From 1861593dad97cc0365a8bc8bc635d725beaa3a86 Mon Sep 17 00:00:00 2001
From: Sasserisop <mail@sasserisop.com>
Date: Wed, 20 Dec 2023 20:51:20 -0700
Subject: [PATCH] fixed cauchy euler

---
 content/Cauchy-Euler equations (lec 10-11).md | 17 +++++++++++------
 1 file changed, 11 insertions(+), 6 deletions(-)

diff --git a/content/Cauchy-Euler equations (lec 10-11).md b/content/Cauchy-Euler equations (lec 10-11).md
index c8c7aa7..7abc81f 100644
--- a/content/Cauchy-Euler equations (lec 10-11).md	
+++ b/content/Cauchy-Euler equations (lec 10-11).md	
@@ -5,7 +5,7 @@ We know how to solve second order equations where a, b, c are constants. Even if
 #cauchy-euler equations are equations in the form:
 $$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)},\ x>0$$
 where $a,\ b,\ c$ are still constants and $\in \mathbb{R}$
-Note: x=0 is not interesting as the derivative terms disappear.
+Note: $x=0$ is not interesting as the derivative terms disappear.
 How to solve? There are two approaches:
 Textbook only use 2nd method, prof doesn't like this. You can find both methods in the profs notes. Btw, do you know Stewart? Multimillionaire, he's living in a mansion in Ontario.
 
@@ -14,12 +14,14 @@ $x=e^t\Rightarrow t=\ln x$ (x is always +)
 (do $x=-e^t$ if you need it to be negative.)
 find derivatives with respect to t now. y is a function of t which is a function of x.
 $\frac{dy}{dx}=\frac{dy}{dt}{\frac{dt}{dx}}=\frac{ dy }{ dt }{\frac{1}{x}}\Rightarrow \underset{ \text{Important} }{ x\frac{dy}{dx}=\frac{dy}{dt} }$
-compute 2nd derivative of y wrt to x:
-$\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2} \frac{dt}{dx}\cdot \frac{dt}{dx}+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$
+using $\frac{dy}{dx}=\frac{dy}{dt}{\frac{dt}{dx}}$ and the chain rule, compute 2nd derivative of y wrt to x:
+$\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2} \frac{dt}{dx}\cdot \frac{dt}{dx}+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{1}{x^2}\frac{dy}{dt}$
 $\underset{ \text{Important} }{ x^2{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$
 plugging those derivatives in we get: #remember
 $$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy(t)=f(e^t)$$
 ^ this is a constant coefficient second order non-homogenous equation now! We can solve it now using prior tools.
+> If you make the substitution $x=-e^t$ and go through the derivation, you get:
+> $a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy(t)=f(-e^t)$ <- Very nice that it's so similar, makes it easy to remember.
 
 ## Example:
 #ex #second_order #second_order_nonhomogenous #cauchy-euler
@@ -28,14 +30,17 @@ $$x^2{\frac{d^2y}{dx^2}}+3x{\frac{dy}{dx}}+y=x^{-1},\ x>0$$
 substitute: $x=e^t$
 transform using the technique we showed just earlier:
 $\frac{d^2y}{dt^2}+2{\frac{dy}{dt}}+y=e^{-t}$
-1) $r^2-2r+1=0$
+1) $r^2+2r+1=0$
 $r_{1,2}=-1$
 $y_{h}(t)=c_{1}e^{-t}+c_{2}te^{-t}$
 2) $y_{p}(t)=At^2e^{-t}$ <- using method of undetermined coefficients
+
+$\underbrace{ \cancel{ At^2e^{-t} }+\cancel{ A 2t(-e^{-t}) }+2Ae^{-t}\cancel{ -2Ate^{-t} } }_{ y_{p}'' }\quad+\underbrace{ \cancel{ 2At^2(-e^{-t}) }+\cancel{ 2A 2te^{-t} } }_{ 2y_{p}' }\quad+\underbrace{\cancel{  At^2e^{-t} } }_{ y_{p} }=e^{-t}$
+$2Ae^{-t}=e^{-t}$
 $A=\frac{1}{2}$
-general solution in terms of t:
+general solution in terms of $t$:
 $y(t)=c_{1}e^{-t}+c_{2}te^{-t}+\frac{1}{2}t^2e^{-t}$
-but we want solution in terms of x:
+but we want solution in terms of $x$:
 $y(x)=c_{1}e^{-\ln(x)}+c_{2}\ln(x)e^{-\ln(x)}+\frac{1}{2}\ln(x)^2e^{-\ln(x)}$ <- This is rather lousy notation, the y here isn't the same as the y above. Conceptually though, it's all oke doke.
 $$y(x)=c_{1}x^{-1}+c_{2}\ln(x)x^{-1}+\frac{1}{2}{\ln(x)^2}x^{-1}$$
 We are done.