Solving $f''(x) = -f(x)$ without using Euler's formula

Question

As part of a proof to Euler's formula, I got to the differential equation $f(x) = -f''(x).$ I already know that the solution to this equation is: $f(x) = c_1 \sin(x) + c_2 \cos(x),$ but beside the fact that both $\sin$ and $\cos$ satisfy this equation, how can I prove this lemma? And preferably without using Euler's formula, so that this could be a valid proof.

Answer 1

You can solve that equation using this: $$\frac {f''}{f}=\left (\frac {f'}{f}\right )'+\left (\frac {f'}{f} \right )^2$$ Substitute $u=\frac {f'}{f}$ $$u'+u^2=-1$$ This is separable.

Answer 2

Say you've computed $f$ numerically (using, say, Euler's method). That is, you don't know a formula for $f$, but you know some values of $f$ to some decimal places. If you graph $f$ you'll see a wave, but let's say you don't recognize that graph.

One thing we can do is plot the configuration space. This will be $\vec r(t)=(f(t),f'(t))$, a point on the plane depending on $t$ (as well as the initial conditions of $f$). If you do this, one thing will surprise you: no matter the initial conditions of $f$, the point $\vec r(t)=(f(t),f'(t))$ appears to move in circles around the origin! Furthermore, the angular velocity seems to always be 1 (equivalently, the velocity equals the radius).

So, inspired by your empirical findings, you make three conjectures: one, that the point is a constant radius away from the origin (aka $\|\vec r(t)\|=C$), and two, that the point is moving at a constant velocity (aka $\|\vec r'(t)\|=C$), and third, the radius and the velocity are equal (aka $\|\vec r\|=\|\vec r'\|$).

To prove the first one, we see that we want to prove: \begin{align} \|r(t)\|&=C\\ \sqrt{(f)^2+(f')^2}&=C\\ (f)^2+(f')^2&=C^2\\ \left((f)^2+(f')^2\right)'&=0 \end{align} Now, we prove this, using $f''=-f$: \begin{align} \left((f)^2+(f')^2\right)'&=2ff'+2f'f''\\ &=2ff'+2f'(-f)\\ &=2ff'-2ff'\\ &=0 \end{align} and the first conjecture is proved. The next two are similar: I leave them as exercises to you.

Now we know that $(f(t),f'(t))$ travels around a circle with constant angular velocity 1, and a little trigonometry tells you that the equation for that is $(R\cos(t+\varphi),R\sin(t+\varphi))$ where $R$ is the radius of the circle and $\varphi$ is the offset (it tells you where on the circle you start). Thus: $$f(t)=R\cos(t+\varphi)$$ And we didn't even need to know what the derivative of cosine is! Some trigonometry will show that this is equivalent to your solution.

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Note that, while our solution and proof is something that could be done on pencil and paper, our conjectures that we made at the start required us to have computed $f$ for some values! So it often goes: to make real progress, sometimes you have to get your hands dirty.

In general, this is how it goes in research math (and, once upon a time, this was research math): you have a problem, you have no idea how to solve it, you do lots of calculating, you spot a pattern, and that lets you get a foothold into solving the problem. (And then you hide all the ugly computations by the time you write a paper, so that it looks like you knew what you were doing all along.)

Answer 3

Thank you everyone for the comments and answers. I wanted to share with you a solution I found out for this problem. After treating the system like a harmonic oscillator (it satisfies this equation when $\frac{k}{m} = \frac{1}{sec^2} $), I realized there is general formula which works for any ordinary differential equation. let $y$ be a function of $x$, then: $$\int_{y_0}^{y(x)} y'' dy = \left.\frac{y'^2}{2}\right\vert_{y'_0}^{y'(x)}$$ which is basically just the work energy theorem after division by the mass (I will prove this result at the end of this answer). Using this formula and setting $y'' = -y$ we get: $$\int_{y_0}^{y(x)} -y dy = \left.\frac{y'^2}{2}\right\vert_{y'_0}^{y'(x)}$$ $$\left.\frac{-y^2}{2}\right\vert_{y_0}^{y(x)} = \left.\frac{y'^2}{2}\right\vert_{y'_0}^{y'(x)}$$ $$y_0^2-y(x)^2 = y'(x)^2 - y_0'^2$$ which is a (relatively) simple separable first order differential equation.

proof of the formula:

1. Evaluate the integral $$\int_{y_0}^{y(x)} y'' dy$$
2. Rewrite $y''$ as $\frac{dy'}{dx}$
3. Use step two to change $y''dy$ to $\frac{dy'}{dx}dy$, then to $dy'\frac{dy}{dx}$
4. Notice that $\frac{dy}{dx} = y'$ so $dy'\frac{dy}{dx} = y'dy' = \frac{(\frac{dy}{dx})^2}{2} + C$ And so $$\int_{y_0}^{y(x)} y'' dy = \left.\frac{y'^2}{2}\right\vert_{y'_0}^{y'(x)}$$ (The change of the limits of integration from $y(x)$ and $y_0$ on the left to $y'(x)$ and $y'_0$ on the right, comes from the fact that we ended up integrating by $dy'$, not by $dy$, and so the limits must change to the limits of $y'$.)