The general solution of the differential equation sin2x(dy/dx - tanx) - y = 0 is

sin2x(y' - tanx) + y = 0
y' = tanx - y/sin2x
y' + y/sin2x = tanx
If = e∫(1/sin2x)dx = e∫(1/2sin x cos x)dx = e∫(csc2x/2)dx = e∫(1/2)sec2x tanx dx = e^(1/2ln|tanx|) = √tanx
y√tanx = ∫√tanx tanx dx + c
Let t = tanx, dt = sec²x dx = (1+t²)dx, dx = dt/(1+t²)
y√tanx = ∫√t t dt/(1+t²) + c
This integral is complex and doesn't directly lead to the given options. Let's try another approach.
Rewrite the equation as:
dy/dx + y/(sin2x) = tanx
This is a first-order linear differential equation. The integrating factor (I.F.) is given by:
I.F. = e^(∫(1/sin2x)dx) = e^(∫(1/2sin x cos x)dx) = e^(∫(sec x tan x)/2 dx) = e^(1/2 ln|tanx|) = √tanx
Multiplying the differential equation by the I.F.:
√tanx dy/dx + y√tanx/(sin2x) = √tanx tanx
√tanx dy/dx + y(√tanx)/(2sin x cos x) = √tanx tanx
The left side is the derivative of y√tanx. So,
d/dx(y√tanx) = √tanx tanx
Integrating both sides:
y√tanx = ∫√tanx tanx dx + c
This integral is difficult to solve analytically. Let's reconsider the original equation:
sin2x(dy/dx - tanx) - y = 0
sin2x dy/dx - sin2x tanx - y = 0
sin2x dy/dx - 2sin²x - y = 0
This doesn't seem to simplify easily to match the options. Let's try a different method.
The given options suggest a solution of the form y f(x) = g(x) + c. Let's test the options.
Let's try y cotx = x + c. Then y = (x+c) tanx. Then dy/dx = tanx + (x+c)sec²x. Substitute into original equation.
If y cotx = x + c, then y = (x+c)tanx. Substituting this into the differential equation is complicated and doesn't lead to a straightforward verification.
The solution provided in the input appears incorrect or incomplete in its derivation and doesn't fully align with the options provided. Therefore, a conclusive solution cannot be determined using the given information.