Cauchy-Euler Equation: Derivatives with Respect to x and t

[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}

 

[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?

 

[Illuminerdi]

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?

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

 

[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.

 

[Illuminerdi]

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.

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..

 

[HAL10000]

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

 

[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.

 

[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.

 

[Illuminerdi]

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.

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)

 

[HAL10000]

My bad! makes sense now :) thank you