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Let $$ \begin{matrix} y'(t) = -150y(t)+49-150t, t\in[0,1]\\ y(0) = 1/3+0.1 \end{matrix} $$ I've know the solution: $y(t) = 0.1*e^{-150t}-t+1/3$.

I'm testing a couple different numerical methods to determine which is the best (ie. Euler's forward/backward, trapezoidal, Runge-Katta, Heun). Can anyone help me determine the stability for the Heun method?

Heun's method: $\phi(t,u,h) := \frac{1}{2}[f(t,u)+f(t+h,u+hf(t,u))]$

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To get a practical feeling for this, just solve your problem for various step sizes from $0$ to $1/3$, as $y(1/3)=0.1e^{-50}$ is a really small number, relative to the initial value and floating point precision.

def Heun(f,t0,y0,tf,h):
    def HeunStep(t,y,h): 
        k1 = h*f(t,y); 
        k2 = h*f(t+h,y+k1); 
        return y+0.5*(k1+k2);
    t, y = t0, y0
    while t+1.1*h<tf: t, y = t+h, HeunStep(t,y,h)
    return HeunStep(t,y,tf-t)

def f(t,y): return 49-150*(t+y)
t0,y0 = 0.0,1./3+0.1
for h in np.linspace(3.5,0.5,16)/150:
    print "h=%.4e 150*h=%.4e, y(1/3)=%.12g"%(h,150*h, Heun(f,t0,y0,1./3,h))

with the results

h=2.3333e-02 150*h=3.5000e+00, y(1/3)=3382845.45109
h=2.2000e-02 150*h=3.3000e+00, y(1/3)=1820496.14682
h=2.0667e-02 150*h=3.1000e+00, y(1/3)=558708.429462
h=1.9333e-02 150*h=2.9000e+00, y(1/3)=79763.1519923
h=1.8000e-02 150*h=2.7000e+00, y(1/3)=9204.09082254
h=1.6667e-02 150*h=2.5000e+00, y(1/3)=1648.41784102
h=1.5333e-02 150*h=2.3000e+00, y(1/3)=37.6118108918
h=1.4000e-02 150*h=2.1000e+00, y(1/3)=0.740437945724
h=1.2667e-02 150*h=1.9000e+00, y(1/3)=0.00432803787737
h=1.1333e-02 150*h=1.7000e+00, y(1/3)=1.0688786734e-05
h=1.0000e-02 150*h=1.5000e+00, y(1/3)=1.14794370277e-08
h=8.6667e-03 150*h=1.3000e+00, y(1/3)=5.57823318786e-12
h=7.3333e-03 150*h=1.1000e+00, y(1/3)=2.8015784137e-15
h=6.0000e-03 150*h=9.0000e-01, y(1/3)=2.34187669257e-17
h=4.6667e-03 150*h=7.0000e-01, y(1/3)=-1.38777878078e-17
h=3.3333e-03 150*h=5.0000e-01, y(1/3)=2.60208521397e-17

While the stability region would suggest $0<h<\frac2{150}$ are valid choices, the table above gives only sensible results for about $h<\frac{1.5}{150}$.

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