Second-Order Linear Equations

Why exponentials: with constant coefficients, $y''$, $by'$ and $cy$ must cancel one another for every $x$, which is only possible if they are the same kind of function. The exponential $e^{rx}$ is the function whose derivatives are multiples of itself, so substituting it reduces the differential equation to an algebraic one for $r$. The three cases are the three ways a real quadratic can have roots, and each supplies two independent solutions — exactly what a second-order equation needs (the proofs below give the details).

Why the discriminant decides: the roots are $r = \dfrac{-b \pm\sqrt{b^2 - 4\omega^2}}{2}$. When $b^2 < 4\omega^2$ they are complex, and the imaginary part produces oscillation. When $b^2 > 4\omega^2$ both roots are real and negative, so the solution is a sum of two decaying exponentials and crosses zero at most once. At $b = 2\omega$ the two roots merge. The figure starts $y'' + by' + 4y = 0$ at $y(0) = 1$, $y'(0) = 0$: $b = 1$ overshoots and rings, $b = 4$ returns fastest without crossing, $b = 6$ creeps back.

How the data pin the constants: differentiate first, $y' = Ae^x + 2Be^{2x}$, then set $x = 0$. The conditions become $A + B = 3$ and $A + 2B = 5$; subtracting gives $B = 2$, then $A = 1$, so $y = e^x + 2e^{2x}$. Two conditions for two constants, always a $2 \times 2$ linear system.

Each term is a force. The spring pulls back in proportion to the stretch ($-kx$, Hooke's law), the damper resists in proportion to the speed ($-cx'$), and Newton's second law sets their sum equal to $mx''$; moving everything to one side gives the equation. The characteristic roots are $\dfrac{-c \pm\sqrt{c^2 - 4mk}}{2m}$, so the sign of $c^2 - 4mk$ decides between complex roots (oscillation) and real ones (none). Critical damping $c = 2\sqrt{mk}$ is the smallest damping that prevents oscillation.

Step 3 — the data: $x(0) = A = 1$. Differentiating, $x' = Be^{-3t} - 3(A + Bt)e^{-3t}$, so $x'(0) = B - 3A = 0$ and $B = 3$.

Step 4 — $x(t) = (1 + 3t)e^{-3t}$, which stays positive for all $t > 0$: the mass returns without overshooting.

Step 5 — verify: $x' = -9te^{-3t}$ and $x'' = (-9 + 27t)e^{-3t}$, so $x'' + 6x' + 9x = e^{-3t}\left[(-9 + 27t) - 54t + (9 + 27t)\right] = 0$ ✓.

Step 3 — the data: $A + B = 1$ and $x'(0) = -A - 9B = 0$, so $B = -\tfrac18$ and $A = \tfrac98$: $x = \tfrac98e^{-t} - \tfrac18e^{-9t}$.

Step 4 — compare with critical damping at $t = 2$: $(1 + 6)e^{-6} \approx 0.017$ m, against $\tfrac98e^{-2} - \tfrac18e^{-18} \approx 0.152$ m. More damping made the return slower, because the slow root $-1$ dominates and the mass creeps back. Critical damping is the fastest return without oscillation, which is why door closers and car suspensions are tuned near it.

Expand the right side with the angle-difference formula: $R\cos(\omega t - \varphi) = R\cos\varphi\cos\omega t + R\sin\varphi\sin\omega t$. Matching coefficients requires $R\cos\varphi = A$ and $R\sin\varphi = B$, so $(R, \varphi)$ are the polar coordinates of the point $(A, B)$. That is why the amplitude is $\sqrt{A^2 + B^2}$, and why $\varphi$ must lie in the quadrant of $(A, B)$ rather than be read off $\arctan\tfrac BA$ blindly.

Step 3 — $y = 5\cos(2t - 0.927)$.

Step 4 — the first maximum is where $2t - \varphi = 0$, at $t = \tfrac\varphi2 \approx 0.464$, with value $5$.

