MATH201/content/Cauchy-Euler equations (lec 10-11).md

We know how to solve second order equations where a, b, c are constants. Even if they're not constant some can be expressed as a linear equation. But not always will they be solvable. However, there is one class of second order equations with non constant coefficients that are always solvable.

# Cauchy-Euler equations
<i>If it has a name in it, its very important, if it has 2 names, its very very important!</i>
#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.
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.

introduce change of variables:
$x=e^t\Rightarrow t=\ln x$ (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.
$\frac{dy}{dx}=\frac{dy}{dt}{\frac{dt}{dx}}=\frac{ dy }{ dt }{\frac{1}{x}}\Rightarrow \underset{ \text{Important} }{ x\frac{dy}{dx}=\frac{dy}{dt} }$
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
Find the general solution for:
$$x^2{\frac{d^2y}{dx^2}}+3x{\frac{dy}{dx}}+y=x^{-1},\ x>0$$
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) $r^2+2r+1=0$
$r_{1,2}=-1$
$y_{h}(t)=c_{1}e^{-t}+c_{2}te^{-t}$
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 $t$:
$y(t)=c_{1}e^{-t}+c_{2}te^{-t}+\frac{1}{2}t^2e^{-t}$
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.
#end of lecture 10 #start of lecture 11

Last lecture we did Cauchy Euler equations:
$$ax^2{\frac{d^2y}{dx^2}+bx{\frac{ dy }{ dx }}+cy=f(x)} \qquad x>0$$
where $a,\ b,\ c$ are constants and $\in \mathbb{R}$
substitute $x=e^t$
$a{\frac{d^2y}{dt^2}}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$ <- lousy notation, the y here isn't quite the same as in the above definition.
substitute: $y=x^r$
after calculating derivatives, plugging in, and simplifying we obtain the polynomial equation:
$ar^2+(b-a)r+C=0$
Three cases:
<b>(i)</b> $r_1\ne r_{2}$ then: 
$y_{h}(t)=c_{1}e^{rt}+c_{2}e^{rt}$
$y_{h}(x)=c_{1}x^{r_{1}}+c_{2}x^{r_{2}}$ (lousy notation, because the two $y_{h}$ do not equal each other)
<b>(ii)</b> $r_{1}=r_{2}=r$ then: 
$y_{h}(t)=c_{1}e^{rt}+c_{2}te^{rt}$
$y_{h}(x)=c_{1}x^r+c_{2}x^r\ln(x)$ (derived by reduction of order.)
<b>(iii)</b> $r_{1,2}=\alpha\pm i\beta$ then: 
$y_{h}=e^\alpha(c_{1}\cos(\beta t)+c_{2}\sin(\beta t))$
$y_{h}(x)=x^\alpha(c_{1}\cos(\beta\ln x)+c_{2}\sin(\beta \ln x))$
Now compute your particular solution, $y_{p}$, and combine with $y_{h}$ to obtain your general solution.