Converting nonlinear second order ODE to first order

Question

The second order ODE is: $$ y'' + ky' + (mg/c)\sin(y) = 0 $$ $k$, $m$, $g$ and $c$ and constants.

What is the first order ODE and how do I do it?

Answer 1

Let $x_1 = y$ and $x_2 = \dot{y}$. Then $\dot{x}_1 = x_2$ and $\dot{x}_2 = -kx_2-\frac{mg}{c}\sin(x_1)$. Now stack $x_1$ and $x_2$ into a column vector, i.e. $x = [x_1 \ x_2]^\top$. Then, $\dot{x} = f(x)$ where $f(x)$ is a vector valued function which follows from what I wrote above. This is the first order form.

Answer 2

If y is "small" then sin y ~y This called linearzation

This is special case .

Answer 3

Assume the independent variable is $x$ :

Let $u=\dfrac{dy}{dx}$ ,

Then $\dfrac{d^2y}{dx^2}=\dfrac{du}{dx}=\dfrac{du}{dy}\dfrac{dy}{dx}=u\dfrac{du}{dy}$

$\therefore u\dfrac{du}{dy}+ku+\dfrac{mg}{c}\sin(y)=0$

This belongs to an Abel equation of the second kind.

Let $t=-ky$ ,

Then $y=-\dfrac{t}{k}$

$\dfrac{du}{dy}=\dfrac{du}{dt}\dfrac{dt}{dy}=-k\dfrac{du}{dt}$

$\therefore-ku\dfrac{du}{dt}+ku+\dfrac{mg}{c}\sin(y)=0$

$u\dfrac{du}{dt}-u=\dfrac{mg}{ck}\sin(y)$

$u\dfrac{du}{dt}-u=-\dfrac{mg}{ck}\sin\left(\dfrac{t}{k}\right)$

This belongs to an Abel equation of the second kind in the canonical form.

Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf