Question

Diff Eq dy/dx= (ycos(x))/(1+y^2) initial condition is y(0)=1 work below ...?

Answer 1

dy/dx = (ycos(x))/(1 + y²)
(1 + y²)/y dy = cos(x) dx
(1/y + y) dy = cos(x) dx
Integrating:
ln(y) + y²/2 = sin(x) + c
ln(1) + 1/2 = sin(0) + c
c = 1/2
Thus,
ln(y) + y²/2 = sin(x) + 1/2