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

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(t)=f(e^t)

^ this is a constant coefficient second order non-homogenous 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 #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: (i) 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) (ii) 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.) (iii) 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.