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Math Help - [SOLVED] seperable equation

  1. #1
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    [SOLVED] seperable equation

    hey there..
    i've got unsolved question here..

    Question:
    Solve the initial value problem.
    2yy’ = x / √x^2 – 16 ; y(5) = 2

    2yy’ = x / √x^2 – 16
    2y dy/dx = x / √x^2 – 16
    ∫2y dy = ∫ x / √x^2 – 16 dx
    y^2 + C = √x^2 – 16 + C
    y = (√x^2 – 16)^1/2





    my answer was wrong. the answer is y = ((√x^2 – 16) + 1)^1/2
    how am i going to get the correct answer..?
    and.. what is the use of y(5) = 2?
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  2. #2
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    Quote Originally Posted by nameck View Post
    hey there..
    i've got unsolved question here..

    Question:
    Solve the initial value problem.
    2yy’ = x / √x^2 – 16 ; y(5) = 2

    2yy’ = x / √x^2 – 16
    2y dy/dx = x / √x^2 – 16
    ∫2y dy = ∫ x / √x^2 – 16 dx
    y^2 + C = √x^2 – 16 + C
    y = (√x^2 – 16)^1/2





    my answer was wrong. the answer is y = ((√x^2 – 16) + 1)^1/2
    how am i going to get the correct answer..?
    and.. what is the use of y(5) = 2?
    You have gotten it to the step

    \int{2y\,dy} = \int{\frac{x}{\sqrt{x^2 - 16}}\,dx} correctly.

    y^2 + C_1 = \frac{1}{2}\int{2x(x^2 - 16)^{-\frac{1}{2}}\,dx}

    y^2 + C_1 = \frac{1}{2}\int{u^{-\frac{1}{2}}\,du} after making the substitution u = x^2 - 16

    y^2 + C_1 = u^{\frac{1}{2}} + C_2

    y^2 + C_1 = \sqrt{x^2 - 16} + C_2

    y^2 = \sqrt{x^2 - 16} + C, where C = C_2 - C_1


    Now use your initial condition to find C

    y(5) = 2

    so 2^2 = \sqrt{5^2 - 16} + C

    4 = 3 + C

    C = 1.


    Therefore the solution is:

    y^2 = \sqrt{x^2 - 16} + 1

    or if you like

    y = \pm \sqrt{\sqrt{x^2 - 16} + 1}.
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