MATH201/content/Second order homogenous linear equations (lec 5-7).md

most of these "models" in EE are based on these DE. You'll see how important DE are in chemical, electrical, mechanical, engphys, civil (very important for civil!), (mining? idk what's in mining :D -prof)
*DE are important* -prof
## Second order linear equations
Second order equations arise from very simple problems many engineers face, for instance a pendulum can be described by a second order equation.
#second_order
### $$a_{2}(t)y''+a_{1}(t)y'+a_{0}(t)y=f(t)$$
To motivate our interest: #fix
![[Drawing 2023-09-15 13.32.48.excalidraw]]
$ma=my''=-by'-ky$ 
Look how a second order equation describes the motion of a mass-spring system!
> Circuits that contains resistors, capacitors and inductors also behaves with this equation as well if you ignore the external magnetic fields around the circuit.

The equation $my''+by'+ky=0$ is a homogenous second order equation. (in this case, it's full name is homogenous second order linear equation with constant coefficients.) 
>Similar pattern with the electrical circuit analogy. This DE ignores external forces on the mass-spring system, it only considers the friction and the spring. If we push the mass then there would be an external force.

It's called second order because we have second derivative in the equation.


#ex #second_order_homogenous
### $$y''-4y'+3y=0$$
(This equation is homogenous as the RHS is equal to 0)
Imagine there's no y' (meaning no friction) you kind want the derivates to equal itself, an exponential!
We guess the solution is of the form $y(t)=e^{rt}$
$y(t)=e^{rt}$
$y'=re^{rt}$
$y''=r^2e^{rt}$
$r^2e^{rt}-4re^{rt}+3e^{rt}=0$ <- Our guess worked!
$r^2-4r+3=0$
$ar^2+br+c=0$
$r_{1,2}=\frac{{-b\pm \sqrt{ b^2-4ac }}}{2a}$
so $r_{1,2}=1,3$
so two possibilities of the equation:
$y_{1}(t)=e^t$ or $y_{2}(t)=e^{3t}$
so the general solution is the sum of the two possibilities (But why? See principle of super position below.)
$$y(t)=c_{1}e^{t}+c_{2}e^{3t}$$
and we're done.

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#end of lec 5 #start of lec 6

#ex #IVP #second_order_homogenous Same equation from last lecture, but now an IVP:

$$y(t)=c_{1}e^{t}+c_{2}e^{3t} \quad c_{1},c_{2}\in\mathbb{R} \quad\text{let } y(0)=0,\ y'(0)=4\quad \text{ What is } c_{1}, c_{2}?$$
> Lemma: if $y(t)=c_{1}e^{t}+c_{2}e^{3t}$ then $y'(t)=c_{1}y_{1}+c_{2}y_{2}$
> proof: let $y_{1}=e^{r_{1}t}\qquad y_{2}=e^{r_{2}t}$
> $y(t)=c_{1}e^t+c_{2}e^{3t}$
> $y'(t)=c_{1}r_{1}e^{r_{1}t}+r_{2}c_{2}e^{r_{2}t}$
> $y'(t)=c_{1}r_{1}e^{r_{1}t}+c_{2}r_{2}e^{r_{2}t}$
> since $c_{1}r_{1}$ is just a product of two arbitrary constants, we can replace them with a new constant.
> $y'(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$
> $y'(t)=c_{1}y_{1}+c_{2}y_{2} \quad \Box$

We are given $y(0)=0$
$c_{1}e^0+c_{2}e^{3*0}=0$
$c_{1}+c_{2}=0$
We are also given $y'(0)=4$
$c_{1}+3c_{2}=4$

Solving the linear system of equations gives: $c_{1}=-2,\ c_{2}=2$ which gives the solution:
$$y'(t)=-2e^t+2e^{3t}$$

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# Principle of super position
Remember from the example above where I said the "general solution is the sum of the two possibilities"? Let's explore and see why that is:
Recap: suppose we have an equation of the form $ay''+by'+cy=0$
$y(t)=e^{rt}$
then $ar^2+br+c=0$
case i) $r_{1},r_{2}=\frac{{-b\pm \sqrt{ b^2-4ac }}}{2a}, {r_{1}}\ne r_{2}$ (over damped case, there are 3 possible cases, see below.)
$y_{1}(t)=e^{r_{1}t}\qquad y_{2}(t)=e^{r_{2}t}$
$y(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$ is also a solution. But why? Principle of super position.
## Principle of super position:
If $y_{1}(t)$ solves $ay''+by'+cy=f_{1}(t)$
and $y_{2}(t)$ solves $ay''+by'+cy=f_{2}(t)$ on an interval $I$.
Then the following function that is a combination: $y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)$
solves $ay''+by'+cy=c_{1}f_{1}(t)+c_{2}f_{2}(t)$
Now we prove it:
Plugging in $y=c_{1}y_{1}(t)+c_{2}y_{2}(t)$ into $ay''+by'+cy=c_{1}f_{1}(t)+c_{2}f_{2}(t)$ gives us:
$a(c_1y_{1}''+c_{2}y_{2}'')+b(c_1y_{1}'+c_{2}y_{2}')+c(c_1y_{1}+c_{2}y_{2})=c_{1}f_{1}(t)+c_{2}f_{2}(t)$
moving terms gives us: $c_{1}(ay_{1}''+by_{1}'+cy_{1})+c_{2}(ay_{2}''+by_{2}'+cy_{2})=c_{1}f_{1}(t)+c_{2}f_{2}(t) \quad \Box$

