2nd order Runge-Kutta method for linear ODE Could someone please help me with the next step of this 2nd order Runge-Kutta method?
I am solving the following initial value problem (IVP)
$$x' = - \frac12 x(t), \qquad x(0)=2$$
I wish to use the second order Runge-Kutta method
$$x(t+h)=x(t)+1/2(F_1+F_2),$$
where \begin{align*}
F_1&=hf(t,x) \\
F_2&=hf(t+h,x+F_1).
\end{align*}
Let us use a spacing of $h=1$.
My working goes like this:
$$F_1 = -\frac{x(0)}{2}=-1.$$
Then
\begin{align*}F_2&=1\times f(0+1,x(0)+F_1) \\
&=-1/2 \times x(1).\end{align*}
But I have no idea what $x(1)$ is.

Integrating the ODE
$$\dot x = -\frac{1}{2} x$$
we get
$$x (t) = x_0 \cdot \exp\left(-\frac{t}{2}\right)$$
Hence,
$$\begin{array}{rl} x (t+h) &= x_0 \cdot \displaystyle\exp\left(-\frac{t+h}{2}\right)\\\\ &= x_0 \cdot \displaystyle\exp\left(-\frac{t}{2}\right) \cdot \exp\left(-\frac{h}{2}\right)\\\\ &= \displaystyle\exp\left(-\frac{h}{2}\right) \cdot x (t)\\\\ &= \displaystyle\left(1 - \frac{h}{2} + \frac{h^2}{8} - \frac{h^3}{48} + \frac{h^4}{384} - \cdots\right) x(t)\end{array}$$
Using the 2nd order Runge-Kutta, we truncate
$$x (t+h) \approx \displaystyle\left(1 - \frac{h}{2} + \frac{h^2}{8}\right) x(t)$$
