# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-3 => Topic started by: christine on October 11, 2019, 02:00:10 PM

Title: tut0602 quiz 3 solution
Post by: christine on October 11, 2019, 02:00:10 PM
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 $W(y_1, y_2)(t)=ccos(t)$