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The general solution of the differential equation $y''+\omega^2y=0$ can be written as: $$y=\alpha\cos{(\omega(t-c))}+\beta\sin{(\omega(t-c))}$$ Is it correct to say that:

  1. $\omega$ and $c$ are constants
  2. $\alpha$ and $\beta$ are arbitrary constants
  3. $t$ is a variable

QUESTION: What is the difference between constants, arbitrary constants and variables?

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  • $\begingroup$ Yes, all three are correct. $\endgroup$
    – almagest
    Apr 23, 2016 at 9:34
  • $\begingroup$ You've got solution of DE. That's function depend of variable. Now exactly you've got family of solutions, because of arbitary constants (if there are initiall conditions, you could find this arbitary constants). So it's easy to say: variable - smth that you could change, arbitary constants - smth, that you could find (if there are initall conditions) and constants - some numbers from DE. $\endgroup$
    – openspace
    Apr 23, 2016 at 9:42
  • $\begingroup$ I don't see why $c$ is not arbitrary in the same sense that $\alpha$ and $\beta$ are. Usually, the solution is given by, either $\alpha \cos (\omega (t-c))$ or $\alpha \cos \omega t + \beta \sin \omega t$. In any case, you have two arbitrary constants, because your equation is second order. $\endgroup$
    – Antoine
    Apr 23, 2016 at 10:09

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It is totally correct.

The solution of the equation, the function, depends on the value of t (called, the variable).

Also, the constants are properties of the system described by the differencial equation (the elasticity of a material or the mass of a pendulum, etc...), while the arbitrary constants give diferent solutions depending on the initial conditions of the system (for example, the initial phase of a mass connected to a spring).

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  • $\begingroup$ How about the comment of @user40085? Why shouldn't $c$ be also an arbitrary constant? Meaning, isn't $c$ part of the initial conditions? $\endgroup$
    – GambitSquared
    Apr 23, 2016 at 10:35
  • $\begingroup$ @ImreVégh In the case of the constant c, it may depend on the nature of the system and its oscilating properties. It is true that sometimes, the c constant is arbitrary because it is the initial phase of the oscilations, but in some scenarios like in control engineering it is not. So it's an interesting question because c may depend on the context. $\endgroup$
    – Jaime_mc2
    Apr 23, 2016 at 10:54

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