I need to apply the reduction of order into differential equation $$\cos^2xy''-6y=0$$where the first solution is of the form $$y_1=(\beta \tan^2x+1)$$ with $\beta \in R$.
I started solving equation with reduction of order, however I don't know how should I use the knowledge of "$\beta \in R$."
$$y_1=(\beta \tan^2x+1)$$ $$y_2=u\cdot y_1=u\cdot(\beta \tan^2x+1)$$ $$y_2'=(u\cdot y_1)'= \dot u(\beta \tan^2x+1)+u(2\beta \cdot \tan x\cdot \sec^2x)$$
$$y_2''= (\dot u(\beta \tan^2x+1)+u(2\beta \cdot \tan x\cdot \sec^2x))'= \ddot u (\beta \tan^2x+1)+ \dot u(4\beta \tan x \cdot \sec^2x) +u(2\beta \sec^4x+4\beta \tan^2x \sec^2x) $$
After inserting $y_2, y_2',y_2''$ into $\cos^2xy''-6y=0$ I got the form where any of the element of the equation want to reduce. Therefore, I would like to ask should I assume do with $\beta$, can I assume that $\beta$ equals for instance 1?