In [1]:

from __future__ import division

import numpy as np
import matplotlib.pyplot as pt

Consider \(y'=-100y+100t + 101\).

Exact solution: \(y(t)=1+t+ce^{-100t}\).

Exact solution derivative: \(y'(t)=1-100ce^{-100t}\).

In [2]:

def f(t, y):
    return -100*y+100*t + 101

In [28]:

t_end = 0.2

def plot_solution(t0, y0):
    c = (y0-1-t0)/np.exp(-100*t0)
    t_mesh = np.linspace(t0, t_end, 1000)
    solution = 1+t_mesh+c*np.exp(-100*t_mesh)
    
    pt.plot(t_mesh, solution, label="exact")
    pt.plot(t0, y0, "ko")

In [29]:

plot_solution(t0=0, y0=1)
plot_solution(t0=0, y0=1.2)
plot_solution(t0=0, y0=-0.5)
plot_solution(t0=0.05, y0=-0.5)

In [30]:

def integrate_ode(step_function, t0, y0, h):
    times = [t0]
    ys = [y0]

    while times[-1] <= t_end + 1e-14:
        t = times[-1]
        ys.append(step_function(t, ys[-1], h))
        times.append(t + h)

    pt.plot(times, ys, label=step_function.__name__[:4])

In [35]:

def forward_euler_step(tk, yk, h):
    return ...

def backward_euler_step(tk, yk, h):
    return ...

t0 = 0.05
y0 = -0.5
h = 0.1  # start this at 0.001, then grow

plot_solution(t0=t0, y0=y0)
integrate_ode(forward_euler_step, t0=t0, y0=y0, h=h)
integrate_ode(backward_euler_step, t0=t0, y0=y0, h=h)
pt.xlim([t0, t_end])
pt.ylim([-1, 2])
pt.legend()

Out[35]:

<matplotlib.legend.Legend at 0x53efd10>

(Edit this cell for solution.)