Thread: Deriving Euler's Equation: from: t2(d2y/dt2) + at(dy/dt) + by = 0 [CHAIN RULE HELP]

1. Deriving Euler's Equation: from: t2(d2y/dt2) + at(dy/dt) + by = 0 [CHAIN RULE HELP]

I'm given that an Euler equation (*) is of the form:

where a and b are constants. Then I'm told, let x=lnt, and calculate dy/dt and d2y/dt2 in terms of dy/dx and d2y/dx2.

I found:

But then I had problems with d2y/dt2.

This is what I got:

then subbing into (*) is good, except for d2y/dtdx term.

Any help would be greatly appreciated

2. Re: Deriving Euler's Equation: from: t2(d2y/dt2) + at(dy/dt) + by = 0 [CHAIN RULE H

$\dfrac{d^2y}{dt^2}=\dfrac{d}{dt}\left(\dfrac{1}{t } \dfrac{dy}{dx}\right)=$

$-\dfrac{1}{t^2}\dfrac{dy}{dx}+\dfrac{1}{t}\dfrac{d} {dt}\left(\dfrac{dy}{dx}\right)=$

$-\dfrac{1}{t^2}\dfrac{dy}{dx}+\dfrac{1}{t}\dfrac{d} {dx}\left(\dfrac{dy}{dt}\right)=$

$-\dfrac{1}{t^2}\dfrac{dy}{dx}+\dfrac{1}{t}\dfrac{d} {dx}\left(\dfrac{dy}{dx}\dfrac{dx}{dt}\right)=$

$-\dfrac{1}{t^2}\dfrac{dy}{dx}+\dfrac{1}{t^2}\dfrac{ d}{dx}\left(\dfrac{dy}{dx}\right)=$

$-\dfrac{1}{t^2}\dfrac{dy}{dx}+\dfrac{1}{t^2}\dfrac{ d^2y}{dx^2}=$

$\dfrac{1}{t^2}\left(\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}\right)$

3. Re: Deriving Euler's Equation: from: t2(d2y/dt2) + at(dy/dt) + by = 0 [CHAIN RULE H

Thank you very much for you help! I understand it now!