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Given the equation $$y'' + y=0$$ A solution is $y=\sin(t)$

Why can't we stop there since we know a way to solve the system? Why should we consider all of the ways to solve the system?

I would really like to see a real world example when having a single solution is inadequate. I know this is asking a lot, but I often find mathematics only becomes easier to understand once I need to use it to solve something and it becomes relate-able to real things.

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    $\begingroup$ If you know that $x=1$ is a solution to $x^2-4x+3=0$, why would it be important to find the other solution? Or would you just be happy with one solution? $\endgroup$
    – imranfat
    Mar 1, 2016 at 22:25
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    $\begingroup$ First, usually, one wants all soluttions, not just one. Second, a differential equation has initial conditions. This one has initial conditions $y(0)=1, \enspace y'(0)=0$. What about the other initial conditions? $\endgroup$
    – Bernard
    Mar 1, 2016 at 22:28
  • $\begingroup$ @imranfat I can relate that equation to throwing a ball and how at two points it will contact the ground. But I cannot relate this idea to the real world yet. $\endgroup$
    – Klik
    Mar 1, 2016 at 22:36
  • $\begingroup$ @Bernard I can understand that one wants all the solutions, only on the basis that, my professor says so. Couldn't we still use the sine function for other initial conditions? Isn't something like $y=Asin(Bt+K)$ a perfectly valid answer for all initial conditions? $\endgroup$
    – Klik
    Mar 1, 2016 at 22:39
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    $\begingroup$ Writing $y=A\sin(t + K)$ is essentially the same as writing $y=\tilde A\sin(x) + \tilde B cos(x)$, as you can write $\sin(x + \alpha) = \cos(\alpha)\sin(x) + \sin(\alpha) \cos(x)$, i.e. $\tilde A = \cos(\alpha), \tilde B = \sin(\alpha)$ $\endgroup$
    – Roland
    Mar 1, 2016 at 22:47

1 Answer 1

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Consider a point mass $m= 1 \ \mathrm{kg}$ attached to one end of a spring with spring constant $k = 1 \ \mathrm{N/m}$. Suppose that the spring is suspended vertically from an immovable support. The oscillations of the mass around its equilibrium position can then be described by the equation $$y''+y = 0$$ where $'$ denotes time derivative, $y$ the displacement and where we measure distance in $\mathrm{m}$ and time in $\mathrm{s}$.

As you have said, $y = \sin t$ is a solution to this equation. For this solution, we have $y(0) = 0$ and $y'(0) = 1$. This means that the mass starts from rest with unit speed. Furthermore, the maximum displacement of the mass is $1$.

But what if the mass doesn't start from rest and what if it doesn't have unit speed? What if we release the mass $1.5 \ \mathrm{m}$ from its equilibrium at $t=0$ and with zero initial speed? Then your solution would be $y = 1.5 \cos t$. This is physically different from $y = \sin t$.

In order to be able to account for different initial conditions, you need the general solution, which can be written in many ways, one of which is $y = A\sin (t) + B\sin (t)$. We could also write this as $y = C\sin(t+ \phi)$, as $C\sin(t+\phi) = C\cos(\phi)\sin(t) + C\sin(\phi)\cos(t)$. However, this is still very different from simply $y = \sin t$.

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    $\begingroup$ Well dang lol you and I had quite nearly the same answer. Since you answered first, I'll delete mine. $\endgroup$ Mar 1, 2016 at 22:38
  • $\begingroup$ @CameronWilliams Too bad :) $\endgroup$ Mar 1, 2016 at 22:39
  • $\begingroup$ I'd like to work through your example to see if I can model this using the sine function with both sets of initial conditions; could you please elaborate more on the math? Also, I think you meant to say $y=A\sin(t) + B\cos(t)$. $\endgroup$
    – Klik
    Mar 1, 2016 at 22:45
  • $\begingroup$ @Klik What do you want me to elaborate on? $\endgroup$ Mar 1, 2016 at 22:50
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    $\begingroup$ Nevermind. Roland cleared it up in the comments. Ok, so the short story is that we need both solutions to form a general solution so that, given any initial condition, we can model it. $\endgroup$
    – Klik
    Mar 1, 2016 at 23:07

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