0
$\begingroup$

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.

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

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 $$8y^2u'u+6yu=0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.