# Determining the variable of interest in a differential equation

Studying Simple Harmonic Motion and looking at the derivation.

$ma = \Sigma F$

$m \frac{d^2x}{dt^2} = -kx$

$\frac{d^2x}{dt^2} + \frac{k}{m}x = 0$

The textbook then says that "we want to determine what function of time, $x(t)$, satisfies the equation".

This makes sense from the SHM perspective since we want to define the displacement of a particle in terms of time. However just by looking at the differential equation, how were they able to tell that they had to solve for $x(t)$? Is it simply because there exists the second time derivative of $x$ ?

FOLLOWUP

Let's say I didn't care about $x$ and wanted to solve for the velocity of the particle instead. Would it make sense to do the following:

$\frac{dv}{dt} + \frac{k}{m}x = 0$

and try finding a solution for $v$? (assuming it exists)

You are getting it the wrong way. We never see the equation and then decide to solve for $x$. We need to find $x$ as a function of time and that is why we formulate the equation (which may be solved for determining $x$). Each and every variable in that equation has a physical significance.
$x$ is the only dependent variable here. $x$ in Simple Harmonic Motion represents the position of the particle at any instant. $t$ stands for the time, which as you know is an independent variable. Meanwhile, $m$ (mass) and $k$ are constants. So we need to solve for $x$ which helps us determine the position of the particle at any time. And also using the equation for $x$ we can determine the velocity and acceleration of the particle at any instant by using differentiation.
• @Carpetfizz For finding velocity, solve for $x$ and then find derivative of $x$ w.r.t.time. Velocity is simple rate of change of $x$ after all. – user220382 Dec 1 '16 at 9:49
• @Carpetfizz The follow up equation you wrote is correct, but you cannot solve it for $v$ directly. There are two dependent variables $v$ and $x$. You need to write $v$ as $\frac{dx}{dt}$ in order to solve it. You can solve for only one dependent variable from a single differential equation like that. – user220382 Dec 1 '16 at 9:55