Question

Determine (at least approximately) the interval in which the solution is defined.

\(\sin(2x)dx+\cos(3y)dy=0,~y(\dfrac{\pi}{2})=\dfrac{\pi}{3}\)

Answer 1

\[\cos \left( 3y \right) dy=-\sin \left( 2x \right)dx\]
integrating
\[\frac{ \sin 3y }{ 3 }=-\frac{ -\cos 2x }{2 }+c \]
\[when x=\frac{ \pi }{2 },y=\frac{ \pi }{ 3 }\]
\[\frac{ \sin \pi }{ 3 }=\frac{ \cos \pi }{2 }+c,0=\frac{ -1 }{2 }+c ,c=\frac{ 1 }{2 }\]
\[\frac{ \sin 3y }{3 }=\frac{ \cos 2x }{2 }+\frac{ 1 }{2 }=\frac{ 1+\cos 2x }{2 }=\frac{ 2\cos ^{2}x }{ 2 }=\cos ^{2}x\]
\[\cos ^{2}x is positive for all values .hence x can have an value.\]
\[L.H.S. is positive for 0\le 3y \le \pi or 0 \le y \le \frac{ \pi }{3 }\]

Answer 2

Many thanks @surjithayer

Answer 3

site crashed so I am completing it now.
addition
\[\left| \sin 3y \right|\le 1\]
\[\cos ^{2}x \le \frac{ 1 }{ 3 },\left| \cos x \right|\le \frac{ 1 }{ \sqrt{3} },\frac{ -1 }{ \sqrt{3} }\le \cos x \le \frac{ 1 }{ \sqrt{3} }\]
\[If x=\theta ,then the interval is \theta \le x \le \pi -\theta \]
I will show with a diagram.


\[ Find the \angle for which cosx=\frac{ 1 }{\sqrt{3} }\]
then the interval is x to pi-x