), where u(t) represents the unit step function. Key formulas provided include L[x''] = s²X(s) - sx(0) - x'(0) and L[x'] = sX(s) - x(0), which are essential for applying the Laplace Transform to the equation. The transformation simplifies the problem, allowing for a systematic approach to finding the solution.
PREREQUISITESStudents studying differential equations, engineers applying mathematical modeling, and anyone interested in mastering the Laplace Transform for solving dynamic systems.
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.
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)].