Wronskian: geometry of the coefficients

The geometric point is about the coefficients ${𝑐}_1,{𝑐}_2,{𝑐}_3$, not about the functions themselves.

Suppose you are testing whether three functions ${𝑦}_1,{𝑦}_2,{𝑦}_3$ are linearly independent. You assume

and you want to know whether this forces

Now imagine evaluating that equation at just one point ${𝑥}_0$. You get

At this point, the numbers

are just fixed scalars. So the equation has the form

This is one linear equation in the three unknowns ${𝑐}_1,{𝑐}_2,{𝑐}_3$.

Geometrically, all triples

that satisfy this equation form a plane through the origin in coefficient space.

For example, suppose at ${𝑥}_0$,

Then the condition becomes

There are infinitely many nonzero solutions. For instance,

works because

And

also works because

So from that single equation alone, you cannot conclude that

That’s the issue.

What you really need is enough independent equations to pin down the coefficient vector

For three unknown coefficients, you ideally want three independent linear constraints.

The Wronskian creates them by differentiating:

Now evaluate all three at the same ${𝑥}_0$:

Now you have three planes in $\left({𝑐}_1,{𝑐}_2,{𝑐}_3\right)$-space.

If those three planes are positioned independently, their only common intersection is the origin:

That is exactly what

is telling you.

So the geometric story is:

for three unknown coefficients.

That is why evaluating the base equation at a single ${𝑥}_0$ is not enough: it gives you only one constraint on three unknown constants.

Key idea: It’s kind of like Taylor expansion, but bounded because the differential equation is a constraint on the highest order derivative (in terms of the lower order derivatives). For example, in an equation like ${𝑦}^{″}+𝑦=0$, the second order derivative of $𝑦$ is bound the by the equation and is not free to assume any form.