Megan L.

asked • 10/11/17

Ordinary Differential Equations Problem

Find the solution of the given initial value problem in explicit form.
sin(2x)dx+cos(4y)dy=0, y(π/2)=π/4
 
I solve to get the general equation:
y=[arcsin(2cos(2x)+C)]/4 but when I plug in the initial values I get:
y=[arcsin(2cos(2x)+2)]/4 with C=2. This is the wrong answer though, and I think it's because of a domain problem, but I'm not really sure what the domain problem is and how to fix it.

1 Expert Answer

By:

Michael J. answered • 10/11/17

Tutor
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Effective High School STEM Tutor & CUNY Math Peer Leader

Megan L.

I get
cosysiny = cos^2(x) + C
Then I solve for C and get 1/2.
 
I tried solving for y, but I don't think I'm doing it correctly:
 
y=sqrt[(cot(x)+sin^-1(cos^-1(1/2))]
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10/11/17

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

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