The algorithm that we have described is known as
Euler’s method. Let us estimate a solution to
(1.4.1)–
(1.4.2) on the interval
\([0, 1]\) with step size
\(h = 0.1\text{.}\) Since
\(y(0) = 1\text{,}\) we can make our first approximation exact,
\begin{equation*}
Y_0 = y(0) = 1.
\end{equation*}
To generate the second approximation, we will construct a tangent line to the solution at \(y(0) = 1\text{.}\) If we use the slope of the solution curve at \(t_0 = 0\text{,}\)
\begin{equation*}
y'(0) = f(y(0), 0) = y(0) + 0 = 1 + 0 = 1,
\end{equation*}
and make use of the fact that
\begin{equation*}
\frac{y(h) - y(0)}{h} \approx y'(0, y(0)) \qquad \text{or} \qquad y(h) \approx y(0) + hy'(0, y(0)),
\end{equation*}
the estimate for our solution at \(t = 0.1\) is
\begin{align*}
Y_1 & = Y_0 + h f(t_0, Y_0)\\
& = Y_0 + h[Y_0 + t_0]\\
& = 1 + (0.1) [1 + 0]\\
& = 1.1000.
\end{align*}
Similarly, the approximation at \(t = 0.2\) will be
\begin{align*}
Y_2 & = Y_1 + h f(t_1, Y_1)\\
& = Y_1 + h[Y_1 + t_1]\\
& = 1.1000 + (0.1) [1.1000 + 0.1]\\
& = 1.2200.
\end{align*}
Our general algorithm is
\begin{equation*}
Y_{k+1} = Y_k + h f(t_k, Y_k) = Y_k + h[Y_k + t_k] = (1.1) Y_k + (0.01)k.
\end{equation*}