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Math Help - first order ODE

  1. #1
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    first order ODE

    so I have the equation: dx/dt = .8x - .004x^2 t is time in seconds x represents concentration of a salt in solution
    the initial condition being at t=0, x=50
    so I used bernoulli's equation substitution then integrating factor I got:

    x = 1/(.025exp(-.8t) - .005)
    which stops giving realistic values after 2 seconds
    also according the the differential equation the concentration should peak around x=200 but with the solution it doesn't so I'm guessing I did something wrong.

    So what should I have done?
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  2. #2
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    Quote Originally Posted by AsSeenOnTV View Post
    so I have the equation: dx/dt = .8x - .004x^2 t is time in seconds x represents concentration of a salt in solution
    the initial condition being at t=0, x=50
    so I used bernoulli's equation substitution then integrating factor I got:

    x = 1/(.025exp(-.8t) - .005)
    which stops giving realistic values after 2 seconds
    also according the the differential equation the concentration should peak around x=200 but with the solution it doesn't so I'm guessing I did something wrong.

    So what should I have done?
    This equation is separable.
    \int \frac{dx}{.8x-.004x^2}=\int dt
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  3. #3
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    which if I did the integration right I get t(x)= 5/4(ln(x) + ln(.8-.004x)) which is OK I guess except I was supposed to get x in terms of t
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  4. #4
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    Quote Originally Posted by AsSeenOnTV View Post
    which if I did the integration right I get t(x)= 5/4(ln(x) + ln(.8-.004x)) which is OK I guess except I was supposed to get x in terms of t
    By integration, we obtain:
    -\frac{5}{4}ln\bigg(\frac{x-200}{x}\bigg)=t+c\rightarrow ln\bigg(\frac{x-200}{x}\bigg)=-\frac{4t}{5}+c_2\rightarrow e^{ln\big(\frac{x-200}{x}\big)}=e^{\frac{-4t}{5}+c_2}

    Think you can take it from here?
    Last edited by dwsmith; April 29th 2010 at 04:29 PM.
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  5. #5
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    Spoiler:
    \frac{x-200}{x}=C_3e^{\frac{-4t}{5}}

    1-200x^{-1}=C_3e^{\frac{-4t}{5}}

    x(t)=\frac{-200}{C_4e^{\frac{-4t}{5}}-1}
    Last edited by dwsmith; April 29th 2010 at 04:30 PM.
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  6. #6
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    wow you're good at this

    I recalculated using the same method I used the first time and got:

    x(t)= 1/(.005+.015exp(-.8t))

    which gave me a realistic answer
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  7. #7
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    Quote Originally Posted by AsSeenOnTV View Post
    wow you're good at this

    I recalculated using the same method I used the first time and got:

    x(t)= 1/(.005+.015exp(-.8t))

    which gave me a realistic answer
    I don't know what the answer is in decimal form since I usually don't use decimals whenever possible but if I solve for C. I am going to obtain:

    \frac{-200}{C_4-1}=50\rightarrow C_4=-3

    x(t)=\frac{200}{3e^{\frac{-4t}{5}}+1}

    Which is equal to your result if you multiple by 1=\frac{200}{200}
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