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revised Solving IVP's using LT

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content/Free vibrations (lec 11-12).md
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# Free vibrations
## Definition
This is a mass-spring system with friction: (imagine the $\beta$ was $b$ instead)
This is a mass-spring system with friction: (imagine the $\beta$ in the image below was $b$ instead)
![mbk.png](drawings/mbk.png)
This mass-spring system with friction can be modelled by the following differential equation:
$$my''+by'+ky=F_{ext}(t)$$
@ -8,7 +8,7 @@ Where:
$$\begin{matrix}F_{ext}(t) & \text{any external forces applied }(N) \\m & \text{mass } (kg) \\b & \text{damping constant } \left( \frac{Ns}{m} \right) \\k & \text{spring constant } \left( \frac{N}{m} \right)\end{matrix}$$
Free vibrations are when there are <u>no external forces</u> applied upon an oscillatory system. ie:
$$F_{ext}(t)=0$$
Free vibrations is really another word for #second_order_homogenous But it puts emphasis that we are modelling a friction-mass-spring system and that we can describe the solution—the position of the mass $y(t)$—in terms of frequency, phase shift, and amplitude: $\omega,\ \phi,\ A$. It's more of a case study on real world implications than anything else.
Free vibrations is really another word for #second_order_homogenous But it puts emphasis that we are modelling a friction-mass-spring system and that we can describe the solution—the position of the mass $y(t)$—in terms of angular frequency, phase shift, and amplitude: $\omega,\ \phi,\ A$. It's more of a case study on real world implications than anything else.
## System derivation
Let's start solving the DE!
$mr^2+br+k=0$ is the characteristic polynomial.
@ -55,8 +55,9 @@ so we can compute $A$ and $\phi$ from $c_1$ and $c_2$.
## Examples
#ex #second_order_homogenous #IVP
Solve the following IVP's when $b=0$, $b=6$, $b=10$, and $b=12$
(This is more of a study than an example problem. We have already solved equations like this so treat it as an exploration. Note the important takeaways at the end of each case)
$$y''+by'+25y=0 \qquad y(0)=1\quad y'(0)=0$$
(This is more of a study than an example problem. We have already solved equations like this so treat it as an exploration. Note the important takeaways at the end of each case)
1.) b=0 -> no friction in the system (undamped)
$b^2-4mk<0$ (under damped)
$\alpha=-\frac{b}{2m}=0$

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content/Solving IVP's using Laplace transform (lec 17-18).md
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#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!
*Lecture starts with a 25 minute long 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.
So, why did we learn all this stuff about Laplace transforms? We will now see how its useful:
# Solving IVP's with Laplace
@ -45,55 +45,57 @@ 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:
$-A+B-3D=0 \qquad \implies B=\frac{1}{8}-\frac{3}{8}=-\frac{1}{4}$
$-A+B-3D=0$
$-A+C+3D=0$
$A=\frac{1}{8}$
$B=-\frac{1}{4}$
$C=\frac{1}{2}$
$D=-\frac{1}{8}$
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
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
#ex #LT #IVP
Solve the following equation using LT:
$$y''+ty'-2y=2 \qquad y(0)=y'(0)=0$$
hit it with the LT!
$\mathcal{L}\{y''\}+\mathcal{L}\{ty'\}-2\mathcal{L}\{y\}=\frac{2}{s}$
$s^2Y(s)-\frac{d}{ds}\mathcal{L}\{y'\}-2Y(s)=\frac{2}{s}$
$s^2Y-\frac{d}{ds}(sY(s))-2Y=\frac{2}{s}$
$s^2Y(s)-s\cancelto{ 0 }{ y(0) }-\cancelto{ 0 }{ y'(0) }-\frac{d}{ds}\mathcal{L}\{y'\}-2Y(s)=\frac{2}{s}$
$s^2Y-\frac{d}{ds}(sY(s)-\cancelto{ 0 }{ y(0) })-2Y=\frac{2}{s}$
apply product rule:
$s^2Y-Y-s\frac{dY}{ds}=\frac{2}{s}$
$s^2Y-3Y-s\frac{dY}{ds}=\frac{2}{s}$
^ Boooo! another differential equation! :(
$\frac{dY}{ds}$ lies in the s "phase space"
$-s\frac{dY}{ds}+s\left( s-\frac{3}{5} \right)=\frac{2}{s}$
$-s\frac{dY}{ds}+s\left( s-\frac{3}{s} \right)Y=\frac{2}{s}$
This is a linear equation!
divide by -s to get it in standard form
$\frac{dY}{ds}-\left( s-\frac{3}{5} \right)y=-\frac{2}{s^2}$
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:
$\mu(s)=e^{-\int (s-3/s) \, ds}=e^{-s^2/2}e^{\ln{s^3}}=s^3e^{-s^2/2}$
$\frac{d}{ds}(s^3e^{-s^2/2}Y)=-2se^{-s^2/2}$
> ^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.
$u=\frac{s^2}{2}$
$-2\int e^{-u} \, du$
$=2e^{-s^2/2}+C$
$Y(s)=2s^3+C \frac{e^{s^2/2}}{s^3}$
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 $C=0$
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!
We just solved a new equation that hasn't fit into our previous equation types using LT. How cool is that!

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content/_index.md
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@ -17,7 +17,7 @@ Good luck on midterms! <3 -Oct 18 2023
[Free vibrations (lec 11-12)](free-vibrations-lec-11-12.html)
[Resonance & AM (lec 13-14)](resonance-am-lec-13-14.html)
[Laplace transform (lec 14-16)](laplace-transform-lec-14-16.html)
[Solving IVP's using Laplace transform (lec 17-18)](solving-ivps-using-laplace-transform-lec-17-18.html) (raw notes, not reviewed or revised yet.)
[Solving IVP's using Laplace transform (lec 17-18)](solving-ivps-using-laplace-transform-lec-17-18.html)
[(Heaviside) Unit step function (lec 18)](heaviside-unit-step-function-lec-18.html) (raw notes, not reviewed or revised yet.)
[Periodic functions (lec 19)](periodic-functions-lec-19.html) (raw notes, not reviewed or revised yet.)
[Convolution (lec 19-20)](convolution-lec-19-20.html) (raw notes, not reviewed or revised yet.)