Consider the following ordinary differential equation: $$8y^2\frac{d^2y}{dx^2}+ 6y\frac{dy}{dx}=0$$

I want change variables so that $x = x(t)$ and my differential equation will be a function of $\frac{dy}{dt}$ and $\frac{d^2y}{dt^2}$

How do I convert $\frac{d^2y}{dx^2}$?

My guess: $$\frac{d^2y}{dt^2}=\frac{d}{dt}\left[\frac{dy}{dt}\right]=\frac{d}{dt}\left[\frac{dy}{dx}\frac{dx}{dt}\right]=\frac{d}{dt}\left[\frac{dx}{dt}\right]\frac{dy}{dx} + \frac{d}{dt}\left[\frac{dy}{dx}\right]\frac{dx}{dt} \\ = \frac{d^2x}{dt^2}\frac{dy}{dx} + ?$$

Is there a way to simplify $\frac{d}{dt}\left[\frac{dy}{dx}\right]$?

It looks like there is: Changing 2nd order homogeneous differential equation to the one with constant coefficients I'm having trouble obtaining the term they have in their formula though.

  • 1
    $\begingroup$ $dy/dx=(dy/dt)/(dx/dt)$ and use the quotient rule to differentiate it. $\endgroup$ – Ian Sep 18 '18 at 11:44

In this case you have not $x$ in equation, so let $y$ be the independent variable and with assumption $u=\dfrac{dy}{dx}$ then $$y''=\dfrac{du}{dx}=\dfrac{du}{dy}~\dfrac{dy}{dx}=u'u$$ so new DE is $$8y^2u'u+6yu=0$$


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