Separable Differential Equations Problem

Question

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!

Answer 1

The initial condition y(π/2)=π/4 leads to

4y = π, so sin^-1(4y) is not defined since the domain of sin^-1(.) is [−π/2, π/2]

2x = π, so cos^-1(2x) is defined since the domain of cos^-1(.) is [0, π]

 

From sin(4y) = 2cos(2x)+2, the solution

x = ½cos^-1[½ sin(4y) − 1] satisfies the given initial condition.

 

A solution y=f(x) can be found by replacing sin(4y) with SQRT(1-cos²(4y)).

 

From cos(4y) = − (1-16cos⁴x)^½, the solution

y = ¼cos^-1[− (1-16cos⁴x)^½] satisfies the given initial condition.