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Thread: 1st order Separable Equation

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    1st order Separable Equation

    $\displaystyle dx/dt-x^3=x$
    $\displaystyle dx/dt=(x^3+x)$
    $\displaystyle dt=(x^3+x)^{-1}dx$

    So I haven't been able to figure out how to do the integral.

    I don't think I can do fraction decomposition. At least when I try I don't get anything that makes a whole lot of sense.

    I've tried integration by parts and I don't see any thing that looks remotely like the answer. Which is $\displaystyle x= \pm(Ce^{2t}/(1-Ce^{2t}))^{1/2} ,C\geq0$
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    Re: 1st order Separable Equation

    Quote Originally Posted by bkbowser View Post
    $\displaystyle dx/dt-x^3=x$
    $\displaystyle dx/dt=(x^3+x)$
    $\displaystyle dt=(x^3+x)^{-1}dx$

    So I haven't been able to figure out how to do the integral.

    I don't think I can do fraction decomposition. At least when I try I don't get anything that makes a whole lot of sense.

    I've tried integration by parts and I don't see any thing that looks remotely like the answer. Which is $\displaystyle x= \pm(Ce^{2t}/(1-Ce^{2t}))^{1/2} ,C\geq0$
    $\displaystyle \frac{1}{x^3 - x} = \frac{1}{x(x + 1)(x - 1)} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x - 1}$

    I get the equation
    $\displaystyle A(x + 1)(x - 1) + Bx(x - 1) + Cx(x + 1) = 1$

    Thus (after some work)
    $\displaystyle A + B + C = 0$

    $\displaystyle -B + C = 0$

    $\displaystyle -A = 1$

    Can you solve it from there?

    -Dan
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    Re: 1st order Separable Equation

    AH, OK. I see it now. Thanks!
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    Re: 1st order Separable Equation

    Quote Originally Posted by topsquark View Post
    $\displaystyle \frac{1}{x^3 - x} = \frac{1}{x(x + 1)(x - 1)} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x - 1}$

    I get the equation
    $\displaystyle A(x + 1)(x - 1) + Bx(x - 1) + Cx(x + 1) = 1$

    Thus (after some work)
    $\displaystyle A + B + C = 0$

    $\displaystyle -B + C = 0$

    $\displaystyle -A = 1$

    Can you solve it from there?

    -Dan
    There is a mistake here. It is $\displaystyle (x^3+x)^{-1}$ it is not $\displaystyle (x^3-x)^{-1}$.

    Fraction decomposition on $\displaystyle (x^3+x)^{-1}$ leads me to $\displaystyle Ax^2+Bx+A=1$ so A has to both equal 0 and 1 which is impossible.


    Can I instead take the log of each side as so;

    $\displaystyle ln dt = ln (x^3+x)^{-1}$

    $\displaystyle ln dt = -1*ln (x^3+x)$

    exponentiation;
    $\displaystyle e^{ln dt}= -1(x^3+x)$

    and then integrating;

    $\displaystyle t= -1\int(x^3+x)dx$

    This doesn't appear to lead to an appropriate answer either.
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    Forum Admin topsquark's Avatar
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    Re: 1st order Separable Equation

    Quote Originally Posted by bkbowser View Post
    There is a mistake here. It is $\displaystyle (x^3+x)^{-1}$ it is not $\displaystyle (x^3-x)^{-1}$.

    Fraction decomposition on $\displaystyle (x^3+x)^{-1}$ leads me to $\displaystyle Ax^2+Bx+A=1$ so A has to both equal 0 and 1 which is impossible.


    Can I instead take the log of each side as so;

    $\displaystyle ln dt = ln (x^3+x)^{-1}$

    $\displaystyle ln dt = -1*ln (x^3+x)$

    exponentiation;
    $\displaystyle e^{ln dt}= -1(x^3+x)$

    and then integrating;

    $\displaystyle t= -1\int(x^3+x)dx$

    This doesn't appear to lead to an appropriate answer either.
    Whoops! Sorry about that. In this case we get:
    $\displaystyle \frac{1}{x^3 + x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$

    So
    $\displaystyle 1 = A(x^2 + 1) + (Bx + C)x$

    etc.

    The logarithm argument isn't going to work. What, for example, is ln(dt)? The only way to get rid of the dt and dx is to integrate over them.

    -Dan
    Last edited by topsquark; Sep 10th 2013 at 09:15 AM.
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    Re: 1st order Separable Equation

    $\displaystyle Ax^2+A+Bx^2+Cx=x^2(A+B)+Cx+A$

    so A=1 and B=-1 while C=0

    $\displaystyle \int x^{-1} dx + \int -1x(x^{2}+1)^-1$

    $\displaystyle ln|x|+ -1*\int \frac{1}{2}u^{-1} du$

    $\displaystyle t= ln|x|+ - \frac{1}{2} ln|x^2+1|+c$

    The text asks me to "solve the equation" which I take to mean solve for the dependent variable "x"

    $\displaystyle e^t= |x| + -\frac{1}{2} |x^2+1| +e^c$
    $\displaystyle e^t+\frac{1}{2}-e^c=x-x^2$ Provided $\displaystyle x\geq0$

    This still looks pretty wrong though.
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    Re: 1st order Separable Equation

    Quote Originally Posted by bkbowser View Post
    $\displaystyle Ax^2+A+Bx^2+Cx=x^2(A+B)+Cx+A$

    so A=1 and B=-1 while C=0

    $\displaystyle \int x^{-1} dx + \int -1x(x^{2}+1)^-1$

    $\displaystyle ln|x|+ -1*\int \frac{1}{2}u^{-1} du$

    $\displaystyle t= ln|x|+ - \frac{1}{2} ln|x^2+1|+c$
    Try this: Let c = ln|A|. Then your equation becomes:
    $\displaystyle t = ln|x| - \frac{1}{2} ln|x^2+1| + ln|A|$

    "Collapsing" the logarithms:
    $\displaystyle t = ln \left | \frac{Ax}{(x^2 + 1)^{1/2}} \right | $

    Now how does it turn out?

    -Dan

    Addendum: This is also a Bernoulli equation. The solution is a bit simpler doing it that way.
    Last edited by topsquark; Sep 10th 2013 at 02:57 PM.
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