Complex Numbers and Constants of Integration

In summary, when deriving the general solution for a second order, linear, homogeneous differential equation with constant coefficients, if the characteristic equation yields two complex roots, it can be shown that the general solution is equal to the expression: y = {C_1}{e^{ax}}\sin (bx) + {C_2}{e^{ax}}\cos (bx), where the constants ##C_1## and ##C_2## are combined into one complex constant if needed. The legality of this operation is confirmed by the fact that the ##C##'s and ##C'##'s are arbitrary complex numbers.
  • #1
squelch
Gold Member
57
1

Homework Statement



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

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

Homework Equations



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

The Attempt at a Solution



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

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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  • #2
squelch said:

Homework Statement



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

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

Homework Equations



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

The Attempt at a Solution



At the end of the proof, I am left with the expression:
[tex]y = i{C_1}{e^{ax}}\sin (bx) + {C_2}{e^{ax}}\cos (bx)[/tex]
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##?
 
  • #3
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.
 
  • #4
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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Related to Complex Numbers and Constants of Integration

1. What are complex numbers?

Complex numbers are numbers that consist of both a real part and an imaginary part. They are typically represented in the form a + bi, where a is the real part and bi is the imaginary part (with i representing the square root of -1). They are commonly used in mathematics and engineering for solving equations that involve imaginary numbers.

2. How are complex numbers used in integration?

Complex numbers are used in integration as part of the process of finding the antiderivative of a given function. When dealing with complex functions, the use of complex numbers and constants of integration allows for more accurate and precise solutions.

3. What is a constant of integration?

A constant of integration is an arbitrary constant that is added to the antiderivative of a function when performing integration. It is necessary because when taking the derivative of a function, the constant term disappears. Therefore, when finding the antiderivative, the constant of integration is added to account for this missing term.

4. How do you handle constants of integration in complex integration?

In complex integration, the constant of integration is treated the same way as in regular integration. It is added to the antiderivative of the complex function to account for the missing term when taking the derivative. However, in some cases, the constant of integration may have an imaginary component, which should be taken into consideration when solving the integral.

5. Can constants of integration be negative?

Yes, constants of integration can be negative. They are arbitrary constants that can take on any value, including negative values. This is necessary because when taking the derivative, the constant term disappears, so the use of a negative constant allows for more accurate solutions when integrating complex functions.

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