Question

initial value problem? Sin2xdx + cos6ydy = 0 , y(pi/2) = pi/6

Answer 1

I subtracted the cos6ydy to the other side and got sin2xdx = -cos6ydy. The I integrated both sides and got: -1/2cos2x +C = -1/6sin6y +C

Answer 2

I had trouble solving in terms of y

Answer 3

and then attempting to find c

Answer 4

\[ Sin2x dx + \cos6y dy = 0\rightarrow \cos 6y dy =- \sin 2x dx\]
\[\int\limits \cos 6y dy =\int\limits- \sin 2x dx\]
\[\frac{ 1 }{ 6 }\sin 6y =\frac{ 1 }{2 }\cos 2x +c\]
now apply initial condition

Answer 5

(pi/6,pi/2)
1/6 sin (6 pi/2) =1/2 cos(2 pi/6 ) +c
0= 1/2 (1/2) +c
so c=-1/4

Answer 6

Okay, so how would I express the equation in terms of y? So far I have sin(6y) = 3cos(2x) -1/4

Answer 7

it is not a function
but if you restrict area
may find y

Answer 8

in terms of y ... so you mean solve for x

just use inverse cos
\[\frac{1}{2}\cos 2x - \frac{1}{4} = \frac{1}{6} \sin 6y\]
\[\cos 2x = \frac{1}{3} \sin 6y + \frac{1}{2}\]
\[x = \frac{\cos^{-1} (\frac{1}{3} \sin 6y + \frac{1}{2})}{2}\]
this will have restricted values for y as said by earlier post

Answer 9

I'm sorry, I meant solve for y in terms of x. So if you solved for y would the equation look like: \[y = (\sin^{-1} (3\cos2x -1/4))/(6) \] ?

Answer 10

yep
wait distribute the 6 .... 3cos(2x) - 3/2 ...

Answer 11

so, \[y=\sin^{-1} (3\cos2x)-3/2\] ?

Answer 12

\[y = \frac{\sin^{-1} (3\cos 2x - \frac{3}{2})}{6}\]

Answer 13

ooh alright. thanks!

Answer 14

\[y=\sin^−1(3\cos2x−\frac{ 3 }{2 }) 6\rightarrow \frac{ -\pi }{ 2 } \le 3\cos2x−\frac{ 3 }{2 } \le \frac{ -\pi }{ 2 } \]

Answer 15

sorry
right side is +pi/2