Trouble understanding which constants are arbitrary and need to be eliminated while forming Differential Equation (DE) from its general solution (In contrast to constants which are fine to be left un-eliminated).

Given general solution** and we need to find it's DE$$y=A sin(\omega t + \phi)$$

We then differentiate it twice $$y'=-A\omega cos(\omega t + \phi)$$ $$y''=-A \omega^2sin(\omega t + \phi)$$

and get the final DE as $$y''=-y\omega^2 \qquad....(1)$$

Problems with Final DE [i.e (1)]

  • It contains $\omega$ which i considered to be an arbitary constant. This goes against my understanding (see below)
  • what is the difference in $\omega$, $A$ and $\phi$ as constants? Why $A$ and $\phi$ were eliminated but not $\omega$?
  • According to my understanding, Final DE should have been of order 3 but that turns out to be wrong. Why?

My Understanding:

If we have a general solution with say, 2 arbitrary constants then we need to differentiate the general solution 2 times and eliminate ALL of the arbitrary constants to get the DE free of arbitrary Constants.
Is this correct?


  1. ** It is taken from equation of SHM (y being the displacement).
  2. But forgetting our knowledge of physics, how can we mathematically justify $\omega$ in (1).
  • $\begingroup$ Reason for downvote? Happy to improve. $\endgroup$ Dec 14, 2022 at 20:34

3 Answers 3


To some extent, determining which symbols represent "parameters", "arbitrary constants" and "variables" is a combination of convention and relation to real-world systems. Technically, there's nothing inherent to $y = A \sin (\omega t + \phi)$ that says that $y$ is the dependent variable, $t$ the dependent, $\omega$ the parameter and $A$ and $\phi$ the arbitrary constants. We could just as easy declare that this is an implicit definition of $\omega$ as a function of $\phi$, with $y, t, A$ as parameters.

Even following the convention that this is describing $y$ as a function of $t$, we can write the most general possible DE for this system, which will look something like $y y''' = y' y''$.

From a physical perspective, the distinction between $A$, $\omega$ and $\phi$ has more to do with the system itself. For example, if we have simple harmonic motion coming from a mass on a spring, then $\omega$ will be the natural frequency of the system which depends on the physical properties of the mass and the spring, whereas $A$ and $\phi$ are related to the initial conditions - you can set the spring in motion in different ways to vary $A$ and $\phi$, but you can't change $\omega$ without changing the system itself, and from a measurement perspective you're probably more interested in understanding $\omega$ than the other variables.

There's also a small argument towards $y'' = -\omega^2 y$ being the "natural" DE that describes the system from the fact that it's a linear differential equation, where $y y''' = y' y''$ is not, and in most commonly encountered DE theory you'll see linear and mostly linear equations much more often than non-linear ones.


Omega is not an arbitrary constant: it is an important parameter in SHM (called the natural frequency). Every SHM is characterized by its natural frequency and an external force (which is not present in your equation).

As a general rule, the solution space of a second order linear DE with constant coefficients has dimension 2: i.e. the general solution has two arbitrary constants.

  • $\begingroup$ Hi, 1) what do you mean by parameter and how is it different from arbitrary Constant? 2)Also see second point of P.S (edited after ur ans.) $\endgroup$ Dec 14, 2022 at 19:30
  • $\begingroup$ By parameter I mean that this is the quantity that distinguishes one system from another. Omega is not arbitrary in the sense that if you change omega, the solution solves a different DE and hence a different system. A and phi are arbitrary in the sense that if you alter their values, you just get another solution to the same system. $\endgroup$
    – xxjman18
    Dec 14, 2022 at 19:45
  • $\begingroup$ @JustCurious . Do you know what a linear differential equation is? This should answer the question why $A$ is different from $\omega$. $\endgroup$
    – Kurt G.
    Dec 14, 2022 at 19:48
  • $\begingroup$ Regarding your second point under PS, I am not sure what you mean by mathematically justify. In terms of mathematical validity, you just derived that there ought to be a term omega squared appearing in the DE. In terms of understanding, I am not aware of any related mathematical concepts that can be helpful. Maybe someone else can answer this. $\endgroup$
    – xxjman18
    Dec 14, 2022 at 19:49
  • $\begingroup$ @xxjman18 I meant to say that if our conclusion of omega not being a arbitrary constant comes from our understanding of SHM then how will we identify such cases for equations whose meaning (from physics point of view is not known). Anyhow, your answer has led to some degree of clarification though. $\endgroup$ Dec 14, 2022 at 19:55

But forgetting our knowledge of physics, how can we mathematically justify 𝜔 in (1).

You can't. It's fair to take $y$ and $t$ to be the dependent and independent variables here, but we don't have any information beyond that to disambiguate this. From experience, or maybe by the ease with which it falls out, you might assume or conclude that $\omega$ is meant to be left as a parameter.

However, there's no mathematical reason that you couldn't ask this same exact question where any combination of $A,\omega,\phi$ are taken as arbitrary or not.

If I didn't immediately recognize this problem, my first instinct would be to take all constants as arbitrary. This would lead to a third-order equation. If we knew that the result should definitely be second-order (or first-order), it would really be a toss-up which constants should be considered arbitrary.

Basically, I would consider this a malformed problem. If this is your first or second brush with differential equations -- the prime time to be asked a question like this -- you likely lack the experience and context to immediately recognize that $\omega$ should be distinguished as a parameter.


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