Can someone show the steps to do the integration in deriving the 2nd order Adams Bashforth method?

For the ODE

$$\frac{dy}{dx} = f(x,y)$$

where $f(x,y)$ is a linear polynomial given by:


and $h = x_i-x_{i-1}$.

Integrating from $x_i$ to $x_{i+1}$ gives:

$$y_{i+1} = y_i + \int_{x_i}^{x_{i+1}}f(x,y)dx$$

and finally ends with the 2nd order Adams Bashforth method:

$$y_{i+1} = y_i + \frac{h}{2}[3f(x_i,y_i)-f(x_{i-1},y_{i-1})]$$

The sources that I've checked always leave out the actual integration of $f(x,y)$. Can someone show me how it's done? Thanks!


I kept trying and I think I finally got it. Hope this helps someone else! $$\begin{split}\int_{x_i}^{x_{i+1}}f(x,y)dx &= \int_{x_i}^{x_{i+1}}\left(f(x_i,y_i) + \frac{f(x_i,y_i)-f(x_{i-1},y_{i-1})}{h}(x-x_i)\right)dx\\&=\int_{x_i}^{x_{i+1}}f(x_i,y_i)dx+\frac{f(x_i,y_i)-f(x_{i-1},y_{i-1})}{h}\int_{x_i}^{x_{i+1}}(x-x_i)dx\\&=f(x_i,y_i)(x_{i+1}-x_i)+\frac{f(x_i,y_i)-f(x_{i-1},y_{i-1})}{h}\left[\frac{(x_{i+1}-x_i)^2}{2}-\frac{(x_i-x_i)^2}{2}\right]\\&=f(x_i,y_i)h+\frac{f(x_i,y_i)-f(x_{i-1},y_{i-1})}{h}\frac{h^2}{2}\\&=\frac{h}{2}[3f(x_i,y_i)-f(x_{i-1},y_{i-1})]\end{split}$$

  • $\begingroup$ You've definitely got it! :) $\endgroup$ – Kyle Apr 14 '18 at 7:03

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