[Math] Suppose that the Wronskian of any 2 solutions of $\frac{d^2y}{dt^2}+p(t)\frac{dy}{dt}+q(t)y=0$ Prove that P(t)=0.

Question

Suppose that the Wronskian of any 2 solutions is constant of $\frac{d^2y}{dt^2}+p(t)\frac{dy}{dt}+q(t)y=0$ Prove that P(t)=0.

So my attempt:

$$W(t)=y_1y_2'-y_1'y_2$$

So what I thought I would do is set W(t) = C, some constant:

$$C=y_1y_2'-y_1'y_2$$
$\frac{C+y_1'y_2}{y_2'}=y_1$ and ${y_1y_2'-C}{y_2}=y_1'$
Then I thought I can substitute these into the original equation and somehow prove p(t)=0

$\frac{d^2y}{dt^2}+p(t)[{y_1y_2'-C}{y_2}]+q(t)\frac{C+y_1'y_2}{y_2'}=0$

I don't think this is quite right though…any thoughts?

Answer 1

First the Wronskian

$$W=y_1y_2'- y_1'y_2$$

has derivative

$$W'= y_1'y_2'+y_1y_2''-y_1''y_2-y_1'y_2' = y_1y_2''-y_1''y_2$$

Since $y_1$ and $y_2$ are solutions to the differential equation, we have

$$y_1''+ p(t)y_1' + q(t)y_1 = 0$$ $$y_2'' + p(t)y_2' + q(t)y_2 = 0$$

Multiplying the first equation by $-y_2$ and the second by $y_1$ and adding gives

$$(y_1y_2''-y_1''y_2) + p(t)(y_1y_2' -y_1'y_2) = 0$$

This can be written as

$$ W' + p(t)W = 0.$$

If $W=c\neq{0}$ then $p(t)c= 0$, imply $p(t)=0$

Note. we have a separable differential equation

$$dW/W = -p(t)dt$$