give the solution of the initial-value problem

Question

dy/dx=(1+2cos^2(x))/y, (y > 0), y= 1 when x = 0.
Thanks.

Answer 1

Is it an answer on my question?

2cos^2(x)=1+cos(2x)
ydy=(2+cos(2x))dx integrating
y^2/2=2x+sin(2x)/2+C Find C
1/2=C

y=sqrt(4x+sin(2x)+1)

Answer 2

Thats right.
Thank you.