Inactive Tutor answered • 10/11/17
Tutor
New to Wyzant
cos(2y)dy = -sin(2x)dx
Integrate both sides of the equation.
∫cos(2y)dy = -∫sin(2x)dx
∫(1 - 2sin2y)dy = -∫(2sinxcosx)dx
Take it from here.
Note, the integral of the right side of equation must contain the arbitrary constant C. You will use the initial value condition to solve for C, once you have your general solution.
Megan L.
Also, how did you get the new integral from original integral? Using double angle identities?
Report
10/11/17
Inactive Tutor
After reviewing this question, you don't really need the double-angle identities. You can integrate both sides using the substitution method.
∫cos(2y)dy and -∫sin(2x)dx
u = 2y z = 2x
du = 2dy dz = 2dx
(1/2)du = dy (1/2)dz = dx
Then you will have
(1/2)∫cos(u)du = - (1/2)∫sin(z)dz
(1/2)sin(u) = - (1/2)(-cos(z))
(1/2)sin(2y) = (1/2)cos(2x) + C
sin(2y) = cos(2x) + 2C
2y = arcsin(cos(2x) +2C)
y = (1/2)arcsin(cos(2x) +2C)
Report
10/11/17
Megan L.
I plugged in the initial condition, and I got C=1, but when entering the explicit equation with 1 plugged into C, I still get the wrong answer.
Report
10/12/17
Megan L.
10/11/17