added lec 24

This commit is contained in:
Sasserisop 2023-11-04 16:36:36 -06:00
parent 42a20e4b22
commit bc3766c009
9 changed files with 1925 additions and 10 deletions

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@ -94,11 +94,11 @@ $A(x-y)=e^{y+2x}$
>$\lim_{ n \to 0 }\ln(n)=y+2x$
>$\lim_{ n \to 0 }\frac{d}{dx}\ln(n)=0=\frac{dy}{dx}+2$
>$\frac{dy}{dx}=-2\quad \Box$
>so from $\frac{dy}{dx}=-2\frac{{x-y-\frac{1}{2}}}{x-y+1}$ we get:
>$-2=-2\frac{{x-y-\frac{1}{2}}}{x-y+1}$
>$x-y+1=x-y-\frac{1}{2}$
>$1=-\frac{1}{2}$
>So what does this all mean? I honestly have no idea. I think it means we assumed that $e^{y+2x}=0$ is defined and because we arrived at a contradiction, our assumption was wrong. That didn't really get us to show if it was a valid solution or not like I imagined.
>plugging into the equation $(2x-2y-1)dx+(x-y+1)dy=0$ yields:
>$1=\frac{2(x-y+1)}{2x-2y-1}$
>$2x-2y-1=2(x-y+1)$
>$-1=2$
>So what does this all mean? I think it means that even if we imagine that $\frac{dy}{dx}$ exists, the equation is not satisfied and $y=x$ is definitely not a solution even when we try to cheat a little.
We can rearrange to our liking, but we have found the general solution to the DE:
#### $$x-y=Ae^{2x+y}$$

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@ -116,7 +116,7 @@ $\mathcal{L}^{-1}\{\alpha F(s)+\beta G(s)\}=\alpha \mathcal{L}^{-1}\{F\}+\beta \
This can be proven rather easily due to the linearity of the forward transform (wasn't done in class unfortunately).
## Examples
#ex #inv_LT I
#ex #inv_LT
Compute this inverse LT:
$$\mathcal{L}^{-1}\left\{ \frac{1}{s^5}+\frac{3}{(2s+5)^2}+\frac{1}{s^2+4s+8}+ \frac{{s+1}}{s^2+2s+10} \right\}$$
Notice that all these terms approach 0 as s approaches inf.

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@ -73,7 +73,7 @@ $\implies a_{n}=0$, $n=0,1,2,\dots$
#end of lec 22 #start of lec 23
Mid terms are almost done being marked!
## Examples
## Solving DE using series
Let's start using power series to start solving DE!
No magic formulas we need to memorize when solving equations using power series (Yay!)
#ex
@ -153,4 +153,47 @@ $z(x)=a_{0}\left( 1+\sum_{k=1}^\infty \frac{(1*4*\dots(3k-2))^2}{(3k)!}x^{3k} \r
$a_{1}\left( x+\sum_{k=1}^\infty \frac{(2\cdot 5\cdot \dots(3k-1))^2}{(3k+1)!} x^{3k+1}\right)$
there we go, $z$ is a linear combination of those two expressions
class done at 1:56 (a lil late but the journey is worth it)
#end of lec 23
#end of lec 23 #start of lec 24
*midterms have been marked and returned today.*
we consider:
$$y''+p(x)y'+q(x)y=0$$
this is in standard form, it's a second order linear equation
Definition:
if $p(x)$ and $q(x)$ are **analytic** functions in a vicinity of $x_{0}$ then $x_0$ is **ordinary**. Otherwise, $x_{0}$ is **singular**.
we expect that the solution y can be represented by a power series. This is true according to the following theorem:
Theorem: If $x_{0}$ is ordinary point then the differential equation above has two linearly independent solution of the form $\sum_{n=0} ^\infty a_{n}(x-x_{0})^n, \qquad\sum_{n=0}^\infty b_{n}(x-x_{0})^n$.
The radius of convergence for them is at least as large as the distance between $x_{0}$ and the closest singular point (which can be real or complex).
![[Drawing 2023-10-30 13.12.57.excalidraw.png]]
## Examples for calculating $\rho$
#ex
$$(x+1)y''-3xy'+2y=0 \quad x_{0}=1$$
put it in standard form:
$y''-\frac{3xy'}{x+1}+\frac{2y}{x+1}=0$
the only singular point for this equation is $x=-1$
so the minimum value of radius convergence is $\rho=2$ (distance between -1 and x_0)
we are guaranteed that the power series will converge *at least* in $(-1,3)$, possibly more. You can try solving for y as a power series.
#ex
$$y''-\tan xy'+y=0 \quad x_{0}=0$$
notice the coefficient beside y is 1, 1 is analytic and differentiable everywhere, obviously!
Same goes for any polynomial, it's obvious that any polynomial is infinitely differentiable but it's important to know.
What about tan x?
$\tan x=\frac{\sin x}{\cos x}$ is not defined on $x=\frac{\pi}{2}\pm n\pi, \qquad n=0,1,2,\dots$
the closest singular points are $\frac{\pi}{2}$ and $\frac{-\pi}{2}$ so our radius of convergence is the minimum distance of x_0 to these two points:
$\rho\geq\mid x_{0}-\frac{\pi}{2}\mid=\frac{\pi}{2}$
convergence could be larger, but we are guaranteed convergence on $x=x_{0}-\rho$ to $x_{0}+\rho$
#ex
$$(x^2+1)y''+xy'+y=0 \qquad x_{0}=1$$
put it in standard form:
$y''+\frac{x}{x^2+1}y'+\frac{y}{x^2+1}=0$
remember singular points can be complex the two singular points are:
$x^2=1=0 \qquad x=\pm i$
now we have to compute the two distances of these singular points to x=1
![[Drawing 2023-11-03 13.40.54.excalidraw.png]]
To calculate distance: $\alpha_{1}+\beta_{1}i, \qquad \alpha_{2}+\beta_{2}i$
$\sqrt{ (\alpha_{1}-\alpha_{2})^2+(\beta_{1}-\beta_{2})^2 }$
$\rho\geq \sqrt{ 1^2+1^2 }=\sqrt{ 2 }$
#end of lec 24

