Posted by Claire on .
Looking for some guidance please, I need to find the initialvalue problem for dy/dx = 1+2cos^2x / y. y = 1 when x = 0.
any help is much appricated

maths 
Reiny,
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