About this point in Tenenbaum and Pollard's Ordinary Differential Equations, my copy starts to fall apart. The Linear Differential Equation of Order Greater than One is the fundamental groundwork for harmonic motion. Oscillators. Elementary electric circuits. I collect here the relationships I review most.
Definition 18.1. A linear differential equation of order n is an equation which can be written in the form
where f0(x), f1(x), . . . , fn(x), and Q(x) are each continuous functions of x defined on a common interval I, and y(n) denotes the nth derivative of y. y and its derivatives each has (ordinary) exponent one. An ordinary differential equation of order n cannot have terms such as y3 or [y(k)]2.
If fn(x) is everywhere zero, the equation is of order n-1.
If Q(x) is not everwhere zero, the equation is nonhomogeneous.
If Q(x) is everywhere zero, we have the
Homogeneous linear differential equation of order n:
This is the case of interest to me. At least until I reintroduce offset and other transfer functions which music tools take care to avoid or remove, but which result from many common forms of motion. Q(x), you shall have your revenge.
SOLUTIONS
If we further assume that each function fk(x) is a constant, then we obtain the following expression {pg 212}:We can cleverly suppose y = emx is a solution, (or let someone else do it, and read their book), and find this is no sooner said than done. The question then remains, for what values of m? Some straightforward differentiation, and at last we have an answer to the problem. Each value of m which satisfies the algebraic expression
will make y = emx a solution of (20.1). (20.14) is the characteristic equation of (20.1).
________________
... skipping Roots of Characteristic Equation Real.....
Alas.
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LESSON 20D. Some or All Roots of the Characteristic Equation Imaginary.
My solutions are not real. They are imaginary. All of them. Oh good! What could possibly go wrong? At this point in the film, he ignores the music, opens the closet, and is killed by the monster.
My plan is to eat the monster. Act casual.
Hey.
So... you eat people in the closet.
I like what you've done with the place. No, really, it's very 1950 East Berlin.
Is that what happened to my cords?
I see you left the lamé. No. I totally disagree. I think the gold would have been a nice accent.
Where do you-- UPSTAIRS? You mean that wasn't the CATS? No, I don't know any teenage gir- look, I'm not having this-- Who's holding the Cthulu talking mask? Huh? That's right, I am. {It's a DQ crown. Don't tell} You can talk when it's your turn, or be eaten by space AIDS.
That's what I thought. Look, I'm not trying to be a bitch here, but I had an imaginary monster in my closet when I was 5.
I was over at Steve's the other day and the All Roots Real in his closet called you a woos. He said you butcher like a baby.
Signal processing books assume knowledge of the equivalence between an exponential solution and one written in sines and cosines. For the second order equation, there are four forms.
Second Order Equation:
If the coefficients in the characteristic equation are real, the imaginary roots must occur in conjugate pairs. For the second order equation, call the two roots α + iβ and α - iβ. Then the general solution is
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; y_c \;&=\; {c_1}'e^{(\alpha + i\beta)x}\; + {c_2}'e^{(\alpha - i\beta)x}\\ &= \; {c_1}'e^{\alpha x}e^{i\beta x} + {c_2}'e^{\alpha x}e^{-i\beta x}\\ \textbf{form 1}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= \; e^{\alpha x}({c_1}'e^{i\beta x} + {c_2}'e^{-i\beta x}) \end{align*} "\small \begin{align*} (20.5)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; y_c \;&=\; {c_1}'e^{(\alpha + i\beta)x}\; + {c_2}'e^{(\alpha - i\beta)x}\\ &= \; {c_1}'e^{\alpha x}e^{i\beta x} + {c_2}'e^{\alpha x}e^{-i\beta x}\\ \textbf{form 1}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= \; e^{\alpha x}({c_1}'e^{i\beta x} + {c_2}'e^{-i\beta x}) \end{align*}")
Hey.
So... you eat people in the closet.
I like what you've done with the place. No, really, it's very 1950 East Berlin.
Is that what happened to my cords?
I see you left the lamé. No. I totally disagree. I think the gold would have been a nice accent.
Where do you-- UPSTAIRS? You mean that wasn't the CATS? No, I don't know any teenage gir- look, I'm not having this-- Who's holding the Cthulu talking mask? Huh? That's right, I am. {It's a DQ crown. Don't tell} You can talk when it's your turn, or be eaten by space AIDS.
That's what I thought. Look, I'm not trying to be a bitch here, but I had an imaginary monster in my closet when I was 5.
I was over at Steve's the other day and the All Roots Real in his closet called you a woos. He said you butcher like a baby.
Signal processing books assume knowledge of the equivalence between an exponential solution and one written in sines and cosines. For the second order equation, there are four forms.
