Consider the ODE $$ y' = \alpha y, \qquad y(0) = 1 $$ whose exact solution is $$ y(t) = \exp(\alpha t) $$ Consider the simple scheme $$ y_0 = 1, \qquad y_1 = \exp(\alpha h), \qquad \frac{y_{i+1} - y_{i-1}}{2h} = \alpha y_i, \qquad i=1,2,\ldots $$ i.e., with $a = 2\alpha h$, iterative scheme is $$ y_{i+1} = a y_i + y_{i-1}, \qquad i=1,2,\ldots $$
from numpy import zeros,exp
from matplotlib.pyplot import plot,xlabel,ylabel,legend
def ode(alpha,h,N):
"""
h = step size
N = number of steps to take
"""
y = zeros(N)
t = zeros(N)
t[0], y[0] = 0, 1
t[1], y[1] = h, exp(alpha*h)
a = 2.0*alpha*h
for i in range(1,N-1):
y[i+1] = a*y[i] + y[i-1]
t[i+1] = t[i] + h
ye = exp(alpha*t)
plot(t,y,'o-',t,ye,'r-')
xlabel('t')
ylabel('y')
legend(('Numerical','Exact'),loc='upper left')
First try with $\alpha=1$.
alpha = 1
h = 0.1
N = 20
ode(alpha,h,N)
Now try with $\alpha=-1$.
alpha = -1
h = 0.1
N = 100
ode(alpha,h,N)