Previous page : LSO - Two equal real roots

Say we got two complex roots p+qi and p–qi. Using the solution for two different roots we get
Using the fact that
we get that
Now we want both C+D and (C–D)i to be real, We can choose C and D to be complex conjugates, say C=U+Vi this gives us
In that way the constants may be any value. Let us write C instead of 2U and D instead of -2V, this will finally give us the solution
An example
Let us check against a special case if this really works. Say the equation has pure complex roots then we must have started with an equation of the form
that will have the characteristic equation
If b is positive we get the solutions
This would give us
as a solution to the differential equation. This is something we already found when we solved the equation
where we found the solution to be
that, with a bit of algebra, can be found to be equivalent to the solution we found now. We can use the formula
and we let t be x and we can let the constant in front of t be ω. This would give us
so q= ω, C=Acosθ and D=–Asinθ.

Previous page : LSO - Two equal real roots
