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

#start of lec 17
summary of all the equations (separable linear exact, ...) available on eclass
(homogenous (2 forms), Bernoulli equation, linear coefficients,)

second order equations:
put it in standard form
solve the homogenous eq, 
with constant coefficients
3 cases:  overdamped critically damped under damped
now find particular solution to get general solution $y=y_{h}+y_{p}$
either use #mouc  or #voparam 

we also have Cauchy-Euler equations (two methods, either guess y=x^r or use the general formula)

reduction of order is a technique when you're given a solution and you have to find the other

mechanical systems $my''+by'+ky=F_{o}\cos(\gamma t)$ (he recommends solving with mouc) that gives you the transient solution.
the particular solution is the steady part.
recall $\mu$=amplitude $\gamma_{r}$=resonance freq $\mu(\gamma_{_{r}})$=amplitude at resonance

Laplace transforms: definition and 4 important properties
That's all that will be covered on the midterm. 25 minutes spent 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:

#ex #LT #second_order_nonhomogenous 
$$y''+y=t^2+2 \qquad y(0)=1 \qquad y'(0)=-1$$
LHS=RHS, so the Laplace transforms of each side must also be equal:
$\mathcal{L}\{y''+y\}=\mathcal{L}\{t^2+2\}$
applying linearity:
$\mathcal{L}\{y''\}+\mathcal{L}\{y\}=\mathcal{L}\{t^2\}+\mathcal{L}\{2\}$
using properties:
$s^2Y(s)-s+1+Y(s)=\frac{2!}{s^3}+\frac{2}{s}$
factor out Y(s)
$Y(s)(s^2+1)=\frac{2(s^2+1)}{s^3}+s-1$
$Y(s)=\frac{2}{s^3}+\frac{s}{s^2+1}-\frac{1}{s^2+1}$
now we take the inverse LT to obtain y(s)
$$y(t)=t^2+\cos(t)-\sin(t)$$
Done!

second example:
#ex #LT #second_order_nonhomogenous 
$$y''-y=(t-2)e^{t-2} \qquad y(2)=0 \qquad y'(2)=0$$
we need y at 0 but its as 2.
so 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 again :) 
where $y(0)=0 \qquad \frac{dy}{dx}(0)=0$
hit it with the LT!
notice 1/s^2 is LT of x so we have to shift it
$s^2Y(s)-Y(s) =\frac{1}{(s-1)^2}$
$Y(s)=\frac{1}{(s-1)^3(s+1)}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)^3}+\frac{D}{s+1}$
$\frac{{A(s-1)^2(s+1)+B(s-1)(s+1)+C(s+1)+D(s-1)^3}}{(s-1)^3(s+1)}$
$\begin{matrix}A+D=0  \\-A-B-3D=0 \\ -A+c+3D=0 \\ A-B+C-D=1\end{matrix}$
$A=\frac{1}{8}$
$B=-\frac{1}{4}$
$C=\frac{1}{2}$
$D=-\frac{1}{8}$
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!