Wronskian ,Second order linear equation

(t-5)y' ' + 6y' + sec^2(t) y = 0

what is the general form of the wronksian , W(y1,y2)(t)

A) Ce^6t
B)Ce^-6t
C)C(t-5)^6
D)C/(t-5)^6

any form of help would be much appreciated

Very clever question Shark. You have shown me a question I had not seen before, so it's quite interesting. The answer is D.
Let u=u(t) and v=v(t) be two linearly independent solutions of the equation above. Let W=W(t) denote the wronskian.Then W=uv'-vu'. We can use the product rule to find W'. After doing so and simplifying we obtain W'=uv''-vu''. Now we know (t-5)v''+ 6v'+ sec^2t v =0 so that v'' = (6v' + sec^2t v) /(5-t). Similarly u'' = (6u' + sec^2 t u)/ (5-t). Then
W ' = u(6v' + sec^2t v) /(5-t) - v(6u' + sec^2 t u)/ (5-t)
This simplifies to W ' = -6(uv'-vu')/(t-5) = -6W/(t-5)
so that W'/W = -6/(t-5)
Integrating we have ln|W| = -6ln|t-5| + D
so that |W| = e^D * e^(ln|t-5|^(-6)) = C * |t-5|^(-6) = C/ |t-5|^6 = C/ (t-5)^6
We can drop the absolute value around W since (t-5)^6>=0 if we specify that C>=0.
Then W = C/(t-5)^6 where C is an constant >=0.