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Math Help - 1st order Separable Equation

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    1st order Separable Equation

    dx/dt-x^3=x
    dx/dt=(x^3+x)
    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 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
    dx/dt-x^3=x
    dx/dt=(x^3+x)
    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 x= \pm(Ce^{2t}/(1-Ce^{2t}))^{1/2} ,C\geq0
    \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
    A(x + 1)(x - 1) + Bx(x - 1) + Cx(x + 1) = 1

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

    -B + C = 0

    -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
    \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
    A(x + 1)(x - 1) + Bx(x - 1) + Cx(x + 1) = 1

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

    -B + C = 0

    -A = 1

    Can you solve it from there?

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

    Fraction decomposition on (x^3+x)^{-1} leads me to 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;

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

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

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

    and then integrating;

    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 (x^3+x)^{-1} it is not (x^3-x)^{-1}.

    Fraction decomposition on (x^3+x)^{-1} leads me to 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;

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

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

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

    and then integrating;

    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:
    \frac{1}{x^3 + x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}

    So
    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; September 10th 2013 at 09:15 AM.
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    Re: 1st order Separable Equation

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

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

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

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

    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"

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

    This still looks pretty wrong though.
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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
    Ax^2+A+Bx^2+Cx=x^2(A+B)+Cx+A

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

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

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

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

    "Collapsing" the logarithms:
    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; September 10th 2013 at 02:57 PM.
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