Question

Looking for some guidance please, I need to find the initial-value problem for dy/dx = 1+2cos^2x / y. y = 1 when x = 0.
any help is much appricated

Answer 1

the hard part is to integrate the cos^2x

we know that cos 2x = 2cos^2x - 1
so replace the 2cos^2x with cos2x + 1

then
dy/dx = 1+2cos^2x
= 1 + cos2x + 1 = 2 + cos2x

then y = 2x + (1/2)sin2x + c
given: when x=0, y = 1
1 = 0 + sin0 + c
c = 1

y = 2x + (1/2)sin2x + 1