Wzory:
y(x) = exp(-x)
y'   = -y

Warunek pocztkowy: y(0) = 1

-------- Euler (explicit) --------
y[0]   = 1
y[n+1] = y[n] - h*y[n]
x[n+1] = x[n] + h

-------- Euler (backward) --------
y[0]   = 1
y[n+1] = y[n] / (1+h)
x[n+1] = x[n] + h

-------- Trapezoidal rule --------
y[0]   = 1
y[n+1] = y[n] * (1 - 0.5*h) / (1 + 0.5*h)
x[n+1] = x[n] + h