Step 5 — check at $t = 0$: $5\cos\varphi = 5\cdot\tfrac35 = 3$, matching $3\cos 0 + 4\sin 0 = 3$ ✓.

Step 3 — $3\cos 3t + 4\sin 3t = 5\cos(3t - \varphi)$ with $\varphi = \arctan\tfrac43 \approx 0.927$.

Step 4 — so $y = 5e^{-t}\cos(3t - \varphi)$: an oscillation with quasi-period $\tfrac{2\pi}{3} \approx 2.09$ between the envelopes $\pm5e^{-t}$.

Step 5 — one quasi-period multiplies the envelope by $e^{-2\pi/3} \approx 0.123$, so each maximum is about $12\%$ of the one before.

Step 6 — verify: $e^{-t}\cos 3t$ and $e^{-t}\sin 3t$ both solve the equation because $-1 \pm 3i$ are the roots, and $y(0) = 3$, $y'(0) = -3 + 12 = 9$ ✓.

Two independent solutions of a second-order linear equation span all solutions (the solution space is $2$-dimensional: a solution is pinned by $y(0), y'(0)$), so distinct roots finish the problem. $\blacksquare$

Substitute: $e^{rx}\left[(2r + r^2x) + b(1 + rx) + cx\right] = e^{rx}\left[(2r + b) + x(r^2 + br + c)\right] = 0$ — both brackets vanish precisely because $r$ is a double root. $\blacksquare$

Real linear combinations — half the sum, and the difference over $2i$ — extract $e^{\alpha x}\cos\beta x$ and $e^{\alpha x}\sin\beta x$, two real independent solutions. Decay times oscillation, straight from algebra. $\blacksquare$

Correspondingly the general solution carries exactly two free constants, and the solution space of the homogeneous equation is two-dimensional: find two independent solutions and every solution is a combination of them. The characteristic equation delivers those two: distinct real roots give two exponentials; a double root gives $e^{rx}$ and $xe^{rx}$; complex roots give a cosine and a sine dressed in an exponential envelope. Applying initial data always ends the same way — a small linear system in the two constants, one equation from $y(0)$, one from $y'(0)$.

Step 1 — characteristic equation: $r^2 - 3r + 2 = (r-1)(r-2) = 0$: roots $1, 2$.

Step 2 — general solution: $y = Ae^{x} + Be^{2x}$.

Step 3 — apply the data: $y(0) = A + B = 3$; since $y' = Ae^x + 2Be^{2x}$, $y'(0) = A + 2B = 5$.

Step 4 — solve the little system: subtracting, $B = 2$, then $A = 1$: $y = e^x + 2e^{2x}$.

Step 5 — evaluate: $e^{\ln2} = 2$ and $e^{2\ln2} = (e^{\ln2})^2 = 4$, so $y(\ln 2) = 2 + 2\cdot4 = 10$.

Step 6 — verify the ODE once: $y'' = e^x + 8e^{2x}$, $3y' = 3e^x + 12e^{2x}$, $2y = 2e^x + 4e^{2x}$; then $y'' - 3y' + 2y = (1 - 3 + 2)e^x + (8 - 12 + 4)e^{2x} = 0$ ✓.

Step 1 — characteristic: $r^2 + 2r + 5 = 0$ gives $r = \dfrac{-2 \pm \sqrt{4 - 20}}{2} = -1 \pm 2i$.

Step 2 — real solution form ($\alpha = -1$, $\beta = 2$): $y = e^{-x}(A\cos 2x + B\sin 2x)$.

Step 3 — data: $y(0) = A = 1$; the product rule gives $y'(0) = -A + 2B = -1$, so $B = 0$: $y = e^{-x}\cos 2x$.

Step 4 — read it: an oscillation of angular frequency $2$ (period $\pi$) inside a decaying envelope $e^{-x}$ — underdamped motion, crossing zero every $\tfrac\pi2$, each swing about $e^{-\pi} \approx 4\%$ the height of the one two crossings before. The root's real part is the decay rate; its imaginary part is the frequency: the algebra is the physics.