Megan L.

asked • 10/12/17

Separable 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:
(1/4)sin(4y)=(1/2)cos(2x)+C then I plug in the initial values to get C=1/2
 
then eventually I get 
sin(4y)=2cos(2x)+2
 
From here, I've been informed that since I will need to take arcsine to solve for the explicit formula, I have to use a identity since there is a domain problem with arcsine. 
 
Without taking into account the arcsine identity/domain, I simply get:
 
y=(1/4)arcsin(2cos(2x)+2) but this is NOT the answer. Please help!

Arturo O.

Megan,
 
I worked this problem and got the same solution as you.  What was the identity that they suggested?  The domain restriction would be that the argument of the arcsine must be between -1 and 1, but I do not think that affects the form of the solution.  Maybe your solution is equivalent to the "book" solution, with the book solution in another form?  Do you happen to have the book solution?  If so, please add it to the problem or post it in a comment.
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10/13/17

Megan L.

I don't have the solution since it's online homework and I know my answer is wrong since I tried submitting it and it still says it's incorrect.
 
This is the explanation my professor gave:
Keep in mind that arcsine is the inverse of sine only on a restricted domain. The sine function is not one-to-one, and as such does not have an inverse. In order to define an inverse for the sine function, we restrict its domain to the interval [−π2,π2]. In this case, you are looking at sin(2x) when x=π2, which means you are evaluating sin at 2x=2⋅π2=π. That means that 2x is outside of the range of arcsin. You need to make some adjustment to your answer to account for this fact.
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10/13/17

1 Expert Answer

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Kris V. answered • 10/14/17

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