Laplace Transform initial value problem to solve Use the Laplace Transform to solve the initial value problem for $y''+3y'+2y=6e^{-t}$, $y(0)=1$, $y'(0)=2$.
 A: Hint:
Ok, lets go slow.


*

*$\displaystyle \mathcal{L} (y'') = s^2 y(s) -s y(0) - y'(0)$

*$\displaystyle \mathcal{L} (y') = sy(s) - y(0)$

*$\displaystyle \mathcal{L} (y) = y(s)$


Now, we substitute all of those into the DEQ and arrive at:
$$s^2 y(s) -s -2 +3(y(s)-1) + 2y(s) = \frac{6}{s+1}$$
$$y(s) = \frac{s^2+6 s+11}{(s+1)^2(s+2)} =  \frac{3}{s+2}-\frac{2}{s+1}+\frac{6}{(s+1)^2}$$
Now, you can do the inverse of each term on the RHS.
$$y(t) = e^{-2 t} (e^t (6 t-2)+3)$$
A: Note that the LT of a derivative of $y$ is $s \hat{y}(s) - y(0)$.  This comes from integration by parts:
$$\int_0^{\infty} dt \, y'(t) e^{-s t} = [y(t) e^{-s t}]_0^{\infty} + s \int_0^{\infty} dt \, y(t) e^{-s t}$$
Similarly, the LT of $y''$ is $s^2 \hat{y}(s) - s y(0) - y'(0)$.
The LT of the RHS is
$$6 \int_0^{\infty} dt \, e^{-(s+1) t} = \frac{6}{s+1}$$
Using the initial conditions specified, the differential equation is now an algebraic one:
$$(s^2+ 3 s + 2) \hat{y} - s-2 - 3 = \frac{6}{s+1}$$
Solving for $\hat{y}(s)$:
$$\hat{y}(s) = \frac{s+5}{(s+2)(s+1)} + \frac{6}{(s+2)(s+1)^2} = \frac{s^2+6 s+11}{(s+2)(s+1)^2}$$
There are a couple of ways to get $y(t)$ from $\hat{y}(s)$.  You may either use partial fractions or compute residues of $\hat{y}(s) e^{s t}$ at the poles $s=-2$ and $s=-1$.  I will illustrate the latter.
The pole $s=-2$ is simple so its residue is straightforward:
$$\text{Res}_{s=-2} \frac{s^2+6 s+11}{(s+2)(s+1)^2} e^{s t} = \frac{4-12+11}{(-2+1)^2} e^{-2 t} = 3 e^{-2 t} $$
The pole at $s=-1$, however, is a double pole and we use a derivative to compute the residue:
$$\begin{align}\text{Res}_{s=-1} \frac{s^2+6 s+11}{(s+2)(s+1)^2} e^{s t} &= \left [ \frac{d}{ds} \frac{(s^2+6 s+11) e^{s t}}{(s+2)}  \right ]_{s=-1}\\ &= \left [ \frac{e^{s t} \left(s^2+(s+2) (s (s+6)+11) t+4 s+1\right)}{(s+2)^2}\right ] _{s=-1}\\ &= (6 t-2 ) e^{-t}\end{align}$$
Therefore
$$y(t) = (6 t-2 ) e^{-t} + 3 e^{-2 t}$$
