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lgbasallote

  • 2 years ago

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

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  1. lgbasallote
    • 2 years ago
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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?

  2. A.Avinash_Goutham
    • 2 years ago
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    wait wat's 1-cos2x?

  3. A.Avinash_Goutham
    • 2 years ago
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    it's 2sin^2x......it's exact

  4. lgbasallote
    • 2 years ago
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    wait what?

  5. A.Avinash_Goutham
    • 2 years ago
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    u kno 1-cos2x = 2sinx*sinx

  6. lgbasallote
    • 2 years ago
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    yeah?

  7. lgbasallote
    • 2 years ago
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    oh i see your point

  8. A.Avinash_Goutham
    • 2 years ago
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    actually um wateva u call it it's root((1-cos2x)/2) = sinx

  9. lgbasallote
    • 2 years ago
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    so it's exact..

  10. lgbasallote
    • 2 years ago
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    \[\int (3+y+2y^2 \sin^2 x)dx\] \[3x + xy + y^2(x - \sin x \cos x)\]

  11. lgbasallote
    • 2 years ago
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    then i take derivative of y

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

  13. A.Avinash_Goutham
    • 2 years ago
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    u kno there's a direct solution for exact equations

  14. lgbasallote
    • 2 years ago
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    really? what?

  15. A.Avinash_Goutham
    • 2 years ago
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    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

  16. lgbasallote
    • 2 years ago
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    oh that..yeah...im more comfortable with this though

  17. lgbasallote
    • 2 years ago
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    im more comfortable when the solution is one thread instead of inconsistent

  18. A.Avinash_Goutham
    • 2 years ago
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    ok

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