# Second order homogeneous Differential EQ with complex coefficients.

1. Mar 28, 2007

### WHOAguitarninja

1. The problem statement, all variables and given/known data
This is the result of a problem from my Quantum class, but I figure it would be best to ask in here as my question is purely a question of how to solve a certain differential equation.

the equation is of the form 0=Y''-i*a*Y' + b*Y
where Y is a function of t
So the coefficient of Y' is imaginary, and the coefficient of Y is real.

The given initial condition is Y(0)=1

There is also another process (closely related in the physics problem) happening that is given by the equation

0=Z'' + i*a*Z' + b*Z

where a and b are the same a and b from the other equation. The initial condition on this is Z(0)=0

2. Relevant equations

The equations given above, and I beleive it is necessary to use the characteristic equation r^2+i*a*r+b

3. The attempt at a solution

I have used the characteristic equation and obtained that for Y, r=-ia/2 +/- i*sqrt(a^2+4b)/2. For convenience sake, let's call this r=iA +/- iB

I assumed this would form a combination of exponentials of the form

Y=C*e^(iA+iB)*t + D*e^(iA-iB)*t

I'm pretty sure this is fine up to this point, the problem arises in that I cannot seem to use the initial conditions to attain an exact solution. Plugging in t=0, we see that C+D =1. Using the Z equation is not helpful, as the exponents have a different sign on the "A" part. My next attempt was to break up the complex exponentials into their respective sine and cosine components. This did not help my situation at all, even when both Y and Z were considered.

My last idea that may work is the following.

I know that characteristic r's of the form a+/- ib give solutions like e^at*(A*cos(b) + B*sin(b)). I see that I can definitely get a exact solution if I break up MY equation as Y= e^(iAt)*(C'*cos(B) + D'*sin(B)).

I have a strong feeling that this is not allowed though, though I can't seem to find definite reasons either way. Any help is ENORMOUSLY appreciated.

2. Mar 29, 2007

### pki15

Well you should be able to solve this equation as usual. (Basically ignoring the imaginary bit)

$$y'' -iay' + by=0$$
$$r^2-iar+b=0$$

$$r = \frac{ia \pm \sqrt{(-ia)^2-4b}}{2} = \frac{ia \pm \sqrt{a^2-4b}}{2}$$

Let the two solutions be A,B.

Then

$$y = c_1 e^{At} + c_2 e^{Bt}$$

Your initial condition y(0)=1 is only enough to solve for one of c_1 or c_2. You get:

$$c_1 + c_2 = 1$$

Last edited: Mar 29, 2007
3. Mar 30, 2007

### HallsofIvy

Staff Emeritus
The set of solutions to a linear, homogeneous, second order differential equation form a two dimensional vector space. One "initial condition" is not sufficient to get a specific solution. That has nothing to do with complex numbers.