Complex Numbers and Constants of Integration

squelch
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Homework Statement



Suppose that the characteristic equation to a second order, linear, homogeneous differential equation with constant coefficients yielded two complex roots:
\begin{array}{l}<br /> {\lambda _1} = a + bi\\<br /> {\lambda _2} = a - bi<br /> \end{array}
This would yield a general solution of:
y = {C_1}{e^{(a + bi)x}} + {C_2}{e^{(a - bi)x}}

I would like to prove that this is equal to the expression:
y = {C_1}{e^{ax}}\sin (bx) + {C_2}{e^{ax}}\cos (bx)

Homework Equations



Euler's identity:
{e^{ix}} = \cos (x) + i\sin (x)

The Attempt at a Solution



At the end of the proof, I am left with the expression:
y = i{C_1}{e^{ax}}\sin (bx) + {C_2}{e^{ax}}\cos (bx)

Can ##i## be "rolled up" into the constant of integration ##C_1## and the whole thing just defined as a single, undetermined constant?
 
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squelch said:

Homework Statement



Suppose that the characteristic equation to a second order, linear, homogeneous differential equation with constant coefficients yielded two complex roots:
\begin{array}{l}<br /> {\lambda _1} = a + bi\\<br /> {\lambda _2} = a - bi<br /> \end{array}
This would yield a general solution of:
y = {C_1}{e^{(a + bi)x}} + {C_2}{e^{(a - bi)x}}

I would like to prove that this is equal to the expression:
y = {C_1}{e^{ax}}\sin (bx) + {C_2}{e^{ax}}\cos (bx)

Homework Equations



Euler's identity:
{e^{ix}} = \cos (x) + i\sin (x)

The Attempt at a Solution



At the end of the proof, I am left with the expression:
y = i{C_1}{e^{ax}}\sin (bx) + {C_2}{e^{ax}}\cos (bx)
If ##C_1## and ##C_2## are the same constants as in ##y = {C_1}{e^{(a + bi)x}} + {C_2}{e^{(a - bi)x}}##, I don't think this is correct...
squelch said:
Can ##i## be "rolled up" into the constant of integration ##C_1## and the whole thing just defined as a single, undetermined constant?
... and if they are differrent constants, why not "roll up" the ##i## into ##C_1##?
 
Samy_A said:
If ##C_1## and ##C_2## are the same constants as in ##y = {C_1}{e^{(a + bi)x}} + {C_2}{e^{(a - bi)x}}##, I don't think this is correct...
I'm just trying to derive a textbook definition, in case the derivation is required on an exam.
At a point in the derivation, coming from the original equation, ##{e^{ax}}\cos (bx)[{C_1} + {C_2}] + i{e^{ax}}\sin (bx)[{C_1} - {C_2}]##.
I combined the constants into a new ##C_1## and ##C_2##, mostly to match the textbook equation. I suppose it'd be more clear (and proper) to call them ##C_1 '## and ##C_2 '##
... and if they are differrent constants, why not "roll up" the ##i## into ##C_1##?
I don't see a reason why I wouldn't be able to, but if it's a legal operation or not is my question. Call it a sanity check.
 
squelch said:
I'm just trying to derive a textbook definition, in case the derivation is required on an exam.
At a point in the derivation, coming from the original equation, ##{e^{ax}}\cos (bx)[{C_1} + {C_2}] + i{e^{ax}}\sin (bx)[{C_1} - {C_2}]##.
I combined the constants into a new ##C_1## and ##C_2##, mostly to match the textbook equation. I suppose it'd be more clear (and proper) to call them ##C_1 '## and ##C_2 '##

I don't see a reason why I wouldn't be able to, but if it's a legal operation or not is my question. Call it a sanity check.
Yes, you can do that. As the ##C##'s and ##C'##'s are arbitrary complex numbers anyway, there is absolutely no reason why you couldn't do it.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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