Laplace transform initial value problem-

[annas425]

Laplace transform initial value problem--need help!

Looking at the solutions to these initial value problems, I am very confused as to how the highlighted steps are derived (both use heaviside step functions). I know the goal is to get the fractions in a familiar form so that one can look them up in a "Common Laplace Transform" Table to take the inverse and get the solution. But did they do this using partial fraction expansion? Or some other Algebra tricks that I am unaware of? Any clarity or help with either of these problems is MUCH appreciated!

First Problem:
Determine the solution to x' - 5x = f(t), where x(0)=0 and f(t) = 0 for t<3; f(t) = t for 3<=t<4; f(t) = 0 for t>=4.
Solution...

Second Problem:
Determine the solution to x'' + 25x = f(t), where x(0)=0 and x'(0)=0 and f(t) = t for 0<=t<1; f(t)= cos(t-1) for t>=1.
Solution...

Thank you so much, in advance! I don't need someone to entirely work out these problems; I just need some light shed on the highlighted steps!

 

[jvicens]

Hello,

I understand your confusion with the highlighted steps in these initial value problems. The key to solving these problems is to use the Laplace transform technique, which involves transforming the differential equation into an algebraic equation in the s-domain. Let me walk you through the steps for each problem.

First Problem:
To solve this initial value problem, we need to take the Laplace transform of both sides of the differential equation. This gives us:

sX(s) - x(0) - 5X(s) = F(s)

Where X(s) is the Laplace transform of x(t) and F(s) is the Laplace transform of f(t). We can substitute the given initial condition x(0)=0 into this equation to get:

sX(s) - 5X(s) = F(s)

Next, we need to find the Laplace transform of the given function f(t). Since f(t) is defined piecewise, we can break it into three parts and use the property of linearity of the Laplace transform. This gives us:

F(s) = 0 for t<3
F(s) = L{t} for 3<=t<4
F(s) = 0 for t>=4

Using the definition of the Laplace transform, we get:

F(s) = 0 for t<3
F(s) = 1/s^2 for 3<=t<4
F(s) = 0 for t>=4

Now, we can substitute this into our previous equation to get:

sX(s) - 5X(s) = 1/s^2

We can then use partial fraction expansion to simplify this equation and get:

X(s) = 1/(s^3 - 5s)

Now, we can use the Laplace transform table to find the inverse transform of X(s) and get the solution for x(t).

Second Problem:
Similarly, we can solve this initial value problem by taking the Laplace transform of both sides of the differential equation:

s^2X(s) - sx(0) - x'(0) + 25X(s) = F(s)

Since we are given the initial conditions x(0)=0 and x'(0)=0, we can simplify this equation to get:

s^2X(s) + 25X(s) = F(s)

Next, we need to find the Laplace transform of the given function