Find the Wronskian of the second-order homogeneous differential equation

Question

Problem:

Show that if $p$ is differentiable and $p(t)>0$, then the Wronskian $W(t)$ of two solutions of two solutions of $[p(t)y']' +q(t) y=0$ is $W(t)=c/p(t)$, where $c$ is a constant.

Consider the second-order homogeneous equation \begin{equation} y''+ p(t) y'+q(t)=0. \end{equation}

As I know, the existence of two solutions is ensured whenever $p(t)$ and $q(t)$ are continuous over some open interval. However, in the problem, we don't know the continuity of $p'(t)$ and $q(t)$. I'm confused whether the Wronskian can be calculated without the existence of solutions. Any help is appreciated! Thank you!

Answer 1

Hint: Since $p$ is differentiable and is different from zero, rewrite the differential equation as : $$y''+\dfrac {p'}py'+\dfrac {q}{p}y=0$$ Use Abel's identity : $$W=c\exp \int -\dfrac {p'}{p}dt$$ You can also easily prove that: $$W'=-\dfrac {p'}pW$$ where $W$ is the wronskian.

Answer 2

If you have two solutions, then for $W(t)=W[y_1,y_2](t)$, as we know the desired result, consider \begin{align} \frac{d}{dt}(p(t)W(t))&=\frac{d}{dt}\det\pmatrix{y_1(t)&y_2(t)\\py_1'(t)&py_2'(t)} \\ &=\det\pmatrix{y_1'(t)&y_2'(t)\\p(t)y_1'(t)&p(t)y_2'(t)}+\det\pmatrix{y_1(t)&y_2(t)\\(p(t)y_1'(t))'&(p(t)y_2'(t))'} \\ &=0+\det\pmatrix{y_1(t)&y_2(t)\\-q(t)y_1(t)&-q(t)y_2(t)} \\ &=0 \end{align} Thus $p(t)W(t)=c$ is constant.

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For the existence of solutions use $v=py'$ as the second state component in the first-order system \begin{align} y'(t)&=\frac1{p(t)}v(t)\\ v'(t)&=-q(t)y(t) \end{align} So for the existence of solutions it is sufficient that the positive function $p$ is continuous.