Second Order Equation:
If the coefficients in the characteristic equation are real, the imaginary roots must occur in conjugate pairs. For the second order equation, call the two roots α + iβ and α - iβ. Then the general solution is
Using Euler's formula,
Let new constants
 \end{align*} "\small \begin{align*} c_1 = &{c_1}' + {c_2}'\\ c_2 = &i({c_1}' - {c_2}') \end{align*}")
Then we have
The imaginary part has been subsumed into c2. The substitution is trivial as far as it goes, but we are heading somewhere with this substitution. In what follows, I must be clear about the operation of complex numbers in the triangle inequality (Pythagorean theorem). I find this part continues to cause me some discomfort, which is part of the reason for this review.
Next we multiply by 1:
\!c_1\! \cos{\beta x} \;+\; \left( \frac {\sqrt{{c_1}^2 + {c_2}^2}} {\sqrt{{c_1}^2 + {c_2}^2}} \right )\! c_2 \sin{\beta x} \\ &= \sqrt{{c_1}^2 + {c_2}^2}\left( \frac {c_1} {\sqrt{{c_1}^2\ + {c_2}^2}} \cos{\beta x} \,+\, \frac {c_2} {\sqrt{{c_1}^2 + {c_2}^2}} \sin{\beta x} \right) \end{align*} "\begin{align*} c_1 \! \cos&{\beta x}\;\; +\;\; c_2 \sin{\beta x}\\ &=\,\left( \frac {\sqrt{{c_1}^2 + {c_2}^2}} {\sqrt{{c_1}^2 + {c_2}^2}} \right )\!c_1\! \cos{\beta x} \;+\; \left( \frac {\sqrt{{c_1}^2 + {c_2}^2}} {\sqrt{{c_1}^2 + {c_2}^2}} \right )\! c_2 \sin{\beta x} \\ &= \sqrt{{c_1}^2 + {c_2}^2}\left( \frac {c_1} {\sqrt{{c_1}^2\ + {c_2}^2}} \cos{\beta x} \,+\, \frac {c_2} {\sqrt{{c_1}^2 + {c_2}^2}} \sin{\beta x} \right) \end{align*}")
....right? Hehehe. Okay, but seriously. Let's build the right triangle with legs of length c1 and c2. It looks like this:
Which gives the identities:
The imaginary part has been subsumed into c2. The substitution is trivial as far as it goes, but we are heading somewhere with this substitution. In what follows, I must be clear about the operation of complex numbers in the triangle inequality (Pythagorean theorem). I find this part continues to cause me some discomfort, which is part of the reason for this review.
Next we multiply by 1:
....right? Hehehe. Okay, but seriously. Let's build the right triangle with legs of length c1 and c2. It looks like this:
Fig. 1 - a truck of c's
Which gives the identities:
Time to take stock. There is not yet any significance to the triangle drawn. The constants have been rearraned into two new quantities: an angle and a magnitude. Also, c2 lies on the imaginary axis. I'm grumpy. I demand that these two new quantities, and their arrangement in space, have physical meaning. I expect
1) the magnitude be a real-world magnitude: a nonnegative scalar quantity which is significant to the solution to a problem, which is equally the sum of component (forces).
2) An angle measured along an imaginary circle perpendicular to the real plane of motion, the sine of which remains measurable in physical space.
O, physic!
I have played into your hand again!
The equations now read:
\\ &= \sqrt{{c_1}^2 + {c_2}^2}\,\sin (\beta x \,+\, \delta) \\ \textbf{form 3} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= c \sin (\beta x \,+ \,\delta) \end{align*} "\begin{align*} c_1 \! \cos{\beta x}\;\; +\;\; c_2 \sin&{\beta x}\\ &= \sqrt{{c_1}^2 + {c_2}^2}\,(\sin \delta\, \cos{\beta x} \,+\, \cos\delta\, \sin{\beta x} )\\ &= \sqrt{{c_1}^2 + {c_2}^2}\,\sin (\beta x \,+\, \delta) \\ \textbf{form 3} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= c \sin (\beta x \,+ \,\delta) \end{align*}")
And the same procedure can be used to construct
\\ &=\; \sqrt{{c_1}^2 + {c_2}^2} \cos (\beta x \,-\, \delta)\\ \textbf{form 4}\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= c \cos (\beta x \,- \,\delta) \end{align*} "\begin{align*} c_1 \! \cos{\beta x}\; +\; c_2 \sin{\beta x}\; &=\; \sqrt{{c_1}^2 + {c_2}^2} (\cos \delta \cos\beta x\,+\, \sin \delta \sin \beta x)\\ &=\; \sqrt{{c_1}^2 + {c_2}^2} \cos (\beta x \,-\, \delta)\\ \textbf{form 4}\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= c \cos (\beta x \,- \,\delta) \end{align*}")
where c is amplitude and δ is phase angle (offset). The value of δ will be different in form 4: in Fig.1, we now measure from the upper right (corresponding) angle. The two functions 3 and 4 are everywhere identical.
O, physic!
I have played into your hand again!
The equations now read:
And the same procedure can be used to construct
where c is amplitude and δ is phase angle (offset). The value of δ will be different in form 4: in Fig.1, we now measure from the upper right (corresponding) angle. The two functions 3 and 4 are everywhere identical.
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