D.E. - Piecewise Laplace Transforms

[nut.the.dutch]

Greetings all/any:

Homework Statement

Take the Laplace transform of the following initial value problem and solve for Y(s) = L{y(t)}
y'' - 3y' -28y = 1 if 0<t<1
0 if 1<t

initial conditions: y'(0) = y(0) = 0

a. Y(s) = ?
b. now find the inverse transform to find y(t). Use step(t-c) for u_c(t)

*Note:
1/(s(s-7)(s+4)) = (-1/28)/s + (1/44)/(s+4) + (1/77)/(s-7)

Homework Equations

not entirely sure. I know random equations for the left hand side of the shifted equation, but I'm lost as to how to even start applying them. Nevertheless:

L{f(t-a) u(t-a)} = F(s)e^-as

L{g(t) - g(t-a)f(t-a)}

...etc? again, I have all of these equations, but I don't know how to use them so even if I found out some way to just plug numbers in, I still wouldn't understand what the concept behind this is.

The Attempt at a Solution

I can solve the left hand side:

y'' - 3y' -28y = f(t); y(0)=y'(0)=0

-3L{y'(t)} = sL{y(t)} - y(0) = -3(sY(s) - 0) = -3sY(s)

L{y''(t)} = s²L{y(t)} -sy(0) -s'(0) = s²Y(s) - 0s - 0 = s²Y(s)

= [s²Y(s)] -3[sY(s)] - 28[Y(s)] = F(s)


= Y(s) [s² -3s -28]= F(s)

...and then I'd have to find the laplase tranform of the right hand side as well, and divide through to isolate and solve for Y(s)

That said. I have no idea where to start on the right hand side. I've looked at the book, (Edwards & Penny 4th ed.), looked at notes, tried finding stuff online - basically what I get is a lot of examples of solutions, but I don't understand what they're doing to find their right hand side equations...

Thank you.

 

[Cyosis]

Your right hand side is 1. What is the Laplace transform of 1?

 

[Defennder]

You can interpret the RHS of 1 to be the unit step function u(t) if you're using a one-sided Laplace transform.