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Math Help - Separation of Variables

  1. #1
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    Separation of Variables

    Hello, I'm a little stuck on this problem:

    Find a solution to the differential equation subject to the initial conditions.

    dz

    dt = tez , through the origin.




    Here is how I tried to solve the problem:


    dz

    dt = tez

    multiply both sides by dt, and then divide both sides by ez to get like variables with like variables:

    dz

    ez = tdt



    Take the integral of both sides:

    (integral sign) 1/ez dz = (integral sign) t dt

    substitution
    w = ez
    dw = ez dz

    (integral sign)1/w dz = (integral sign) t dt

    ln|w| = t2/2 + C

    ln|ez| =
    t2/2 + C

    exponentiate both sides to get rid of the natural log:
    eln|ez| = et2/2 + C

    the e and the natural log cancel out
    ez = e(t^2)ec

    ez = Ce(t^2) ----(where C is just a constant)



    Now this is where I am stuck. Have I done it correctly so far? What do I need to do next? Please help. Thank you.
    Last edited by NeedsHelpPlease; August 7th 2012 at 09:45 AM.
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  2. #2
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    Re: Separation of Variables

    Hmm...
    \int e^{-z} dz = -e^{-z}

    You can't do that w substitution as you will get  dw =e^{z} dz but there is no e^{z} in the integral to cancel it out
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  3. #3
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    Re: Separation of Variables

    I would agree that you should just use the result given by jgv115 but you can use the substitution.

    You should end up with \displaystyle \int \frac{1}{w^2}dw
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  4. #4
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    Re: Separation of Variables

    Quote Originally Posted by a tutor View Post
    I would agree that you should just use the result given by jgv115 but you can use the substitution.

    You should end up with \displaystyle \int \frac{1}{w^2}dw
    You can use the result IF you mutliply the top and bottom by \displaystyle \begin{align*} e^z \end{align*} first.
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  5. #5
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    Re: Separation of Variables

    Quote Originally Posted by Prove It View Post
    You can use the result IF you mutliply the top and bottom by \displaystyle \begin{align*} e^z \end{align*} first.
    I see, you need to multiply by 1 before it works..
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  6. #6
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    Re: Separation of Variables

    Just in case a picture also helps...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case t), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).




    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; August 7th 2012 at 01:58 AM.
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    Re: Separation of Variables

    I'm sorry. I am still not sure what substitution to use. How would I end up with my substitution as 1/w2 ​??
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  8. #8
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    Re: Separation of Variables

    Quote Originally Posted by NeedsHelpPlease View Post
    I'm sorry. I am still not sure what substitution to use. How would I end up with my substitution as 1/w2 ​??
    \displaystyle \int \frac{1}{w^2}dw is not your substitution it is the result of doing the substitution you started.

    You had w=e^z

    so

    dw=e^z \ dz

    and then

    \displaystyle \int \frac{1}{e^z}dz=\int \frac{1}{w}\frac{dw}{w}=\int\frac{1}{w^2}dw
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  9. #9
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    Re: Separation of Variables

    Quote Originally Posted by a tutor View Post
    \displaystyle \int \frac{1}{w^2}dw is not your substitution it is the result of doing the substitution you started.

    You had w=e^z

    so

    dw=e^z \ dz

    and then

    \displaystyle \int \frac{1}{e^z}dz=\int \frac{1}{w}\frac{dw}{w}=\int\frac{1}{w^2}dw
    \displaystyle = \int{w^{-2}\,dw}, which should be easy to integrate...
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