I can't understand this fact about the wronskian determinant in differential equations.

Consider a homogeneous second order linear differential equation with constant coefficients.

Taken two solutions $y_1$ and $y_2$ on a interval $I$ and their wronskian determinant $W(x)$ then

$y_1$ and $y_2$ are l.i. $\implies$ $W(x)=0 \forall x \in I$

And that's ok.

My doubts are about this: which is the right implication?

$\exists x_0\in I \mid W(x_0)=0 \implies y_1 $and $y_2$ are l.i.


$\forall x\in I \mid W(x)=0 \implies y_1 $and $y_2$ are l.i.?

Maybe is it true that

$\exists x_0\in I \mid W(x_0)=0 \implies \forall x\in I \mid W(x)=0$

If so I can't understand this last implication.

Can anyone help me?

Thanks in advice


Maybe is it true that
$\exists x_0\in I \mid W(x_0)=0 \implies \forall x\in I \mid W(x)=0$

Yes, that is correct (even for homogeneous second order linear differential equations with non-constant coefficients).

If $y_1, y_2$ are two solutions of the homogeneous second order linear differential equation $$ y'' + p(x) y' + q(x) y = 0 $$ on an interval $I$ then the Wronskian $W(x) = y_1(x) y_2'(x) - y_2(x) y_1'(x)$ satisfies the linear differential equation $$ W'(x) = -p(x) W(x) $$ which has the solution $$ W(x) = W(x_0) \exp \bigl(-\int_{x_0}^x p(t)dt \bigr) $$

It follows that

  • either $W(x) = 0$ for all $x \in I$,
  • or $W(x) \ne 0$ for all $x \in I$.

$(y_1, y_2)$ are linearly dependent in the first case and linearly independent in the second case.

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  • $\begingroup$ hi, can you show how you get $$W(x) = W(x_0) \exp \bigl(-\int_{x_0}^x p(t)dt \bigr)$$ from $$W'(x) = -p(x) W(x)$$? $\endgroup$ – Salihcyilmaz Oct 25 '16 at 11:41
  • $\begingroup$ @Salihcyilmaz: That is a first order linear differential equation, and the solutions can be determined explicitly, see e.g. en.wikipedia.org/wiki/Linear_differential_equation. $\endgroup$ – Martin R Oct 25 '16 at 11:46
  • $\begingroup$ how did you decide arbitrary constant $c_1 = w_0$ and where does limits of integration come from? $\endgroup$ – Salihcyilmaz Oct 25 '16 at 20:49
  • $\begingroup$ @Salihcyilmaz: With $w_0 := W(x_0)$, $W$ is the solution of the initial value problem $w' = -p(x)w, w(x_0) = w_0$. $\endgroup$ – Martin R Oct 25 '16 at 20:51

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