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
