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

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!

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## 1 Expert Answer

Kris V. answered • 10/14/17

Tutor

5
(36)
Experienced Mathematics, Physics, and Chemistry Tutor

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-cos

^{2}(4y)).From cos(4y) = − (1-16cos

^{4}x)^{½}, the solutiony = ¼cos

^{-1}[− (1-16cos^{4}x)^{½}] satisfies the given initial condition.## Still looking for help? Get the right answer, fast.

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Arturo O.

10/13/17