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Math Help - trouble with the constant term of a separable DE

  1. #1
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    trouble with the constant term of a separable DE

    I have a really simple separable DE from an input output problem and I'm having trouble with the constant. (I think)

    \frac{dx}{dt}=0.6-\frac{3x(t)}{500} and x(0)=0

    \frac{dx}{dt}=\frac{300-3x(t)}{500}

    \frac{dx}{300-3x(t)}=\frac{dt}{500}

    \int\frac{1}{300-3x(t)}dx(t)=\int\frac{1}{500}dt

    \int\frac{1}{300-3x(t)}dx(t)=\frac{t}{500}+c let u=300-3x(t) then du=0-3dx(t)

    \frac{-1}{3}\int\frac{1}{u}du(t)=ln|u|=ln|300-3x(t)|

    \frac{-1}{3}ln|300-3x(t)|=\frac{t}{500}+c

    ln|300-3x(t)|=-3\frac{t}{500}+c

    So I'm guessing that my problem starts here. I know I have to exponentiate both sides to solve for the dependant variable. And I'm assuming that after exponentiation the equation looks like this;

    300-3x(t)=e^{-3\frac{t}{500}+c}

    x(t)=100-\frac{-1}{3}e^{-3\frac{t}{500}+c}

    Even if I were to evaluate the problem with the initial value provided this doesn't seem to match the provided answer, x(t)=100(1-e^{-3t/500}), at all.
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  2. #2
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    Re: trouble with the constant term of a separable DE

    Quote Originally Posted by bkbowser View Post
    300-3x(t)=e^{-3\frac{t}{500}+c}

    x(t)=100-\frac{-1}{3}e^{-3\frac{t}{500}+c}

    Even if I were to evaluate the problem with the initial value provided this doesn't seem to match the provided answer, x(t)=100(1-e^{-3t/500}), at all.
    You have too many negative signs in the second equation. Let's try it out with the negative sign corrected:

    x(t)=100-\frac{1}{3}e^{-3t/500+c}

    x(0) = 100 - \dfrac{1}{3}e^{0+c} = 0

    e^c = 300

    Going back to the original equation:

    x(t)=100-\frac{1}{3}e^{-3t/500+c} = 100 - \frac{1}{3}e^{-3t/500}e^c = 100-100e^{-3t/500}

    Looks like you got the right answer to me.
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