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Math Help - Solve DE of the form dx/dt = f(x).

  1. #1
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    Solve DE of the form dx/dt = f(x).

    The Question:
    Solve the differential equation:
    \frac{dx}{dt}\ =\ -\frac{1}{5}(x\ +\ 1)^{\frac{1}{2}}, given that x\ =\ 80 when t\ =\ 0.
    Express your answer in the form x\ =\ f(t).

    My Attempt:
    Integrate both sides:
    \int\frac{dx}{dt}\ =\ \int{-\frac{1}{5}(x+1)^{\frac{1}{2}}.
    Therefore:
    \int\frac{dx}{\sqrt{x\ +\ 1}}\ dx\ =\ \int{-\frac{1}{5}}.
    So:
    \int\frac{1}{\sqrt{x+1}}\ dx\ =\ -\frac{1}{5}t.

    Let u\ =\ x\ +\ 1.
    \frac{du}{dx}\ =\ 1, so du\ =\ dx.
    Substituting:
    \int{\frac{1}{\sqrt{u}}}}\ du\ =\ -\frac{1}{5}t.
    Simplifying:
    2\sqrt{x\ +\ 1}\ +\ C\ =\ -\frac{1}{5}t.
    -t\ =\ 10\sqrt{x\ +\ 1}\ +\ 5C.
    t\ =\ -10\sqrt{x\ +\ 1}\ -\ 5C..
    Substitute x\ =\ 80, t\ =\ 0:
    0\ =\ -10(9)\ -5C, so C\ =\ 18.

    Therefore:
    t\ =\ -10\sqrt{x\ +\ 1}\ +\ 90..

    However, the answer given is:
    x\ =\ (9\ -\ \frac{1}{10}t)^{2}\ -\ 1.

    Please provide a hint as to where I may have gone wrong.
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  2. #2
    A Plied Mathematician
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    Quote Originally Posted by isx99 View Post
    Express your answer in the form x\ =\ f(t).
    Why don't you do this first, and see if your answer is really wrong?
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  3. #3
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    Thank you for your reply.

    Subtracting 90 from both sides:
    t\ -\ 90\ =\ -10\sqrt{x\ +\ 1}.
    Therefore:
    x\ +\ 1\ =\ \frac{t^2}{100}\ +\ 81.
    Subtracting 1 from both sides:
    x\ =\ \frac{t^2}{100}\ +\ 80.

    But, when I expand the answer given, I get:
    9^2\ -\ 2(9)(\frac{1}{10}t)\ +\ \frac{t^2}{100}\ -\ 1.
    Which equals:
    \frac{t^2}{100}\ -\ \frac{9t}{5}\ +\ 80.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by isx99 View Post
    Subtracting 90 from both sides:
    t\ -\ 90\ =\ -10\sqrt{x\ +\ 1}.
    Therefore:
    x\ +\ 1\ =\ \frac{t^2}{100}\ +\ 81.
    (a + b)^2 = a^2 + 2ab + b^2, not a^2 + b^2.

    -Dan
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