austinL
  • austinL
Determine (at least approximately) the interval in which the solution is defined. \(\sin(2x)dx+\cos(3y)dy=0,~y(\dfrac{\pi}{2})=\dfrac{\pi}{3}\)
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\cos \left( 3y \right) dy=-\sin \left( 2x \right)dx\] integrating \[\frac{ \sin 3y }{ 3 }=-\frac{ -\cos 2x }{2 }+c \] \[when x=\frac{ \pi }{2 },y=\frac{ \pi }{ 3 }\] \[\frac{ \sin \pi }{ 3 }=\frac{ \cos \pi }{2 }+c,0=\frac{ -1 }{2 }+c ,c=\frac{ 1 }{2 }\] \[\frac{ \sin 3y }{3 }=\frac{ \cos 2x }{2 }+\frac{ 1 }{2 }=\frac{ 1+\cos 2x }{2 }=\frac{ 2\cos ^{2}x }{ 2 }=\cos ^{2}x\] \[\cos ^{2}x is positive for all values .hence x can have an value.\] \[L.H.S. is positive for 0\le 3y \le \pi or 0 \le y \le \frac{ \pi }{3 }\]
austinL
  • austinL
Many thanks @surjithayer
anonymous
  • anonymous
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. |dw:1378396441927:dw| \[ Find the \angle for which cosx=\frac{ 1 }{\sqrt{3} }\] then the interval is x to pi-x

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