Q3 TUT 0701

[Victor Ivrii]

Find the Wronskian of two solutions of the given differential equation without solving the equation.
$$
\cos(t)y''+\sin(t)y'-ty=0.
$$

[Tzu-Ching Yen]

Reformat equation into form $y'' + p(t)y' + q(t) = 0$
We can see that
$p(t) = \frac{\sin(t)}{\cos(t)}$
By the equation for wronskian
$$W(t) = \exp \bigl(\int p(t) dt\bigr) = \exp \bigl(\int \frac{\sin(t)}{\cos(t)} dt\bigr) = \exp (-\ln(\cos(t))+\ln (c_0)) = \frac{c_0}{\cos(t)}$$
where $c_0$ depends on choice of $y_2$ and $y_2$

[Michael Poon]

Thomas, your solution seems mostly correct, but I think you are missing something small. If I remember correctly, Abel's theorem has a negative factor you forgot.

Instead, the solution should be:

$$W(t)=e^{∫-p(t)dt}=e^{∫-\tan(t)dt}=e^{\ln(\cos(t))+c}=c(\cos(t))$$, for some constant c.

[Wei Cui]

First, we divide both sides of the equation by $\cos(t)$ and we get:
                                                                                                $y''+\tan(t)y'-\frac{t}{\cos(t)}y = 0$

Then the equation in the form of $L(y) = y'' +p(t)y' +q(t)y=0$, then in this case $p(t) = tan(t)$
According to Abel's Theorem, then the Wronskian                         $W(y_1,y_2)(t) = Ce^{-\int \tan(t)dt}$
                                                                                                                      $=Ce^{\int \frac{1}{\cos(t)}d\cos(t)}$
                                                                                                                      $= Ce^{\ln(\cos(t))}$
                                                                                                                      $= C\cos(t)$                   


Therefore, the solution is $W(y_1,y_2)(t) = C\cos(t)$.