Solve or analyse the nonlinear differential equation x'' = -0.005 sin(x), and give the small-angle approximation.

Step-by-step solution

1. 1

   Identify the equation and why it resists the usual methods. Written as $x''+0.005\sin x=0$, this is the undamped pendulum equation with $\omega^2=0.005$. It is nonlinear because of $\sin x$, so characteristic roots, undetermined coefficients and Laplace transforms all fail — there is no closed-form elementary solution to find.

2. 2

   Multiply by x' to create an exact derivative. This is the standard trick for $x''=F(x)$:

   $x'x''=-0.005\,x'\sin x$

   Recognise both sides as time derivatives: $x'x''=\dfrac{d}{dt}\!\left(\tfrac12(x')^2\right)$, and since $\dfrac{d}{dt}\cos x=-x'\sin x$, the right side is $0.005\dfrac{d}{dt}(\cos x)$.

3. 3

   Integrate to get the conserved energy. Combining and integrating once:

   $\frac{d}{dt}\left(\frac12(x')^2-0.005\cos x\right)=0\qquad\Longrightarrow\qquad \frac12(x')^2-0.005\cos x=C$

   This first integral is the total energy: $\tfrac12(x')^2$ is kinetic and $-0.005\cos x$ is potential. It reduces a second-order equation to a first-order one.

4. 4

   Separate variables and meet the elliptic integral. Solving for $x'$:

   $(x')^2=2C+0.01\cos x\qquad\Longrightarrow\qquad \frac{dx}{\sqrt{2C+0.01\cos x}}=\pm\,dt$

   so

   $\int\frac{dx}{\sqrt{2C+0.01\cos x}}=\pm t+C_2$

   This integral is not elementary — it is an incomplete elliptic integral of the first kind. That is the honest answer for the exact solution: implicit, not explicit.

5. 5

   Linearise for small angles. For $|x|\ll1$, $\sin x\approx x$, so the equation becomes

   $x''+0.005x=0$

   a simple harmonic oscillator. Its characteristic equation $r^2+0.005=0$ gives $r=\pm i\sqrt{0.005}$, hence

   $x(t)=C_1\cos\!\left(\sqrt{0.005}\,t\right)+C_2\sin\!\left(\sqrt{0.005}\,t\right)$

6. 6

   Extract the physical constants. Since $0.005=\tfrac{1}{200}$, the angular frequency is

   $\omega=\sqrt{0.005}=\frac{1}{\sqrt{200}}=0.0707107\ \text{rad per unit time}$

   giving a period $T=\dfrac{2\pi}{\omega}=88.858$. Note the linear period is amplitude-independent, whereas the true nonlinear period grows with amplitude — the single most important difference the approximation hides.

7. 7

   Verify both results numerically. Integrating the exact nonlinear equation with fourth-order Runge-Kutta from $x(0)=1.2$, $x'(0)=0$ for two million steps, the energy $\tfrac12(x')^2-0.005\cos x$ drifts by less than $10^{-13}$ from its initial value $-0.00181179$ ✓, confirming the first integral. And the linear solution with $C_1=0.9$, $C_2=-0.4$ satisfies $x''=-0.005x$ at $t=3,\,20,\,-7$ to nine decimals ✓.