# Uniqueness of general solution to SHO

This may be a duplicate, though I have searched and not found this question answered, and it may also belong more on Mathematics Stack exchange than here -- in which case I'll transfer.

My question is: how does one prove (both intuitively and rigorously) that the solution to the SHO, being a linear combination of a sine and cosine, is the most general and unique solution?

The way it is most often solved is by simply suggesting $x(t) = \exp(\Omega t)$, then solving $\Omega = \pm ik/m$, and ending up with $x(t) = A\cos(\omega t) + B\sin(\omega t)$ with $\omega = |\Omega|$.

Suppose I am going trough this derivation with a high-school physics enthusiast, and he/she asks me "You've simply supposed $x(t)$ to be exponential, and showed that if it is, the solution is $\dots$, how do you know this is $\textit{the}$ solution?". I've done a differential equations class, and even though I passed it, I seem to have missed this crucial aspect.

EDIT

Since the posting of this question, two answers have been posted only answering the question of $\textit{how}$ the SHO should be solved. A question I did not ask.

My question has boiled down to this; how do I show that the space of solutions of the SHO (and any 2nd order ODE) is two-dimensional? This would answer my question.

• The 'true' solution depends on the initial value, but Euler's formula says the exponential function is related to sine+cosine terms. May 23, 2018 at 12:15
• For example, see math.stackexchange.com/q/823470 May 23, 2018 at 12:23
• The fundamental theorem of linear differential equations tells you that a second order ODE will have two basis functions; for SHO these are Sin and Cos. Linear combinations of these make up the general solution to the homogeneous case. The complex exponential ansatz leads you to the general solution. May 23, 2018 at 12:26
• @KyleKanos, I'm not looking for a solution to an IVP per se, the general solution of the SHO is a linear combination of those two sines and cosines. Also, I know how to make the sines and cosines out of the complex exponentials - that is not what this question is about at all. Assuming the solution showed above is found, how do I know this is the most general solution? I'd prefer both an intuitive and rigorous proof if possible. Preferably one that a high-school physics enthusiast can understand - not implying I am one, because I'm not. May 23, 2018 at 12:28
• @StijnD'hondt so you're looking for the uniqueness & existence theorem for 2nd order Diff Eqs?? May 23, 2018 at 12:31

Simple harmonic motion corresponds to a 2nd order differential equation. Such equations have two linearly independent solutions, and the general solution is some linear combination of these two.

A more 'thorough' but tedious algebraic treatment considers the so-called auxiliary equation of the general second order DE

$ay'' + by' + cy = 0$,

and the three possible cases (distinct real roots, repeated real roots, and conjugate complex roots) for the auxiliary function. For more detail see any introductory book that includes a section on second-order ordinary differential equations.

The general solution for the conjugate complex roots is

$y=e^{\alpha x}(c_1\cos(\beta x) + c_2\sin(\beta x))$.

SHO corresponds to the case of two conjugate complex roots (m = $\pm i\omega$), with $\alpha = 0$ and $\beta=\omega$.

Answering your question in another way: if you guess a solution, substitute it in the DE and see that it solves it, that is a confirmation that it is a solution. Euler's formula does the rest.

EDIT: As per the comment of @KyleKanos, you may be missing the relevance of the relevant uniqueness theorem here.

• Hello, and thank you for your answer. Though I must say it's not an answer to my question. As stated, I already know how to solve these equations, I simply need to know how to prove that the space of solutions to a 2nd order ODE is two-dimensional. I'll edit my question to make this more clear. May 23, 2018 at 12:44
• Sure. If you're looking for a proof of that you'd probably be better off asking on math se. It's a generic result that n'th order differential equations have n linearly independent solutions. May 23, 2018 at 12:46

Consider Newton's second law for a simple Hooke spring

$$m \ddot x = -kx.$$

Below I show two ways of arriving at the general solution for this system. There are other ways of solving this. I will also note that these are not unique solutions until initial conditions have been applied.

The Intuitive Solution

Our solution to the differential equation is a function $f(t)$ that is proportional to its second derivative by a negative constant, i.e.

$$f(t) = -\alpha \frac{d^2f}{dt^2}.$$

Both sine and cosine satisfy this, so you have to add them together to find the general solution. This approach could also let you arrive at the equally-valid solution $x(t) = c_1 e^{i\omega t} + c_2 e^{-i\omega t}$.

In my opinion, the most basic intuitive reason for superposing the solutions is that you don't know where $\sin\omega t$, $\cos\omega t$, or both is the best model until you apply your initial conditions, which provide phase information.

The Rigorous Solution

We can rearrange the differential equation to be

$$\ddot x + \frac{k}{m}x = 0 .$$

Defining $\omega = \sqrt\frac{k}{m}$, we have $\ddot x + \omega^2x = 0$. We can apply an characteristic equation to solve our differential equation by replacing our derivatives with polynomial powers:

$$u^2 + \omega^2 = 0$$ $$u = \pm i\omega.$$

The solution to a characteristic equation $u = \alpha \pm i\beta$ gives us a solution to our differential equation of

$$x(t) = c_1e^{(\alpha + i\beta)t} + c_2e^{(\alpha - i\beta)t} = c_1 e^{\alpha t}\cos\beta t + c_2 e^{\alpha t}\sin\beta t.$$

Noting that in our case $\beta = \omega$ and $\alpha = 0$, we arrive the general solution

$$x(t) = c_1 e^{i\omega t} + c_2 e^{-i\omega t} = A\sin\omega t + B\cos\omega t.$$

• Thank you for your answer, and I'm sorry if the way I asked my question was unclear, but this is not an answer to my question. I already know how to solve these equations, in fact, in my original question the exact same solution as you have given is described in words -- making an exponential ansatz and going from there... My question is about proving that the space of solutions to this equation is two-dimensional, basically. May 23, 2018 at 12:43
• Ah. Well you know that $\sin\omega t$ and $\cos\omega t$ are linearly independent because the equation is only zero if $A=B=0$. They also obviously span the solution set, so they form a basis.
– zh1
May 23, 2018 at 13:04