Fundamental Sets and General Solutions

The fundamental set theorem is the reason engineers can write "the general solution of a damped spring-mass system is $c_1e^{r_1t}+c_2e^{r_2t}$" and mean it exactly, not approximately — this underlies vibration analysis, circuit theory (RLC circuits), and control systems, all covered under Homogeneous Equations with Constant Coefficients. In quantum mechanics, the time-independent Schrödinger equation for simple potentials is a second-order linear ODE, and identifying a fundamental set of solutions is precisely how bound-state and scattering solutions are classified. In numerical analysis, the theorem justifies why solving an $n$th-order linear ODE numerically only ever requires tracking $n$ independent trajectories (or an $n$-dimensional state vector), which is the basis of state-space methods and the transition from scalar ODEs to first-order systems discussed under Higher-Order Linear Equations. In algebraic and differential Galois theory, the dimension of the solution space of a linear ODE (a $\mathcal{D}$-module) is a basic invariant used to classify special functions; hypergeometric and Bessel equations are studied precisely by exhibiting fundamental sets and their monodromy. Finally, in statistics and signal processing, the state-space (Kalman filter) formulation of linear time-varying systems relies on exactly this vector-space structure of the solution set to propagate uncertainty consistently through time.

:::quiz[If $y_1,y_2$ solve a linear homogeneous 2nd-order ODE with continuous coefficients on $I$ and $W[y_1,y_2](t_1)=0$ for one $t_1\in I$, what can you conclude about $W$ at other points of $I$?]

• It must be nonzero everywhere else in $I$
• It is identically zero on all of $I$ ✓
• It could be zero at some points and nonzero at others
• Nothing can be concluded without more information

:::quiz[The Cauchy–Euler equation $2t^2y''-ty'+y=0$ has solutions $y_1=t$, $y_2=t^{1/2}$ on $(0,\infty)$. What is $W[y_1,y_2](4)$?]

• $0$
• $-1$ ✓
• $2$
• $\sqrt{4}=2$, so undefined sign

:::quiz[For the repaired fundamental set $\{e^{2t}, te^{2t}\}$ of $y''-4y'+4y=0$, what is $W[e^{2t},te^{2t}](0)$?]

• $0$
• $1$ ✓
• $4$
• $e^{4\cdot0}=1$ but with a minus sign, so $-1$

Ask the tutor: Why does a nonzero Wronskian at just one point guarantee it's nonzero everywhere on the interval, and how does that fact depend on the equation being linear and homogeneous?

Think of the solution set of $(\ast)$ as a landscape you are trying to map. Each individual solution is a single point discovered by trial, substitution, or luck. The fundamental set theorem says this landscape is flat and two-dimensional (for a second-order equation): once you have two independent "directions" — two solutions that are not scalar multiples of each other — you can reach every point in the landscape by walking a straight combination of those two directions. There is no hidden third direction, no exotic solution lurking outside the span. This is a strong structural claim, and it is exactly what makes linear ODEs solvable in a systematic way rather than an art of clever guesses. The Wronskian is the instrument that detects whether two solutions are truly two independent directions or secretly the same direction in disguise. Geometrically, $(y_1(t), y_1'(t))$ and $(y_2(t), y_2'(t))$ are two points in the phase plane $\mathbb{R}^2$ at each time $t$; the Wronskian is (twice) the signed area of the parallelogram they span. If that area is ever nonzero, the two vectors are not parallel, and — as proved under The Wronskian and Linear Independence — this non-parallelism persists for all $t \in I$ whenever $y_1, y_2$ solve $(\ast)$. The fundamental set theorem then says: two non-parallel initial vectors are enough to hit any target initial vector $(y_0, y_0')$ by scaling and adding, and each such target initial vector corresponds to exactly one solution curve by uniqueness. Combining these two facts is the whole proof.

Example 1 (Cauchy–Euler equation with an asymmetric interval and non-integer exponents). Consider on $I = (0,\infty)$ the equation

Try $y = t^r$: then $y' = rt^{r-1}$, $y'' = r(r-1)t^{r-2}$, and substitution gives

Factor: $2r^2 - 3r + 1 = (2r-1)(r-1)$, with roots $r = 1$ and $r = \tfrac12$. So

are both solutions on $(0,\infty)$. Compute the Wronskian directly:

For any $t_0 > 0$, $W[y_1,y_2](t_0) = -\tfrac12\sqrt{t_0} \neq 0$. First put the equation in standard form to confirm continuity hypotheses: dividing by $2t^2$,

so $p(t) = -\tfrac{1}{2t}$ and $q(t) = \tfrac{1}{2t^2}$ are continuous on $I=(0,\infty)$ (the singularity at $t=0$ is excluded from $I$, which is exactly why we work on $(0,\infty)$ rather than all of $\mathbb{R}$). All hypotheses of the theorem hold, so $\{y_1,y_2\}$ is a fundamental set on $(0,\infty)$ and the general solution is

By the theorem, any solution of the original equation on $(0,\infty)$ — however it was found, by series methods, numerically, or otherwise — must be expressible this way. For instance, one can now immediately solve the IVP $y(4) = 10$, $y'(4) = 3$: substituting, $4c_1 + 2c_2 = 10$ and $c_1 + \tfrac14 c_2 = 3$, giving (solving) $c_2 = 2$, $c_1 = 5/2$, so $y(t) = \tfrac52 t + 2\sqrt t$ is the unique solution — uniqueness guaranteed by the theorem, not just a hopeful guess.

Example 2 (verifying two solutions are NOT a fundamental set, and repairing it). Consider $y'' - 4y' + 4y = 0$ on $I = \mathbb{R}$ (standard form already, $p=-4$, $q=4$ continuous everywhere). Two students propose

Both are genuine solutions (check: $y_1'' - 4y_1' + 4y_1 = 4e^{2t}-8e^{2t}+4e^{2t}=0$, and $y_2$ is a constant multiple so it also satisfies the linear homogeneous equation). Compute the Wronskian:

The Wronskian is identically zero, so the hypothesis of the theorem fails: $\{y_1, y_2\}$ is not a fundamental set (indeed $y_2 = 2y_1$, manifestly dependent). The theorem gives no conclusion here — and correctly so, since $c_1y_1 + c_2y_2 = (c_1+2c_2)e^{2t}$ only ever produces multiples of $e^{2t}$, never reaching, e.g., the solution $t e^{2t}$ (readily checked: $(te^{2t})'' - 4(te^{2t})' + 4(te^{2t}) = 0$). To repair this, use reduction of order or the characteristic-root method to find the second independent solution $y_2^{\text{true}}(t) = te^{2t}$. Now

for every $t \in \mathbb{R}$, so $\{e^{2t}, te^{2t}\}$ is a fundamental set, and the general solution is $y(t) = c_1e^{2t} + c_2te^{2t}$, matching the standard repeated-root theory previewed under Homogeneous Equations with Constant Coefficients.