MATH201/content/Cauchy-Euler equations (lec 10).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
*If it has a name in it, its very important, if it has 2 names, its very very important!*
#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.
$\frac{dy}{dx}=\frac{dy}{dt}{\frac{dt}{dx}}=\frac{ dy }{ dt }{\frac{1}{x}}\Rightarrow \underset{ \text{Important} }{ x\frac{dy}{dx}=\frac{dy}{dt} }$
compute 2nd derivative of y wrt to x:
$\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2} \frac{dt}{dx}\cdot \frac{dt}{dx}+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$
$\underset{ \text{Important} }{ x^2{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$
plugging those derivatives in we get:
$$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy=f(e^t)$$
^ this is a constant coefficient equation now! We can solve it now using prior tools.

## 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
$A=\frac{1}{2}$
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