Question

Find the general solution:

\[(3+y+2y^2\sin^2 x)dx + (x+ 2xy -y\sin 2x)dy=0\]

Answer 1

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

Answer 2

wait wat's 1-cos2x?

Answer 3

it's 2sin^2x......it's exact

Answer 4

wait what?

Answer 5

u kno 1-cos2x = 2sinx*sinx

Answer 6

yeah?

Answer 7

oh i see your point

Answer 8

actually um wateva u call it it's root((1-cos2x)/2) = sinx

Answer 9

so it's exact..

Answer 10

\[\int (3+y+2y^2 \sin^2 x)dx\]
\[3x + xy + y^2(x - \sin x \cos x)\]

Answer 11

then i take derivative of y

Answer 12

\[x + 2yx - 2y\sin x\cos x + g'(y) = x+2xy - y\sin 2x\]
\[g'(y) = 2y\sin x \cos x - y\sin 2x\]
\[g(y) = 2\sin x \cos x - \sin 2x\]
so the G.S. would be \[3x + xy + y^2(x - \sin x \cos x) + 2\sin x \cos x - \sin 2x + C\]
correct?

Answer 13

u kno there's a direct solution for exact equations

Answer 14

really? what?

Answer 15

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

Answer 16

oh that..yeah...im more comfortable with this though

Answer 17

im more comfortable when the solution is one thread instead of inconsistent

Answer 18

ok