Cauchy-Euler Equation

1. Mar 21, 2012

HAL10000

Let x=e^t. Then, assuming x>0, we have t=ln(x) and

$\frac{dy}{dx}$=$\frac{dy}{dt}$*$\frac{dt}{dx}$ = $\frac{1}{x}$*$\frac{dy}{dt}$,

$\frac{d^{2}y}{dx^{2}}$= $\frac{1}{x}$*($\frac{d^{2}y}{dx^{2}}$*$\frac{dt}{dx}$) - $\frac{1}{x^{2}}$*$\frac{dy}{dt}$ = $\frac{1}{x^{2}}$*($\frac{d^{2}y}{dt^{2}}$-$\frac{dy}{dt}$)

I don't understand why the derivative with respect to x of $\frac{dy}{dt}$ is $\frac{d^{2}y}{dx^{2}}$*$\frac{dt}{dx}$

2. Mar 21, 2012

HAL10000

I might have just explained it to myself... or not...

What you would really get by taking the derivative of $\frac{dy}{dt}$ is $\frac{d^{2}y}{dtdx}$ but the book wrote $\frac{d^{2}y}{dt^{2}}$*$\frac{dt}{dx}$ instead... which is really the same thing?

3. Mar 22, 2012

Illuminerdi

I'm a bit confused as to why your book would even derive the Cauchy-Euler equation using that method at all.

4. Mar 22, 2012

HAL10000

they use this method because dt/dx is 1/x... this times 1/x allows you to factor out a 1/x^2.... which cancels out the coefficient in the first term of the second order D.E. when you substitute it into the original equation.... and the same idea works for all terms, cancelling out the coefficients.

5. Mar 22, 2012

Illuminerdi

Right, but are you implying that the coefficients are inverse powers (1/x^2, 1/x)? Because there's a much more intuitive derivation if it's just x^2, x, etc..

6. Mar 22, 2012

HAL10000

Yeah, they are x^2 and x, that's why they cancel

7. Mar 23, 2012

Illuminerdi

I don't like using the math type on these sites, so I just wrote it out and scanned it. Here's the derivation I've used for the Cauchy-Euler equation. Again, not sure this is what you're looking for, but let's see if it helps.

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8. Mar 23, 2012

HAL10000

This one's a little different from my book but it uses the same ideas, I think you meant to write 1/x when substituting for the dx/dt.
Thanks, for some reason I have trouble understanding differentiation notation, that was my only problem. Now it's clear.

9. Mar 23, 2012

Illuminerdi

No, you're doing dt/dx, which isn't necessary in the derivation I used.

I think this is what you mean:
t=ln(x)
dt/dx=1/x

But this is what I mean:
x=e^t
dx/dt=e^t=x (by the definition above)

10. Mar 24, 2012

HAL10000

My bad! makes sense now :) thank you