# Question concerning 2nd order homogeneous linear diff eqs

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1. Aug 4, 2015

### kostoglotov

1. The problem statement, all variables and given/known data

Regarding the case where the auxillary (characteristic) equation has complex roots, we solve the quadratic in the usual way using $i$ to get the general solution

$$y(x) = e^{\alpha x}\left(C_1 \cos{\beta x} + i C_2 \sin{\beta x}\right)$$

And the textbook shows

$$y(x) = e^{\alpha x}\left(C_1 \cos{\beta x} + C_2 \sin{\beta x}\right)$$

without the imaginary number $i$ in the equation.

At first I just assumed that the $i$ has been subsumed into the constant $C_2$, but then what is happening when we solve an initial value problem of this form, and find that $C_2$ is actually a real number? Where has the $i$ gone?

2. Aug 4, 2015

### fzero

If the initial conditions were such that $C_2$ was real in the first solution, then if we had used the 2nd solution, the initial condition would imply that $C_2'$ was imaginary. The important fact is that sin and cos are linearly independent solutions, so we can use any linear combination of them.

3. Aug 4, 2015

### SteamKing

Staff Emeritus
This link provides as clear an explanation as any:

http://tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx

4. Aug 5, 2015

### Ray Vickson

From
$$y(x) = e^{\alpha x}\left(C_1 \cos{\beta x} + i C_2 \sin{\beta x}\right)$$
we can write
$$y(x) = e^{\alpha x}\left(A_1 \cos{\beta x} + A_2 \sin{\beta x}\right).$$
where $A_1 = C_1$ and $A_2 = i C_2$. It is just notation, nothing more.

If $y(x) = e^{\alpha x}\left(C_1 \cos{\beta x} + i C_2 \sin{\beta x}\right)$ and $y(0), y'(0)$ are given (for example), then we need
$$C_1 = y(0) \\ \alpha C_1 + \beta i C_2 = y'(0)$$
so
$$C_1 = y(0), \; C_2 = \frac{i}{\beta} \left( \alpha y(0) - y'(0) \right)$$
If course, $iC_2 = (y'(0) - \alpha y(0))/\beta$ is real if $y(0), y'(0)$ are real.