$y''+2\tan x y'-y=0$ Question: Use the variation of parameter method to find the general solution of the following differential equation
$$(\cos x) y''+(2\sin x) y'-(\cos x) y =0\;\;\;\;,\;\;\;\;0<x<1$$
My Try:
I think the question is wrong, since the right hand side term is 0, so the particular integral will also be zero. Thus, the general solution will be equal to homogenous solution. So, I think no use of using the variation of parameter formulas since $y_p(x)=0$  always.
I reduced the equation as in the subject or title and then used integrating factor $$y=(\cos x )z$$ to eliminate the term $y'$ but I got another difficult DE as $z''-2\sec^2xz=0$

Please help with any suggestions or do you think question is correct. Is there a way to solve it ?

 A: By trial and error, you can show that one of the fundamental solutions is $y_1(x) = c\sin(x)$.  Now, Abel's Theorem tells us that the Wronskian of solutions is given by
$$W(x) = c\text{exp}\left(-\int 2\tan(x) dx\right) = c\cos^2(x).$$
But, $W$ is also equal to the determinant of the Wronskian matrix:
$$W(x) = (\cos(x))y_2(x)-(\sin(x))(y_2)'(x).$$
So, solve the equation
$$(\sin(x))(y_2)'(x)-(\cos(x))y_2(x) = \cos^2(x),$$
using VOP (or just integrating factor) to find the other fundamental solution, $y_2(x)$.
I find that $y_2(x) = c\sin(x)-x\sin(x)-\cos(x)$.
Then, the general solution would be
$$y(x) = c_1\sin(x)+c_2(x\sin(x)+\cos(x)).$$
Notice that we do not need to use Abel's theorem.  You could also sub in $z(x) = \sin(x)y(x)$ and obtain a first-order equation in the same spirit of what you were trying.  This approach is know as Reduction of Order.
A: Note that $y_1(x)=\sin x$ is one solution of the second order linear ODE
$$y''+2\tan x y'-y=0$$
If $y_1$ is one solution of ODE
$$y''+P(x)y'+Q(x)y=0.$$
Then the other solution $y_2(x)$ is given as
$$y_2=y_1C_2\int \frac{\exp[-\int P(x) dx]}{y_1^2}$$
See Finding one solution of second order DE using another solution using Wronskian
So here in this case
$$y_2=C_2 \sin x \int \frac{\exp[-2\int \tan x dx]}{\sin^2 x} dx=C_2 \sin x \int \cot^2 x ~dx=C_2 \sin x (-x-\cot x)$$ $$=-C_2(x\sin x+ \cos x)$$
So total solution of the ODE (*) is
$$y(x)=C_1\sin x+C_3(x \sin x+\cos x)$$
