Substituting in \(k_{1}\) and \(k_{2}\), we have \[x_{n+1}=x_{n}+a \Delta t f\left(t_{n}, x_{n}\right)+b \Delta t f\left(t_{n}+\alpha \Delta t, x_{n}+\beta \Delta t f\left(t_{n}, x_{n}\right)\right) ....Substituting in \(k_{1}\) and \(k_{2}\), we have \[x_{n+1}=x_{n}+a \Delta t f\left(t_{n}, x_{n}\right)+b \Delta t f\left(t_{n}+\alpha \Delta t, x_{n}+\beta \Delta t f\left(t_{n}, x_{n}\right)\right) . \nonumber \] We Taylor series expand using \[\begin{aligned} f\left(t_{n}+\alpha \Delta t, x_{n}+\beta \Delta t f\left(t_{n}, x_{n}\right)\right) & \\ =f\left(t_{n}, x_{n}\right)+\alpha \Delta t f_{t}\left(t_{n}, x_{n}\right)+\beta \Delta t f\left(t_{n}, x_{n}\right) f_{x}\left(t_{n}, x_{n}\right)…
The differential equation (3.1) gives us the slope \(f(x_0, y_0)\) of the tangent line to the solution curve \(y = y(x)\) at the point \((x_0, y_0)\). With a small step size \(∆x = x_1 − x_0\), the in...The differential equation (3.1) gives us the slope \(f(x_0, y_0)\) of the tangent line to the solution curve \(y = y(x)\) at the point \((x_0, y_0)\). With a small step size \(∆x = x_1 − x_0\), the initial condition \((x_0, y_0)\) can be marched forward to \((x_1, y_1)\) along the tangent line using Euler’s method (see Fig. \(\PageIndex{1}\)) \[y_1=y_0+\Delta xf(x_0, y_0).\nonumber\]
