Need help with expressing 2nd order ODE as system of 1st order ODE’s

[eternal sand]

I need to rewrite the equation in the picture as a system of 1st order ODEs; z’ = F(t, z). I get how you do it with the typical ay” + by’ + cy = 0, but what’s throwing me of is the division by y, how should I think? I need to be able to use it later in matlab as a function. Thanks in advance for any help❤️

[lean barn]

Maybe if you multiply by y?

[eternal sand]

What would I then do with the y*y” ?

[lean barn]

If you substitute z = z(x) = y'

Then we get y'' = z * z'

Then you'd get:

$\ z \cdot z' + \frac{1}{y} \cdot 2z^2 - \frac{1}{2}y = 0\$

$\ \Rightarrow z \cdot z' = - \frac{1}{y} \cdot 2z^2 + \frac{1}{2}y \$

$\ \Rightarrow z' = - \frac{1}{y} \cdot 2z + \frac{1}{2z}y \$

$\ \Rightarrow z' = \frac{y}{2}\cdot z^{-1} - \frac{2}{y} \cdot z = F(y,z) \$

[sage dune]

adonhs

[lean barn]

Basically rewriting the equation as:

$\ z' = \frac{y}{2}\cdot z^{-1} - \frac{2}{y} \cdot z \$

$\ = \frac{-2}{y} \cdot z + \frac{y}{2}\cdot z^{-1} = f(y) \cdot z + g(y) \cdot z^{-1}\$

n = -1 so this is a Bernoulli Differentialequation

[sage dune]

adonhs

[lean barn]

Whenever you face a Bernoulli Differentialequation as:

$\ y' = f(x) \cdot y + g(x) \cdot y^n$ where $n \in \mathbb{Z}$ \backslash {1}\$

Then we can use the substitution:

$\z=y^{1-n}$ with $y \neq 0$