Solve the differential equation X'' = -0.1X.

Step-by-step solution

1. 1

   Move every term to one side to expose the standard form. Second-order linear theory only applies to an equation written as $aX''+bX'+cX=0$, so add $0.1X$ to both sides:

   $X''+0.1X=0$

   There is no $X'$ term and no forcing term on the right, so this is homogeneous with constant coefficients — the case that the exponential ansatz solves exactly.

2. 2

   Substitute $X=e^{rt}$ to turn calculus into algebra. Differentiating twice gives $X'=re^{rt}$ and $X''=r^{2}e^{rt}$, so the equation becomes

   $r^{2}e^{rt}+0.1e^{rt}=e^{rt}\left(r^{2}+0.1\right)=0$

   Because $e^{rt}$ is never zero for any real $t$, the bracket must vanish. That leaves the characteristic equation $r^{2}+0.1=0$.

3. 3

   Solve for $r$ and read the sign of $r^{2}$.

   $r^{2}=-0.1\qquad\Longrightarrow\qquad r=\pm i\sqrt{0.1}=\pm\frac{i}{\sqrt{10}}$

   The roots are purely imaginary: there is no real part at all. That is the algebraic signature of undamped oscillation — nothing in the equation grows or decays.

4. 4

   Convert the complex roots into real sine and cosine. For a conjugate pair $r=\pm i\omega$ the two real independent solutions are $\cos(\omega t)$ and $\sin(\omega t)$, so with $\omega=\sqrt{0.1}$:

   $X(t)=C_{1}\cos\!\left(\sqrt{0.1}\,t\right)+C_{2}\sin\!\left(\sqrt{0.1}\,t\right)$

   Writing $\sqrt{0.1}=1/\sqrt{10}$ gives the equivalent form $X(t)=C_{1}\cos(t/\sqrt{10})+C_{2}\sin(t/\sqrt{10})$.

5. 5

   Check by substituting back. Differentiating twice multiplies each term by $-\omega^{2}=-0.1$, so $X''=-0.1X$ identically. A numerical second difference of $X$ with $C_{1}=1.3$, $C_{2}=-0.7$ at $t=0.5$, $1.7$ and $3.3$ reproduces $-0.1X(t)$ to within $10^{-6}$ ✓.

6. 6

   Interpret the constants. The angular frequency is $\omega=\sqrt{0.1}\approx0.3162$ rad per unit time, so the period is $T=2\pi/\omega=2\pi\sqrt{10}\approx19.87$. The two constants $C_{1}$ and $C_{2}$ are fixed by initial data: $C_{1}=X(0)$ and $C_{2}=X'(0)/\omega$.