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@ -36,7 +36,7 @@ $\frac{d^2y}{dx^2}-y=xe^x$ <- notice the y here is not the same as the y above,
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 $xe^{\alpha x}$
$s^2Y(s)-Y(s) =\frac{1}{(s-1)^2}$
$s^2Y(s)-s\underbrace{ y(0) }_{ =0 }-\underbrace{ y'(0) }_{ =0 }-Y(s) =\frac{1}{(s-1)^2}$
Isolate $Y(s)$ :
$(s^2-1)Y(s)=\frac{1}{(s-1)^2}$
$Y(s)=\frac{1}{(s-1)^2(s^2-1)}$
@ -45,6 +45,16 @@ Partial fraction time:
$Y(s)=\frac{1}{(s-1)^3(s+1)}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)^3}+\frac{D}{s+1}$
$\frac{{A(s-1)^2(s+1)+B(s-1)(s+1)+C(s+1)+D(s-1)^3}}{(s-1)^3(s+1)}$
$\begin{matrix}A+D=0 \\A-2A+B-3D=0 \\ A-2A+B-B+C+3D=0 \\ A-B+C-D=1\end{matrix}$
$\begin{bmatrix}1 & 0 & 0 & 1 & 0 \\-1 & 1 & 0 & -3 & 0 \\-1 & 0 & 1 & 3 & 0 \\ 1 & -1 & 1 & -1 & 1\end{bmatrix}$
$-A+B-3D=0$
$-A+C+3D=0$
$A=\frac{1}{8}$
$B=-\frac{1}{4}$
$C=\frac{1}{2}$

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@ -22,7 +22,7 @@ Good luck on midterms! <3 -Oct 18 2023
[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.)
[Dirak δ-function (lec 21)](dirak-δ-function-lec-21.html) (raw notes, not reviewed or revised yet.)
[Power series (lec 22-23)](power-series-lec-22-23.html) (raw notes, not reviewed or revised yet.)
[Power series (lec 22-24)](power-series-lec-22-24.html) (raw notes, not reviewed or revised yet.)
</br>
[How to solve any DE, a flow chart](Solve-any-DE.png) (Last updated Oct 1st, needs revision. But it gives a nice overview.)
[Big LT table (.png)](drawings/bigLTtable.png)

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excalidraw-plugin: parsed
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