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Math Help - Antidifferentiation: -Find equation of curve

  1. #1
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    Antidifferentiation: -Find equation of curve

    The points (-1,3) and (0,2) are on a curve, and at any point (x,y) on the curve d^2y/dx^2 = 2 - 4x. Find an equation of the curve.
    Hint: Let d^2y/dx^2 = dy'/dx' and obtain an equation involving y', x and an arbitrary constant C1. From this equation obtain another equation involving y, x, C1, and C2. Compute C1 and C2 from the conditions:

    Answer from book: 3y = -2x^3 + 3x^2 +2x + 6

    THanks so much!
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    The points (-1,3) and (0,2) are on a curve, and at any point (x,y) on the curve d^2y/dx^2 = 2 - 4x. Find an equation of the curve.
    Hint: Let d^2y/dx^2 = dy'/dx' and obtain an equation involving y', x and an arbitrary constant C1. From this equation obtain another equation involving y, x, C1, and C2. Compute C1 and C2 from the conditions:

    Answer from book: 3y = -2x^3 + 3x^2 +2x + 6...

    Hello,

    dy/dx = ∫(dy/dx)dx therefore:

    dy/dx = ∫(2-4x)dx = 2x - 2x +C

    y = ∫(dy/dx)dx therefore:

    y = ∫(2x - 2x + C)dx = x - 2/3*x + Cx + D

    Plug in the coordinates of the points you know:

    (-1, 3): 1 + 2/3 - C + D = 3
    (0, 2): 0 - 0 + 0 + D = 2 Plug in this value into the first equation:

    1 + 2/3 - C + 2 = 3 and solve for C. C = 2/3

    the complete equation is:

    y = -2/3*x + x + 2/3*x + 2

    EB



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