### General solution

Let’s assume we are trying to solve the following differential equation:

.

As it’s a simple, linear ODE, it can be solved by using as ansatz

.

If we replace by it, we get

.

As is nonzero, lambda must be a root of the characteristic polynomial,

.

Using the well-known “quadratic formula”,

.

If we call the two values of and , assuming they are different, the general solution will be a linear combination:

.

### Specific example

If we take the following specific values for the parameters,

(assuming appropriate units)

(assuming appropriate units),

we get the following values for :

.

Now we have to use the given boundary conditions:

By replacing:

Substituting:

Then the full solution will be:

.

This solution will quickly diverge for values of not much smaller than one.

Even if the initial conditions were finely tuned for getting , any small numerical error would give a quickly diverging solution.