Michael J. answered • 10/11/17
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Effective High School STEM Tutor & CUNY Math Peer Leader
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?
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10/11/17
Michael J.
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)
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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.
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10/12/17
Megan L.
10/11/17