Initial value problem using Laplace

Question

First time posting, looking for help with a previous exam paper with a question about Laplace transform for initial value problems. The questions is as follows:

The function y(t) satisfies the initial value problem 
y" + 2y' + 10y = r(t), y(0)=2, y'(0)=3
where
r(t) = 0 if t<0
r(t) = t if 0<=t<1 and
r(t) = 0 if t>=1. 

Demonstrate the laplace transform of y(t) is:

Y(s) = 2s+7/s^2+2s+10 - e^-s/s(s^2+2s+10) + 1/s^2(s^2+2s+10) - e^-s/s^2(s^2+2s+10)

Answer 1

I like to tackle these problems by first taking the Laplace transform of every component.

$ \mathscr{L}${y''}(s) = $ s^2Y(s)- sy(0) - y'(0) $ = $ s^2Y(s) - 2s - 3$

$ \mathscr{L}${y'}(s) = $ sY(s) - y(0)$ = $ sY(s) - 2$

$ \mathscr{L}${y}(s) = $ Y(s) $

Now, there are a couple methods to tackle the right hand side.

1. Use the definition of the Laplace transform, adding the individual integrals with the given boundaries.

2. Use the window function, $\Pi_{(a,b)}$, and the unit step function $u(t)$

I'm going to use the window and unit step functions here. Simply put, the window function returns 0 for t < a, 1 for a < t < b, and 0 for t > b. The unit step function is 0 for t < 0 and 1 for t > 0.

Now,

$ r(t) = t\Pi_{(0, 1)} = tu(t-0) - tu(t-1) = t - tu(t-1)$

Using the following equation:

$\mathscr{L}${g(t)u(t-a)}(s) = $e^{-as}\mathscr{L}${g(t+a)}(s)

$ => \mathscr{L}${r(t)}(s) = $\mathscr{L}${t} $ -e^{-s}\mathscr{L}${t+1}

$=> \mathscr{L}${r(t)}(s) = $\frac{1 - e^{-s}}{s^2} - \frac{e^{-s}}s$

From here, you can put everything together and solve traditionally.