What is Wronskian?

Example 16.7 (Liouville checked on an Euler equation)

On $\intoo{0}{\infty}$, the equation $t^2y'' + ty' - y = 0$ has the solutions $y_1(t) = t$ and $y_2(t) = \frac1t$ (substitute). Their wronskian:

never zero: a fundamental system. Now check Liouville: in normalized form $y'' + \frac1t\,y' - \frac{1}{t^2}\,y = 0$, the companion matrix $A(t) = \begin{pmatrix} 0 & 1\\ \frac{1}{t^2} & -\frac1t\end{pmatrix}$ has trace $-\frac1t$, so

Closing insight: Liouville predicts the shape of the wronskian before any solution is known — here, that $W$ must be $\frac{c}{t}$; this is what powers the reduction-of-order method (Proposition 16.15), where knowing $y_1$ and the wronskian’s form determines $y_2$ by one quadrature.

Example 16.5 (The evaluation isomorphism, concretely)

For $y'' + y = 0$, viewed as the system $X' = \begin{pmatrix} 0 & 1\\ -1 & 0\end{pmatrix}X$ with $X = (y, y')$: the theorem says the solution space is a plane, and that $X \mapsto X(0) = (y(0), y'(0))$ is an isomorphism onto $\R^2$. The solutions $\cos$ and $\sin$ evaluate to $(1, 0)$ and $(0, 1)$ — the canonical basis of $\R^2$ — so they form a basis of the solution space, and every solution is

with the coefficients read directly off the initial data, no linear system to solve. Closing insight: choosing the fundamental system whose initial values are the canonical basis (here $\cos, \sin$) is exactly choosing the columns of $\eu^{tA}$; the evaluation isomorphism is why initial conditions parametrize trajectories — the geometric content of “deterministic dynamics” for linear equations.