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Math Help - Differential equation problem

  1. #1
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    Differential equation problem

    Differential equation problem:

    Question;

    show any function x=x(t) which satisfies xe^{tx} = C is a solution of the differential equation (1+tx)x' = x^2

    This is what I have so far;

    C' = x^2e^{tx} W.R.T. (t)
    x' = t

    so I have;

    C' = (xt * xt)e^{tx}

    -x^2 = c'e^{tx}

    -x^2=x^2 * e^{tx}

    therefore;

    -x^2 = [[xt*xt]e^{tx}]e^{tx}

    then I end up with

    -x^2 = xt(t)e^{2tx}

    -x^2 = xt(x')

    What happens to the e^{2tx} ???
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Differential equation problem

    Your troubles start way before your last line. C is a constant, not a function of x or t. See if that helps.

    -Dan
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  3. #3
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    Re: Differential equation problem

    No it did not help, can you offer another tip to push me in the right direction?

    Thanks
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  4. #4
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    Re: Differential equation problem

    Quote Originally Posted by entrepreneurforum.co.uk View Post
    Differential equation problem:

    Question;

    show any function x=x(t) which satisfies xe^{tx} = C is a solution of the differential equation (1+tx)x' = x^2

    This is what I have so far;

    C' = x^2e^{tx} W.R.T. (t)
    x' = t

    so I have;

    C' = (xt * xt)e^{tx}

    -x^2 = c'e^{tx}

    -x^2=x^2 * e^{tx}

    therefore;

    -x^2 = [[xt*xt]e^{tx}]e^{tx}

    then I end up with

    -x^2 = xt(t)e^{2tx}

    -x^2 = xt(x')

    What happens to the e^{2tx} ???
    \displaystyle \begin{align*} x\,e^{t\,x} &= C \\ \frac{d}{dt} \left( x\, e^{t\, x} \right) &= \frac{d}{dt} \left( C \right) \\ x \, \frac{d}{dt} \left( e^{t\, x} \right) + e^{t\, x} \, \frac{d}{dt} \left( x \right) &= 0 \\ x \left[ e^{t\, x} \, \frac{d}{dt} \left( t\, x \right)  \right] + e^{t\, x} \, \frac{dx}{dt} &= 0 \\ x \, e^{t\, x} \left( t\, \frac{dx}{dt} + x \right) + e^{t\, x} \, \frac{dx}{dt} &= 0 \\ t\, x\, e^{t\, x} \, \frac{dx}{dt} + x^2\, e^{t\, x} + e^{t\, x}\, \frac{dx}{dt} &= 0 \\ e^{t\, x} \left( 1 + t\, x \right) \frac{dx}{dt} + x^2\, e^{t\, x} &= 0 \\ e^{t\, x} \left( 1 + t\,x \right) \frac{dx}{dt} &= -x^2\, e^{t\, x} \\ \left( 1 + t\,x \right) \frac{dx}{dt} &= -x^2 \end{align*}

    I assume this is what your original equation was supposed to be...
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  5. #5
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    Re: Differential equation problem

    Where can I find a good place with a lot of differential equation examples with solutions?





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