In my high school project, I am solving differential equations using the forward Euler numerical method and this is because the equations were too hard to solve analytically.
To find the local error in the approximation. Generally, you would take the actual value and subtract the approximate value and take the modulus. However, I don't know what the actual value is. But I have read that the local error is proportional to the step size squared and in my paper, the step size is 1.
$|T_i|≤\frac{Mh^2}{2}$
However, I dont know what the constant M means or the steps to calculate the constant.
These are the differential equations that I am working with. Note: All symbols that are not functions of t are constants.
\begin{equation} \frac{dS}{dt} = \Lambda - \mu S(t) - \beta S(t) \frac{I(t)}{N(t)} - \zeta \lambda S(t) \end{equation}
\begin{equation} \frac{dE}{dt} = \beta S(t) \frac{I(t)}{N(t)} + \beta V(t) \frac{I(t)}{N(t)} - (\mu + \epsilon) E(t) \end{equation}
\begin{equation} \frac{dI}{dt} = \epsilon E(t) - (\gamma + \mu + \alpha) I(t) \end{equation}
\begin{equation} \frac{dR}{dt} = \gamma I(t) + \zeta V(t) - \mu R(t) \end{equation}
\begin{equation} \frac{dV}{dt} = \zeta \lambda S(t) - \beta V(t) \frac{I(t)}{N(t)} - \zeta V(t) - \mu V(t) \end{equation}
\begin{equation} \frac{dD}{dt} = \alpha I(t) \end{equation}
\begin{equation} \frac{dN}{dt} = \frac{dS}{dt} + \frac{dE}{dt} + \frac{dI}{dt} + \frac{dR}{dt} + \frac{dV}{dt} + \frac{dD}{dt} \end{equation}