# Is it possible to solve initial value problem with central difference scheme?

Lets say I have an ordinary differential equation:

$$\frac{dy(x)}{dx}=a \cdot e^x$$

I would like to solve this equation numerically. I could use for example Euler method or other explicit or implicit scheme to solve initial value problem (having one initial condition). To solve boundary value problem (having two conditions) I could use central difference scheme:

$$\frac{dy(x)}{dx}=\frac{y_{i+1}-y_{i-1}}{2 \Delta x}$$

and create system of equations, which after solution would provide $y(x_i)$ values and $a$ parameter value.

However if I want to solve initial value problem with central difference scheme there is one equation missing. I have too many unknown function values.

Is there some trick, or other equation I could add to state the problem in the right way, so that number of unknown values is equal to the number of equations?

A centered difference in the sense you have set up is possible, but isn't really what you want: you end up with parallel tracks for even and odd indices and end up being a factor of two inefficient. What you really should do is simply evaluate the RHS of the equation at the midpoint of a grid. To wit, define $x_n = x_0 + n \Delta x$. Then the difference equation for the IVP looks like
$$\frac{y_{n+1}-y_n}{\Delta x} = a \cdot e^{(x_n+x_{n+1})/2}$$
This is equivalent to a numerical quadrature using the midpoint rule. With your initial condition, you may solve for all $y_n$.