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

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

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

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

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

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

    But, when I expand the answer given, I get:
    $\displaystyle 9^2\ -\ 2(9)(\frac{1}{10}t)\ +\ \frac{t^2}{100}\ -\ 1.$
    Which equals:
    $\displaystyle \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:
    $\displaystyle t\ -\ 90\ =\ -10\sqrt{x\ +\ 1}.$
    Therefore:
    $\displaystyle x\ +\ 1\ =\ \frac{t^2}{100}\ +\ 81.$
    $\displaystyle (a + b)^2 = a^2 + 2ab + b^2$, not $\displaystyle a^2 + b^2$.

    -Dan
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