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Ordinary differentiation equations (ODE)

  1. Mar 24, 2010 #1
    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 = √x^2 – 16
    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?
     
  2. jcsd
  3. Mar 24, 2010 #2

    Mark44

    Staff: Mentor

    In the step above, where's your constant of integration?
     
  4. Mar 24, 2010 #3
    ok...
    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
     
  5. Mar 24, 2010 #4

    Mark44

    Staff: Mentor

    It wouldn't be the same constant on both sides. If it were, you could subtract C from both sides to get back to what you had.
    You could do this:
    y^2 + A = √(x^2 – 16) + B

    Note that I added parentheses, which are necessary here.
    Or you could just put the constant on one side as

    y^2 = √(x^2 – 16) + C, where C = B - A. This is how it's usually done.

    Now use your initial condition, y(5) = 2, to find C.
     
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