# How do I get d^2 y/dx^2 for a Cauchy-Euler, differential equation?

1. Feb 23, 2014

### s3a

1. The problem statement, all variables and given/known data
How do I get d^2 y/dx^2 for a Cauchy-Euler, differential equation?

Basically, how do I derive d^2 y/dx^2, as given in the following link (since I don't want to just memorize that equation)?:
http://www.sosmath.com/diffeq/second/euler/euler.html

2. Relevant equations
* Cauchy-Euler, differential Equation.
* x = e^t
* Differentiating.
* Chain rule

3. The attempt at a solution
I do get how to derive dy/dx.

When I try to derive d^2 y/dx^2, I get a very close result, but it's different.

Could someone please tell me where my mistake is, or if I'm doing something completely wrong, tell me what it is I am doing wrong?

Here is my work.:
x = e^t

dy/dx = 1/x dy/dt

d/dx (dy/dx) = d/dx(1/x dy/dt)

d^2 x/dx^2 = d/dx (1/x) dy/dt + (1/x) d/dx (dy/dt)

d^2 x/dx^2 = d/dx (1/x) dy/dt + (1/x) d/dt (dy/dx)

d^2 x/dx^2 = d/dx (1/x) dy/dt + (1/x) d/dt (1/x dy/dt)

d^2 x/dx^2 = d/dx (1/x) dy/dt + (1/x) d/dt (e^(-t) dy/dt)

d^2 x/dx^2 = (-1/x^2) dy/dt + [e^(-t)] d/dt (e^(-t) dy/dt)

d^2 x/dx^2 = [-e^(-2t)] dy/dt + [e^(-t)] (-[e^(-t)] [dy/dt] + [e^(-t)] [d^2 y/dt^2])

d^2 x/dx^2 = -e^(-2t) dy/dt – e^(-2t) dy/dt + e^(-2t) d^2 y/dt^2

d^2 x/dx^2 = -2e^(-2t) dy/dt + e^(-2t) d^2 y/dt^2

d^2 x/dx^2 = e^(-2t) [d^2 y/dt^2 - 2 dy/dt]

Any help in deriving d^2 y/dx^2 would be GREATLY appreciated!

2. Feb 24, 2014

### pasmith

I assume you mean $d^2 y/dx^2$ on the left, but your error is in assuming
$$\frac{d}{dx}\left(\frac{dy}{dt}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right)$$
This doesn't work here:
$$\frac{d}{dx}\left(\frac{dy}{dt}\right) = \frac1x \frac{d^2 y}{dt^2}$$
but
$$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(e^{-t} \frac{dy}{dt}\right) = e^{-t}\frac{d^2 y}{dt^2} - e^{-t} \frac{dy}{dt} = \frac1x \frac{d^2 y}{dt^2} - \frac1x \frac{dy}{dt}$$

I would start with the chain rule in the form
$$\frac{d}{dt} = x \frac{d}{dx}$$
and then
$$\frac{d^2 y}{dt^2} = x\frac{d}{dx}\left( x \frac{dy}{dx}\right) = x \frac{dy}{dx} + x^2 \frac{d^2 y}{dx^2} = \frac{dy}{dt} + x^2 \frac{d^2 y}{dx^2}$$
so that
$$\frac{d^2 y}{dx^2} = \frac1{x^2} \left(\frac{d^2 y}{dt^2} - \frac{dy}{dt}\right) \\ = e^{-2t} \left(\frac{d^2 y}{dt^2} - \frac{dy}{dt}\right)$$

3. Feb 25, 2014

### s3a

Thank you very much!