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I don't seem to be getting anywhere. The differential equation is $$\frac{d^2x}{dt^2}=- \omega ^2x.$$

So, $$\frac{1}{x}dx^2=- \omega ^2dt^2$$

I integrated this equation twice but I'm not getting the general solution $x=A(\sin{( \omega t+\phi)})$. Please help.

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closed as off-topic by JMac, ZeroTheHero, sammy gerbil, stafusa, Gert Dec 27 '17 at 22:39

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    $\begingroup$ Have you had a course in differential equations? $\endgroup$ – Chet Miller Dec 27 '17 at 15:53
  • $\begingroup$ Try to think intuitively what this differential equation implies: $x$ is a solution where its second derivative is equal to itself (scaled by a factor $-w^2$). Which functions have this property? $\endgroup$ – nluigi Dec 27 '17 at 15:54
  • $\begingroup$ @nluigi I can think of $e^{iwt}$ which is not the sinusoid solution given in my book. Also I'd like a way to derive it instead of guessing. $\endgroup$ – Dove Dec 27 '17 at 16:17
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    $\begingroup$ $d^2 x\ne dx^2$. $\endgroup$ – ZeroTheHero Dec 27 '17 at 16:22
  • $\begingroup$ @Dove I don't think there is any way to "derive" the solution to the differential equation: there is going to be guesswork, or more politely, experience at play at some stage of solving a differential equation in closed form, unless you have a first order separable or exact equation. I would put this is an answer, but I'm afraid I might be wrong and there actually are special cases besides separable and exact equations that don't involve trying known solutions at all. $\endgroup$ – Styg Dec 27 '17 at 16:29
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You are misinterpreting the equation. It is

$$ \frac{{\rm d}}{{\rm d}t}\left( \frac{{\rm d}x}{{\rm d}t}\right) = -\omega^2 \,x $$

so you cannot separate the variables. You can however use the substitution ${\rm d}x = v\, {\rm d}t$ together with the chain rule $ \frac{{\rm d}v}{{\rm d}t} = \frac{{\rm d}v}{{\rm d}x} \frac{{\rm d}x}{{\rm d}t} =\frac{{\rm d}v}{{\rm d}x} v $

$$ \left. \frac{{\rm d}}{{\rm d}t} (v) = -\omega^2 \,x \right\} v\, \frac{{\rm d}v}{{\rm d}x} = -\omega^2 \,x$$

And now the separation of variables can take place, by moving the ${\rm d}x$ to the right-hand side

$$ \int v\,{\rm d}v =\int (-\omega^2 \,x) {\rm d}x +\mathtt{C}$$

$\Rightarrow$

$$ \left. \frac{v^2}{2} = \mathtt{C} - \frac{\omega^2 x^2}{2} \right\} v = \sqrt{2 \mathtt{C}-\omega^2 x^2} $$

Consider the initial condition $x=0$ and $v=v_0$ and use it to find $\mathtt{C}=\frac{v_0^2}{2}$.

$$ v = \sqrt{v_0^2-\omega^2 x^2} $$

The next integration finds the time dependency

$$ t = \int {\rm d}t = \int \frac{{\rm d}t}{{\rm d}x} {\rm d}x = \int \frac{1}{v}\,{\rm d}x =\int \frac{1}{\sqrt{v_0^2-\omega^2 x^2}}\,{\rm d}x $$

You can use an integration technique like substitution, or an integration table to find

$$ \left. t =\frac{1}{\omega}\, \sin^{-1} \left( \frac{\omega x}{v_0} \right) \right\} x = \frac{v_0}{\omega} \sin(\omega t) $$

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The general method for solving 2nd order equations requires you to make an ansatz (or a guess) as to the form of the function, and refine this guess so it matches the details of the equation and the boundary conditions.

The equation $$ \ddot{x}(t)=-\omega^2 x(t) \tag{1} $$ implies that the second derivative is proportional to the function itself, and this proportionality factor is negative. There are two types of functions that do this: the exponentials of the for $C_\pm e^{\pm i\lambda t}$ and the trigonometric $A\sin(\lambda t+\phi)$ or $B\cos(\lambda t+\phi)$. Here, $\lambda$ is to be determined, as are $C_\pm, A, B$ and $\phi$.

Insert these in turn into (1) to find the connection between $\lambda$, the other constants and $\omega$. The general solution will be a sum of all those that fit the bill. Since this is a 2nd order equation, you ought to be able to manipulate your general solution so that only two unknown constants remain.

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Separation of variables is not the way to go here. Instead, use the auxiliary equation method, so that you'll have $m^2 = - \omega ^2$, where $m$ is the number of derivatives. That will get you to the solution $A\sin( \omega t) + B\cos( \omega t)$.

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As being mentioned in the comments $d^2x \neq dx^2 $,
$d^2x$ itself doesn't mean anything, and $dx^2 $ means $(\Delta x)^2$, where $\Delta x$ is a very small segment.

A hand-waving "proof":

$$\frac{d^2x}{dt^2} = \frac{d}{dt}[\frac{x(t+\Delta t)-x(t)}{\Delta t}]\approx \frac{x(t+2\Delta t)-2x(t+\Delta t)+x(t)}{(\Delta t)^2}$$

All the $x$-terms are all linear, hence there is no $(\Delta x)^2$

Now return to the ODE, a usually good way to solve a ODE is to guess.

We guess the solution is $e^{at}$, then plug in and find out the value of $a$:

$$\frac{d^2}{dt^2}e^{at}=-\omega^2 e^{at} \implies a^2=-\omega^2 \implies a=\pm i\omega $$

hence, the solutions are either $x=Ae^{i\omega t}$or $x=Be^{-i\omega t}$

where $i=\sqrt {-1}$, that pretty much just means oscillation occuring.

Now if we want a complete solution, then we have: $$x(t)=Ae^{i\omega t} + Be^{-i\omega t}$$

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  • $\begingroup$ "The best way to solve a ODE is to guess." I'll have to disagree on that one. I repressed a lot of what I learned in DE and the courses following; but I'm pretty sure if guessing were the ideal method we would spend a lot less time covering them. $\endgroup$ – JMac Dec 27 '17 at 19:39

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