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## 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

-3

*L*{y'(t)} = s

*L*{y(t)} - y(0) = -3(sY(s) - 0) = -3sY(s)

*L*{y''(t)} = s

^{2}

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

^{2}Y(s) - 0s - 0 = s

^{2}Y(s)

= [s

^{2}Y(s)] -3[sY(s)] - 28[Y(s)] = F(s)

= Y(s) [s

^{2}-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.