Explicit solution of -2x^2y+y^2=1; 2xydx+(x^2-y)dy=0?

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How would I go about solving this??? I did find the second problem is an implicit solution of the firsto ne
Update : I'm finding explicit solution, not implicit. Implicit solution is ...show more
Update 2: again please help me out for the explicit solution...not the implicit...i ...show more
Best Answer
-2x^2y + y^2 = 1

do implicit derivation,

-4xy - 2x^2 (dy/dx) + 2y (dy/dx) = 0

-2xy dx - (x^2 - y) dy = 0

or

2xy dx + (x^2 - y) dy = 0
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  • Cherry answered 4 years ago
    2xydx+x^2dy-ydy=0
    on integration

    2y.x^2/2 + x^2y - y^2/2 = 0
    => 2 x^y - y^2/2 = 0
    => 4x^2y - y^2 = 0
    => 4 x^2 - y = 0
    => x^2 = y /4
    putting this value in 1st giving equetion
    -2. y/4.y + y^2 = 1
    => y^2/2 = 1
    => y = 1/ sqrt 2
    so x^2 = 1/ 4 sqrt 2
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  • InF!N!ty answered 4 years ago
    Given -2x^2y+y^2=1

    Differentiating Both Sides

    -2(2xy + x^2dy/dx) + 2ydy/dx = 0

    -4xy - 2x^2dy/dx + 2ydy/dx = 0

    - 2x^2dy/dx + 2ydy/dx = 4xy

    dy/dx = 4xy/(2y - 2x^2) = 2xy/(y-x^2)

    Equation 2) 2xydx+(x^2-y)dy=0

    Checking For dy/dx

    2xydx = -(x^2-y)dy

    dy/dx = 2xy/(y-x^2) = Previously Obtained dy/dx

    Hence, Second Equation Is The Implicit Solution Of The First
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