MATH201/content/Solving IVP's using Laplace transform (lec 17-18).md

#start of lec 17
But first, midterm review!
He spends most of the time giving us a summary of all the equations (separable, linear, homogenous, exact, ...) The review is available on the eclass page. 25 minutes spent in total covering what will be on the midterm!

So, why did we learn all this stuff about Laplace transforms? We will now see how its useful:
# Solving IVP's with Laplace
## Example:
#ex #LT #second_order_nonhomogenous #IVP
Solve the following IVP using Laplace transforms:
$$y''+y=t^2+2 \qquad y(0)=1 \qquad y'(0)=-1$$
LHS=RHS, so the Laplace transforms of each side must also be equal.
We take the Laplace transform of both sides:
$\mathcal{L}\{y''+y\}=\mathcal{L}\{t^2+2\}$
applying linearity:
$\mathcal{L}\{y''\}+\mathcal{L}\{y\}=\mathcal{L}\{t^2\}+\mathcal{L}\{2\}$
using properties:
$s^2Y(s)-sy(0)-y'(0)+Y(s)=\frac{2!}{s^3}+\frac{2}{s}$
$s^2Y(s)-s+1+Y(s)=\frac{2!}{s^3}+\frac{2}{s}$
Now we need to isolate $Y(s)$.
So we add $s-1$ to both sides and factor out $Y(s)$:
$Y(s)(s^2+1)=\frac{2(s^2+1)}{s^3}+s-1$
Divide both sides by $(s^2+1)$ :
$Y(s)=\frac{2}{s^3}+\frac{s}{s^2+1}-\frac{1}{s^2+1}$
Now we take the inverse LT of both sides to obtain $y(t)$
$$y(t)=t^2+\cos(t)-\sin(t)$$
Done. That was pretty fast!

second example:
#ex #LT #second_order_nonhomogenous 
Solve the following IVP using Laplace transforms:
$$y''-y=(t-2)e^{t-2} \qquad y(2)=0 \qquad y'(2)=0$$
We are provided $y(2)$ but usually we need it in the form $y(0)$ What do we do?
We make a substitution: $x=t-2$
$\frac{d^2y}{dx^2}-y=xe^x$ <- notice the y here is not the same as the y above, lousy notation. This graph is shifted 2 units to the left.
where $y(0)=0 \qquad \frac{dy}{dx}(0)=0$
Hit it with the LT!
$\frac{1}{s^2}$ is LT of $x$. Using the shifting property, $\frac{1}{(s-\alpha)^2}$ is the LT of $xe^{\alpha x}$ 
$s^2Y(s)-Y(s) =\frac{1}{(s-1)^2}$
Isolate $Y(s)$ :
$(s^2-1)Y(s)=\frac{1}{(s-1)^2}$
$Y(s)=\frac{1}{(s-1)^2(s^2-1)}$
$Y(s)=\frac{1}{(s-1)^2(s-1)(s+1)}$
Partial fraction time:
$Y(s)=\frac{1}{(s-1)^3(s+1)}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)^3}+\frac{D}{s+1}$
$\frac{{A(s-1)^2(s+1)+B(s-1)(s+1)+C(s+1)+D(s-1)^3}}{(s-1)^3(s+1)}$
$\begin{matrix}A+D=0  \\A-2A+B-3D=0 \\ A-2A+B-B+C+3D=0 \\ A-B+C-D=1\end{matrix}$
$A=\frac{1}{8}$
$B=-\frac{1}{4}$
$C=\frac{1}{2}$
$D=-\frac{1}{8}$
plug into expression then take inv LT to obtain y(t):
final solution: $y(x)=\frac{1}{8}e^x-\frac{1}{4}xe^x+\frac{1}{4}x^2e^x-\frac{1}{8}e^-x$
where x=t-2
all done!
#end of lec 17 #start of lec 18

#ex 
$$y''+ty'-2y=2 \qquad y(0)=y'(0)=0$$
hit it with the LT!
$\mathcal{L}\{y''\}+\mathcal{L}\{ty'\}-2\mathcal{L}\{y\}=\frac{2}{s}$

$s^2Y(s)-\frac{d}{ds}\mathcal{L}\{y'\}-2Y(s)=\frac{2}{s}$
$s^2Y-\frac{d}{ds}(sY(s))-2Y=\frac{2}{s}$
apply product rule:
$s^2Y-Y-s\frac{dY}{ds}=\frac{2}{s}$
^ Boooo! another differential equation! :(
$\frac{dY}{ds}$ lies in the s "phase space"
$-s\frac{dY}{ds}+s\left( s-\frac{3}{5} \right)=\frac{2}{s}$
This is a linear equation!
divide by -s to get it in standard form
$\frac{dY}{ds}-\left( s-\frac{3}{5} \right)y=-\frac{2}{s^2}$
compute integrating factor:
$\mu(s)=e^{-\int (s-3/s) \, ds}=e^{-s^2/2}e^{\ln{s^3}}=s^3e^{-s^2/2}$
$\frac{d}{ds}(s^3e^{-s^2/2}Y)=-2se^{-s^2/2}$

$s^3e^{-s^2/2}Y=-2\int se^{-s^2/2} \, ds$
use u sub.
$u=\frac{s^2}{2}$
$-2\int e^{-u}  \, du$
$=2e^{-s^2/2}+C$
$Y(s)=2s^3+C \frac{e^{s^2/2}}{s^3}$
is this even a legitimate thing to take an inverse of?
the lim of the expression approaches inf as s approaches inf
So what do we do? Well we have that C term. We have to set $C=0$
then:
$Y(s)=\frac{2}{s^3}$
$$y(t)=t^2$$
we just solved a new equation that hasn't fit into our previous equation types using LT. How cool is that!