How to Solve the Initial Value Problem with Laplace Transforms?

[fufufu]

Homework Statement

Solve IVP
y'' + 2y' + 2y = u_pi(t) + u_2pi(t)
with IC
y(0) = 0 and y'(0) = 1.

Homework Equations

L{f''(t)} = s^2Y(s) - sy(0) - y'(0)
u_c(t) = u(t-c) -->Laplace--> e^-cs/s

The Attempt at a Solution

y'' + 2y' + 2y = u_pi(t) + u_2pi(t)
y'' + 2y' + 2y = u(t-pi) + u(t-2pi)
L{y'' + 2y' + 2y} = e^(-pis)/s + e^(-2pis)/s

s^2Y(s) - sy(0) - y'(0) + 2[sY(s)-y(0)] +2Y(s) = e^(-pis)/s + e^(-2pis)/s
s^2Y(s) - 1+ 2sY(s) +2Y(s) = e^(-pis)/s + e^(-2pis)/s
Y(s)(s^2 + 2s +2) = e^(-pis)/s + e^(-2pis)/s + 1
Y(s)((s+1)^2) = e^(-pis)/s + e^(-2pis)/s + 1
Y(s) = e^(-pis)/(s(s+1)^2) + e^(-2pis)/(s(s+1)^2) + 1/((s+1)^2)


before going any further though, I have hunch that the laplace of RHS is incorrect.. it looks correct when i compare it to the equation I am using (above) but I am wondering when or under what circumstances the denominator of e^-cs/s can change (ie, become squared or something like that)?
thanks for any help

 

[mighty2000]

!

it is important to have a strong understanding of mathematical concepts and their applications in order to accurately solve problems. In this case, the Laplace transform is a powerful tool for solving differential equations, and your approach is correct.

To answer your question about the Laplace transform of the right-hand side (RHS), the Laplace transform of u(t-a) is e^(-as)/s. This is because u(t-a) is a step function that is 0 for all values of t less than a, and 1 for all values of t greater than or equal to a. So, when you take the Laplace transform, you are essentially multiplying the function by e^(-as) and then integrating from 0 to infinity.

In this case, you have two step functions, u(t-pi) and u(t-2pi), and their Laplace transforms are e^(-pi*s)/s and e^(-2pi*s)/s, respectively. So, your RHS is correct.

Now, to solve the equation, you can use partial fractions to split the first two terms in your Y(s) into two separate fractions. Then, you can use the inverse Laplace transform to get the solution in the time domain. Finally, you can use the initial conditions to determine the values of the constants and get the final solution.

I hope this helps! Keep up the good work in your mathematical studies.