Consider the IVP
(1.8.1).
Euler’s method with stepsize \(h\) consists of applying the recurrence relation
\begin{equation*}
y_{n+1}=y_n+hf(t_n,y_n), \quad n=0,1,2,\ldots
\end{equation*}
to calculate the successive approximations \(y_1, y_2, \ldots\) to exact solution values \(y(1_1),y(t_2),\ldots\) at the points \(t_1, t_2, \ldots\text{,}\) respectively.
Consider the equally spaced points
\(t_n:=t_0+nh, \quad n=0,1,2,\ldots\text{.}\) At the point
\((t_0,y_0)\text{,}\) the slope of the solution to
(1.8.1) is
\(f(x_0,y_0)\text{.}\) The corresponding tangent line at
\((t_0,y_0)\) is
\(y=y_0+(t-t_0)f(t_0,y_0)\text{.}\) Using this line to estimate
\(y(t_1)\) where
\(t_1=t_0+h\text{,}\) we have
\begin{equation*}
y(t_1) \approx y_1 := y_0 + h f(t_0,y_0).
\end{equation*}
Similarly, we can use the new line to approximate \(y(t_2)\) where \(t_2=t_0+2h=t_1+h\) to get
\begin{equation*}
y(t_2)\approx y_2:=y_1+hf(t_1,y_1).
\end{equation*}
And so on. Thus, we have the recurrence relations for \(n=0,1,2,\ldots\)
\begin{equation*}
t_{n+1}=t_n+h,
\end{equation*}
and
\begin{equation*}
y_{n+1}=y_n+hf(t_n,y_n),
\end{equation*}
which are seeded with the initial value problem’s given data \(y(t_0)=y_0\text{.}\)
