Convert $\ddot{y}(t) = \dot{y}(t)-y(t)^2$ with $y(0)=y_0, \dot{y}(0) = y_1$ into a first order IVP

Question

We're given the following 2. order IVP

$$\ddot{y}(t) = \dot{y}(t)-y(t)^2$$ with initial values $y(0)=y_0, \dot{y}(0) = y_1$. We're asked to convert it into a first Order IVP.

Answer 1

We set $z_0= y(t), z_1 = \dot{y}(t)$

$\begin{bmatrix} z_0(t) \\ z_1(t) \end{bmatrix} : = \begin{bmatrix} y(t) \\ \dot{y}(t) \end{bmatrix}$

Thus $\frac{dz}{dt}(t)= \begin{bmatrix} z_1(t) \\ z_1(t) - z_0(t)^2 \end{bmatrix}$ with initial values

$z(t_0=0)= \begin{bmatrix} y_0 \\ y_1 \end{bmatrix} $