Okay but none of that makes sense, how do we use the proof?
Let the following:
$y_{1}(t)=e^{r_{1}t}$ solves $ay''+by'+cy=0$
$y_2(t)=e^{r_{2}t}$ solves $ay''+by'+cy=0$
$f_{1}(t)=f_{2}(t)=0$
This following can be concluded:
$y(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$ must solve $ay''+by'+cy=0$ by principle of super position.
> Yay! Note this is only true when $f_{1}(t)=f_{2}(t)=0$ aka your RHS in the second order equation must be 0. aka the equation is homogenous.

case ii) $r_{1}=r_{2}=\frac{-b}{2a}$ if $b^2-4ac=0$ (critically damped)
if we assume $y_{1}=e^{r_{1}t}, y_{2}=e^{r_{1}t}$ like before then we get:
$y(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{1}t}=ce^{r_{1}t}$ <- this doesn't seem like it works! We need two integration constants for a second order equation.
$y_{1}(t)=e^{-bt/2a}, y_{2}(t)=te^{-bt/2a}$ for time being we take this as true, we can prove it later.
$y(t)=c_{1}e^{-\frac{bt}{2a}}+c_{2}te^{-\frac{bt}{2a}}$
we can check later at home, but also, how was the idea for this found? He will tell us later.

### linear algebra 101: linear independence makes unit vectors, which forms a basis.
definition: if $y_{1}, y_{2}$ are solutions to $a(t)y''+b(t)y+c(t)=0$ on some interval $I_{1}$
then they are called linearly independent if none of them is a constant multiple of the other.
Theorem: If $y_{1}(t), y_{2}(t)$ are linearly independent solutions to $ay''+by'+cy=0$ then any other solution can be written as $y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)$ (due to super position)

how do we know the two solutions are linearly independent? Test for linear independence:
$y_{1}, y_{2}$ are solutions to $a(t)y''+b(t)y+c(t)y=0$ on some interval $I_{1}$
then they are called linearly independent iff:
$W(y_{1},y_{2})(t)=\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}' & y_{2}'\end{pmatrix}\ne 0$
case i) $b^2-4ac>0 \Rightarrow r_{1}\ne r_{2}$
$y_{1}=e^{r_{1}t}, y_{2}=e^{r_{2}}t$
$W(y_{1},y_{2})=\det\begin{pmatrix}e^{r_{1}t} & e^{r_{2}t} \\r_{1}e^{r_{1}t} & r_{2}e^{r_{2}t}\end{pmatrix}$
$=e^{t(r_{1}+r_{2})}(r_{2}-r_{1})\ne 0$ (because this can only be 0 when $r_{1}= r_{2}$ but we know $r_{1}\ne r_{2}$)
case ii) $b^2-4ac=0\Rightarrow$
$r_{1}=r_{2}=-\frac{b}{2a}=r$
$y_{1}(t)=e^{rt}, y_{2}(t)=te^{rt}$
$W(y_{1},y_{2})=\det\begin{pmatrix}e^{rt} & te^{rt} \\re^{rt} & e^{rt}(1+rt)\end{pmatrix}$
$=e^{2rt}(1+rt)-rte^{2rt}$
$=e^{2rt}\ne 0$
> But what about case iii) ? it wasn't covered here. But now we proven that for case i and ii their two respective solutions are linearly independent and therefore we know we can safely apply the principle of super position on them and obtain their general solutions.

#ex #IVP #second_order_homogenous
$$y''-2y'+y=0 \qquad y(0)=1,\ y'(0)=0$$

$e^{rt}({ r^2-2r+1 })=0$
$(r-1)^2=0\Rightarrow r_{1}=r_{2}=1$ (this is case ii, critically damped)
$y(t)=c_{1}e^t+c_{2}te^t$
$y(0)=c_{1}=1$
$y(t)=e^t+c_{2}te^t$


$y'(0)=0$
$=e^0+c_{2}(0e^0+1\cdot e^0)$
$=1+c_{2}$
$\Rightarrow c_{2}=-1$
$$y(t)=e^t-te^t$$
#end of lecture 6 #start of lecture 7 (sept 20)

Last lecture we covered case i) and case ii).
however there's a third option:
$b^2-4ac<0$
Then we have complex roots:
$r_{1,2}=-\frac{b}{2a}\pm\frac{i\sqrt{ 4ac-b^2 }}{2a}=\alpha\pm i\beta$ <- Complex conjugates. And due to fundamental theorem of algebra, there are only 2 roots in this degree 2 polynomial equation.

