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

#start of lec 17
<i>Lecture starts with a 25 minute long midterm review.</i>
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.

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 $xe^{\alpha x}$ 
$s^2Y(s)-s\underbrace{ y(0) }_{ =0 }-\underbrace{ y'(0) }_{ =0 }-Y(s) =\frac{1}{(s-1)^2}$
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:
$Y(s)=\frac{1}{(s-1)^3(s+1)}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)^3}+\frac{D}{s+1}$
$\frac{{A(s-1)^2(s+1)+B(s-1)(s+1)+C(s+1)+D(s-1)^3}}{(s-1)^3(s+1)}$
$\begin{matrix}A+D=0  \\A-2A+B-3D=0 \\ A-2A+B-B+C+3D=0 \\ A-B+C-D=1\end{matrix}$
We can express this linear system as a matrix:
$\begin{bmatrix}1  & 0 & 0 & 1  & 0 \\-1 & 1 & 0 & -3 & 0 \\-1 & 0 & 1 & 3 & 0 \\ 1 & -1 & 1 & -1 & 1\end{bmatrix}$
Solving this matrix is a PITA. you could use row reduction or Cramer's rule.
However, I happen to know that adding all four gives an equation for $C$ alone. We are lucky, this is not always the case.
$0+0+2C+0=1 \qquad \implies C=\frac{1}{2}$
Add equations 1 and 3:
$C+4D=0 \qquad\qquad\ \ \ \implies D=\frac{-1}{8}$
equation 3:
$-A+C+3D=0 \qquad \implies A=\frac{1}{2}-\frac{3}{8}=\frac{1}{8}$
equation 2:
$-A+B-3D=0 \qquad \implies B=\frac{1}{8}-\frac{3}{8}=-\frac{1}{4}$

Plug into expression then take inv LT to obtain y(x):
$y(x)=\frac{1}{8}e^x-\frac{1}{4}xe^x+\frac{1}{4}x^2e^x-\frac{1}{8}e^{-x}$
substitute back $x=t-2$
$$y(t)=\frac{1}{8}e^{t-2}-\frac{1}{4}(t-2)e^{t-2}+\frac{1}{4}(t-2)^2e^{t-2}-\frac{1}{8}e^{-(t-2)}$$
all done!
#end of lec 17 #start of lec 18
#ex #LT #IVP
Solve the following equation using LT:
$$y''+ty'-2y=2 \qquad y(0)=y'(0)=0$$
hit it with the LT!
$\mathcal{L}\{y''\}+\mathcal{L}\{ty'\}-2\mathcal{L}\{y\}=\frac{2}{s}$
$s^2Y(s)-s\cancelto{ 0 }{ y(0) }-\cancelto{ 0 }{ y'(0) }-\frac{d}{ds}\mathcal{L}\{y'\}-2Y(s)=\frac{2}{s}$
$s^2Y-\frac{d}{ds}(sY(s)-\cancelto{ 0 }{ y(0) })-2Y=\frac{2}{s}$
apply product rule:
$s^2Y-3Y-s\frac{dY}{ds}=\frac{2}{s}$
^ Boooo! another differential equation! :(
$\frac{dY}{ds}$ lies in the s "phase space"
$-s\frac{dY}{ds}+s\left( s-\frac{3}{s} \right)Y=\frac{2}{s}$
This is a linear equation!
divide both sides by $-s$ to get it in standard form
$\frac{dY}{ds}-\left( s-\frac{3}{s} \right)Y=-\frac{2}{s^2}$
compute integrating factor:
$\mu(s)=e^{-\int (s-3/s) \, ds}=e^{-s^2/2}e^{\ln{s^3}}=s^3e^{-s^2/2}$
> ^interesting, why no abs() when integrating $\frac{1}{s}$ ?

recall, $(\mu Y)'=\mu(s)Q(s)$ according to linear equation solving technique.
However, you could also continue from here using the formula $Y(s)=\frac{1}{\mu(s)}\int \mu(s) Q(s) \, ds$
$\frac{d}{ds}(s^3e^{-s^2/2}Y)= \underbrace{ -2se^{-s^2/2} }_{  \mu(s)Q(s)  }$
integrate both sides:
$s^3e^{-s^2/2}Y=-2\int se^{-s^2/2} \, ds$
use u sub:
$u=\frac{s^2}{2} \qquad du=sds$
$=-2\int e^{-u}  \, du$
$s^3e^{-s^2/2}Y=2e^{-s^2/2}+C$
$Y(s)=\frac{2}{s^3}+C \frac{e^{s^2/2}}{s^3}$
Is this even a legitimate thing to take an inverse of?
The lim of the expression approaches inf as s approaches inf due to the exponential.
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!