Question concerning 2nd order homogeneous linear diff eqs

[kostoglotov]

Homework Statement

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?

 

[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.

 

[SteamKing]

Homework Statement

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?

This link provides as clear an explanation as any:

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

 

[Ray Vickson]

Homework Statement

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?

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) \\<br /> \alpha C_1 + \beta i C_2 = y&#039;(0)
so
C_1 = y(0), \; C_2 = \frac{i}{\beta} \left( \alpha y(0) - y&#039;(0) \right)
If course, $iC_2 = (y'(0) - \alpha y(0))/\beta$ is real if $y(0), y'(0)$ are real.