Find the Wronskian of two solutions of the given differential equation without solving the
equation.
$$
\cos (t) y^{\prime \prime}+\sin (t) y^{\prime}-t y=0
$$
First, we divide both sides of the equation by $\cos (t):$
$$
y^{\prime \prime}+\tan (t) y^{\prime}-\frac{t}{\cos (t)} y=0
$$
Now the given second-order differential equation has the form:
$$
L[y]=y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0
$$
Noting if we let $p(t)=\tan (t)$ and $q(t)=-\frac{t}{\cos (t)},$ then $p(t)$ is continuous
everywhere except at $\frac{\pi}{2}+k \pi,$ where $k=0,1,2, \ldots$ and $q(t)$ is also continuous
everywhere except at $t=0$.
Therefore, by Abel's Theorem: the Wronskian $W\left[y_{1}, y_{2}\right](t)$ is given by
$$
\begin{aligned} W\left[y_{1}, y_{2}\right](t) &=cexp\left(-\int p(t) d t\right) \\ &=cexp\left(-\int \tan (t) d t\right) \\ &=c e^{\ln |\cos (t)|} \\ &=c\cos (t) \end{aligned}
$$
