Question: $cos(t)y''+sin(t)y'-ty=0$
Find the Wronskian of two solutions of the given differential equation without solving the equation.
Solution:
Divide both sides by $cos(t)$ and we get: $y''+ tan(t)y'-\frac{t}{cos(t)}y=0$
By Abel's theorem, we have: $W(y_1, y_2)(t)=ce^{-\int{p(t)}dt}$
$W(y_1, y_2)(t)=ce^{-\int{tan(t)}dt}=ce^{-(-ln|cos(t)|)}$
$W(y_1, y_2)(t)=ce^{ln|cos(t)|}=ccos(t)$
Hence, the Wronskian of any pair of solutions of the given equation is $W(y_1, y_2)(t)=ccos(t)$
