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Thread: Stuck at e^(e^2x)..

  1. #1
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    Stuck at e^(e^2x)..

    Hi, I am a computer engineering student and Maths is one of my subject for this university year.

    I was solving some paper and I cant figure out how to solve the following step. (Its based on Linear Differential Equations.)

    This is the step I am stuck at,
    $\displaystyle
    e^{-2x} e^{e^{2x}}
    $

    The answer of this step is
    $\displaystyle
    e^{4x}
    $

    How?
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  2. #2
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    Quote Originally Posted by enli View Post
    Hi, I am a computer engineering student and Maths is one of my subject for this university year.

    I was solving some paper and I cant figure out how to solve the following step. (Its based on Linear Differential Equations.)

    This is the step I am stuck at,
    $\displaystyle
    e^{-2x} e^{e^{2x}}
    $

    The answer of this step is
    $\displaystyle
    e^{4x}
    $

    How?
    You don't even say what you're trying to do with this expression.

    Please post the entire question.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    You don't even say what you're trying to do with this expression.

    Please post the entire question.
    Hi, sorry for the misunderstanding.

    I thought the whole problem is not important, so i didnt post it.

    In the problem, it is shown that those two expressions are equal. I wish to know how it is done.

    To reframe my question, how do you show,

    $\displaystyle e^{-2x} e^{e^{2x}} = e^{4x}
    $
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  4. #4
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    Quote Originally Posted by enli View Post
    Hi, sorry for the misunderstanding.

    I thought the whole problem is not important, so i didnt post it.

    In the problem, it is shown that those two expressions are equal. I wish to know how it is done.

    To reframe my question, how do you show,

    $\displaystyle e^{-2x} e^{e^{2x}} = e^{4x}$
    Clearly the expressions are not equal in general. If you have obtained this expression while attempting to solve your original problem, then you have made a mistake. Which is why I asked you to post the original question.
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  5. #5
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    The original problem is to solve the following (Linear differential equations)
    $\displaystyle
    (D^2+6D+8)y = e^{e^{2x}}
    $

    The book has solved all steps correctly prior to the following step, where I cant follow how it is done
    $\displaystyle
    \frac{1}{D+4} [e^{-2x}\frac{1}{2}e^{e^{2x}}]
    $

    And the next step is,
    $\displaystyle
    =\frac {1}{2}e^{-4x}\int{ e^{4x}e^{-2x}e^{e^{2x}} dx}
    $
    Last edited by enli; May 8th 2009 at 09:14 PM. Reason: added the problem statement
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  6. #6
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    Quote Originally Posted by enli View Post
    The original problem is to solve the following (Linear differential equations)
    $\displaystyle
    (D^2+6D+8)y = e^{e^{2x}}
    $

    The book has solved all steps correctly prior to the following step, where I cant follow how it is done
    $\displaystyle
    \frac{1}{D+4} [e^{-2x}\frac{1}{2}e^{e^{2x}}]
    $

    And the next step is,
    $\displaystyle
    =\frac {1}{2}e^{-4x}\int{ e^{4x}e^{-2x}e^{e^{2x}} dx}
    $
    To solve $\displaystyle \int e^{4x} e^{-2x} e^{e^{2x}} \, dx = \int e^{2x} e^{e^{2x}} \, dx $ make the substitution $\displaystyle u = e^{2x}$.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    To solve
    $\displaystyle \int e^{4x} e^{-2x} e^{e^{2x}} \, dx = \int e^{2x} e^{e^{2x}} \, dx $ make the substitution $\displaystyle u = e^{2x}$.
    Excuse me for not being clear enough about the problem,

    I wish to know following two are equal
    $\displaystyle
    \frac {1}{D+4} [e^{-2x} \frac {1}{2}e^{e^{2x}}] = \frac {1}{2}e^{-4x}\int { e^{4x}e^{-2x}e^{e^{2x}} dx}
    $

    The formula is,
    $\displaystyle
    \frac {1}{D-m}f(x) = e^{mx}\int {e^{-mx} f(x)dx}
    $



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  8. #8
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    Quote Originally Posted by mr fantastic View Post
    To solve
    $\displaystyle \int e^{4x} e^{-2x} e^{e^{2x}} \, dx = \int e^{2x} e^{e^{2x}} \, dx $ make the substitution $\displaystyle u = e^{2x}$.
    Yeah right,

    The formula is,
    $\displaystyle
    \frac {1}{D+m}f(x) = e^{-mx}\int {e^{mx} f(x)dx}
    $

    Now if we follow the formula, that means
    $\displaystyle
    m=4$
    $\displaystyle f(x)=e^{-2x}e^{e^{2x}}
    $

    And hence now it makes sense why I was trying to compare, which was of course wrong
    $\displaystyle e^{-2x} e^{e^{2x}} = e^{4x}$

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