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Math Help - Separable Equations

  1. #1
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    Separable Equations

    1. Solve
    dP/dt = P(a-bP) where a and b are constants
    dP/[P(a-bP)] = dt
    rewrote the integral as
    -dP/[P(bP-a)] = dt
    then used this list of integrals [HTML]http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions[/HTML]
    1/a[ln((bP-a)/P)] = t + C
    multiply both sides by a and raise everything to e
    (bP-a)/P = Ce^a^t

    I'd appreciate it if someone could check this for me and tell me what I might have done wrong.



    2. Find a solution that passes through the points (0, 0) and (0, 1)
    x dy/dx = y^2 - y
    1/ (y^2 - y) = 1/x dx
    used partial fractions on left side to get
    [-1/y + 1/(y-1)] dy = 1/x dx
    -ln y + ln(y - 1) = ln x + C
    ln[(y-1)/y] = ln x + C
    (y-1)/y = Cx

    I'm pretty sure I did something wrong, most likely with the partial fractions b/c if you put the given points into my answer it yields 0=0 and DNE so I'd appreciate if someone could tell me what I did wrong.

    Thanks in advance.
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  2. #2
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    Could you show the partial fractions of the expression

    <br />
\displaystyle<br />
\frac{1}{P(bP-a)}<br />

    and then integration here ?
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Naples View Post
    -dP/[P(bP-a)] = dt
    1/a[ln((bP-a)/P)] = t + C
    multiply both sides by a and raise everything to e
    (bP-a)/P = Ce^a^t
    The integral is written wrong. It should be:
    (1/a)[ln(P/(bP-a))] = t + C

    Also, you are raising the arbitrary constant to the e-th power, which is fine. However you have left it as an arbitrary constant in your final answer. (ie. you are calling it "C.") Please note that this new C you are talking about can never be negative, so it is not quite an arbitrary constant. I would suggest stating your solution and adding the comment that C > 0.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Naples View Post
    2. Find a solution that passes through the points (0, 0) and (0, 1)
    x dy/dx = y^2 - y
    1/ (y^2 - y) = 1/x dx
    used partial fractions on left side to get
    [-1/y + 1/(y-1)] dy = 1/x dx
    -ln y + ln(y - 1) = ln x + C
    ln[(y-1)/y] = ln x + C
    (y-1)/y = Cx
    For starters there is no way we can specify a solution of a first order ODE to pass through two arbitrary points. So the problem is automatically over-constrained from the start.

    I did this problem as a Bernoulli equation and got a different solution.
    \displaystyle y(x) = \frac{1}{Ax + 1}

    I did it your way and got your solution. I am not certain why the two forms do not match up. Anyway, my solution can be passed through the point (0, 1), but not (0, 0).

    -Dan

    Wait. They do match up, but they aren't quite in the same form. Both forms can pass through (0, 1), but not (0, 0).
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