Reducing the Order of a Cauchy-Euler Equation

[Juanriq]

Homework Statement

Reduce the order of a Cauchy-Euler Equation

Homework Equations

x = e^t \mbox{ and } \ln x = t

The Attempt at a Solution

\displaystyle \frac{d y}{d x} = \displaystyle \frac{d y}{d t} \displaystyle \frac{d t}{d x} = \displaystyle \frac{d y}{d t} \cdot \displaystyle \frac{1}{x}<br />
and thus
<br /> \displaystyle \frac{d^2 y}{d x^2} = \displaystyle \frac{d y}{d t} \cdot \displaystyle \frac{-1}{x^2} + \displaystyle \frac{1}{x} \displaystyle \frac{d}{d x} \Bigl ( \displaystyle \frac{d y}{d t} \Bigl )<br />


Here is where I am getting stuck, specifically on \displaystyle \frac{d}{d x} \Bigl ( \displaystyle \frac{d y}{d t} \Bigl )<br /> this step. I know what I should get...
<br /> \displaystyle \frac{1}{x} \Bigl ( \displaystyle \frac{d^2 y}{d t^2} \cdot \displaystyle \frac{1}{x} \Bigl )<br />
But uhhh not getting it. Thanks in advance!

 

[vela]

You apply the chain rule again, except instead of applying it to y, you're applying it to dy/dt. It's a bit clearer if you look at it in terms of operators:

\frac{d}{dx} f = \frac{dt}{dx}\frac{d}{dt} f = \frac{1}{x}\frac{d}{dt} f

so

\frac{d}{dx} = \frac{1}{x} \frac{d}{dt}

 

[Juanriq]

Ahhh...very nice! Thanks a bunch!