calculus

integrate the following
(3y-7x-3)dx+(7y-3x-7)dy=0

asked by waqas
  1. rearrange things a it to get

    (3y-7x-3)+(7y-3x-7)y'=0
    now integrate term by term
    3xy-7/2x^2-3x + 7/2y^2-3xy-7y=0

    the 3xy's go away; multiply by 2/7 to get

    -(x^2+6/7)+(y^2-2y) = 0
    -(x+3/7)^2+(y-1)^2 = 58/49

    an hyperbola opening up and down

    posted by Steve

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