Say we want to solve numerically $y'(x) = f(x) \cdot y$, with $y_0 = y(x=0) = 0$ and applying RK4 method with step $dx = h$:
\begin{align} k_1 &= f(0) \cdot y(0) \cdot h = 0\\ k_2 &= f(0+h/2) \cdot (y_0 + k_1/2) \cdot h = f(h/2) \cdot 0 = 0\\ k_3 &= f(h/2) \cdot (y_0 + k_2/2) \cdot h = 0\\ k_4 &= f(c+h) \cdot (y_0 + k_3) \cdot h = 0 \end{align}
Hence: $y(h) = 0 + (k_1+2 k_2+2 k_3+k_3)/6 = 0...$
In the end: $y_i = y(0+i\cdot h) = 0$, no matter what $f(x)$ looks like!
Any suggestions on this?
y = 0 is the solution of the equation with $y(0) = 0$. You get exact solution for this.