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Find the general solution: \[(3+y+2y^2\sin^2 x)dx + (x+ 2xy -y\sin 2x)dy=0\]

Mathematics
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\[\frac{\partial M}{\partial y} = 1 + 4y\sin^2 x\] \[\frac{\partial N}{\partial x} = 1 + 2y - 2y\cos 2x\] so i guess tis isnt exact? or did i do something wrong?
wait wat's 1-cos2x?
it's 2sin^2x......it's exact

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Other answers:

wait what?
u kno 1-cos2x = 2sinx*sinx
yeah?
oh i see your point
actually um wateva u call it it's root((1-cos2x)/2) = sinx
so it's exact..
\[\int (3+y+2y^2 \sin^2 x)dx\] \[3x + xy + y^2(x - \sin x \cos x)\]
then i take derivative of y
\[x + 2yx - 2y\sin x\cos x + g'(y) = x+2xy - y\sin 2x\] \[g'(y) = 2y\sin x \cos x - y\sin 2x\] \[g(y) = 2\sin x \cos x - \sin 2x\] so the G.S. would be \[3x + xy + y^2(x - \sin x \cos x) + 2\sin x \cos x - \sin 2x + C\] correct?
u kno there's a direct solution for exact equations
really? what?
i dont remember xactly but it's smthin llike integrate m(x,y) keeping y/x as a constant = integrate n(x,y) not containing terms of x/y + c
oh that..yeah...im more comfortable with this though
im more comfortable when the solution is one thread instead of inconsistent
ok

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