$e^{r_{1}t}=e^{(\alpha+i\beta)t}=e^{\alpha t}+e^{i\beta t}$
side note: there are no numbers that are more than two components that are "useful", even quaternions
let $e^{i\beta t}=e^{i\theta}$
expand into power series:
>Recall from math 101: $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$

$=1+\frac{i\theta}{1!}+\frac{{(i\theta)^2}}{2!}+\frac{{(i\theta)^3}}{3!}+\dots$
$=1+\frac{i\theta}{1!}-\frac{\theta^2}{2!}-\frac{i\theta^3}{3!}+\frac{\theta^4}{4!}+\frac{i\theta^5}{5!}-\dots$
$=\left( 1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\dots \right)+i\left( \frac{\theta}{1!}-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\dots\right)$
>Recall from math 101: $\cos(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}\quad\text{and}\quad \sin(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}$

$e^{i\theta}=\cos(\theta)+i\sin(\theta) \quad \Box$ We have proven the Euler formula.

we guess the solution is of the form: $y(t)=e^{rt}=e^{(\alpha+i\beta)t}=e^{\alpha t}(\cos \beta t+i\sin \beta t)$
Lemma: If $u(t)+iv(t)$ solves $ay''+by'+cy=0$ then $u(t),\ v(t)$ are also solutions.
Proof:
$a(u+iv)''+b(u+iv)'+c(u+iv)=0$
$\underbrace{ { (au''+bu'+cu) } }_{ =0 }+i\underbrace{ (av''+bv'+cv) }_{ =0 }=0$ <- since the RHS is zero, both the real and imaginary parts must also equal zero.
$y_{1}(t)=e^{\alpha t}\cos(\beta t),\ y_{2}(t)=e^{\alpha t}\sin(\beta t)$
$\alpha=-\frac{b}{2a},\quad \beta={\frac{\sqrt{ 4ac-b^2 }}{2a}}$
then by principle of super position:
$y(t)=c_{1}y_{1}+c_{2}y_{2}$
now we have to test the two solutions are linearly independent
$W[y_{1},y_{2}]=\det\begin{pmatrix}y_{1}  & y_{2}  \\ y_{1}' & y_{2}'\end{pmatrix}\ne0$
$=\det \begin{pmatrix}e^{\alpha t}\cos(\beta t)& e^{\alpha t}\sin(\beta t)\\ -\beta\sin(\beta t)e^{\alpha t}+\cos(\beta t)\alpha e^{\alpha t} & \beta\cos(\beta t)e^{\alpha t}+\sin(\beta t)\alpha e^{\alpha t}\end{pmatrix}$
$=e^{2\alpha t}[\cos\beta t(\beta \cos(\beta t)+\alpha\sin(\beta t))-\sin \beta t(-\beta \sin(\beta t)+\alpha\cos(\beta t))]$
$=e^{2\alpha t}(\beta \cos^2(\beta t)+\beta \sin^2(\beta t))$
$=e^{2\alpha t}\beta\ne 0$
> Yay! We have shown the two solutions are indeed always linearly independent.

#ex #IVP #second_order_homogenous  
$$y''-2y'+5y=0 \quad y(0)=0 \quad y'(0)=2$$
$r^2-2r+5=0$<-characteristic equation
$r_{1,2}=\frac{{-b\pm \sqrt{ b^2-4ac }}}{2a}$
$r_{1,2}={1\pm \frac{\sqrt{ -4b }}{2}}=1\pm2i$
$y_{1}=e^t\cos(2t)$ $y_{2}=e^t\sin(2t)$
general solution: $y(t)=e^t(c_{1} \cos 2 t +c_{2}\sin 2 t)$
$y(0)=0=c_{1}$ <- this helps us calculate y' easier as we can cross out the $\cos(2t)$ term before taking the derivative.
$y'(t)=\frac{d}{dt}(e^tc_{2}\sin(2t))=e^tc_{2}2\cos(2t)+c_{2}\sin(2t)e^t$
$y'(0)=2=c_{2}2$
therefore: $c_{1}=0,\ c_{2}=1$ and the solution to the IVP is:
$$y(t)=e^t\sin(2t)$$
The solution has a nice graph, where if it was a circuit it would blow up,
or if it was a bridge it would oscillate and eventually collapse.

## In the next lecture:
something more difficult now:
$ay''+by'+cy=f(t)$ Again, a mass-spring system without any external force.
if f(t)=0 we can find the solution easily and use superposition to get the general solution
$ay''+by'+cy=0$
-> general solution is $y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)$

If we can find just one solution in $ay''+by'+cy=f(t)$
let it be $y_{p}(t)$
then the sum of the solutions
$y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)+y_{p}(t)$ must solve $ay''+by'+cy=f(t)$

Theorem: If $a(t),\ b(t),\ c(t)$ are continuous on $I$ , then IVP: $a(t)y''+b(t)y'+c(t)y=f(t)$ ; 
where $y(t_{o})=y_{o}$ , $y'(t_{o})=y_{1}$ has a unique solution.
we will do the proofs next class.
#end of lecture 7