Second-Order Linear Equations

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.