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Math Help - First Order DE-Separating the Variables

  1. #1
    Senior Member bugatti79's Avatar
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    First Order DE-Separating the Variables

    All,

    Going through a book question. Getting stuck right at the end given the IC's See attached... Are the very last two terms equivalent? If so...how?


    Any tips appreciated.

    Thanks
    Attached Thumbnails Attached Thumbnails First Order DE-Separating the Variables-imag0126.jpg  
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  2. #2
    A Plied Mathematician
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    Could you please type up the DE and the IC? I'm having a very hard time reading that scan.
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  3. #3
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    Quote Originally Posted by bugatti79 View Post
    All,

    Going through a book question. Getting stuck right at the end given the IC's See attached... Are the very last two terms equivalent? If so...how?


    Any tips appreciated.

    Thanks
    \displaystyle\int\frac{1}{1+e^{-x}}dx=\ln{(e^x+1)

    You have x+\ln{(e^x+1)
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  4. #4
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by dwsmith View Post
    \displaystyle\int\frac{1}{1+e^{-x}}dx=\ln{(e^x+1)

    You have x+\ln{(e^x+1)
    No, its x+\ln{(e^{-x}+1). It works out the same towards the end when you take log of both sides. I keep getting the same answer and including your answer. I wonder is the book wrong? See attached..
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  5. #5
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Could you please type up the DE and the IC? I'm having a very hard time reading that scan.
    Sorry, I have just read you post after sending it. I will do the next time. Sorry
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  6. #6
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    Quote Originally Posted by bugatti79 View Post
    No, its x+\ln{(e^{-x}+1). It works out the same towards the end when you take log of both sides. I keep getting the same answer and including your answer. I wonder is the book wrong? See attached..
    Oh you have a negative.
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  7. #7
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    Quote Originally Posted by dwsmith View Post
    Oh you have a negative.
    \displaystyle\frac{\cos{y}}{1+e^{-x}}=Ae^x

    \displaystyle\frac{\cos{\frac{\pi}{4}}}{1+1}=A

    \displaystyle\frac{\sqrt{2}}{4}=A
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  8. #8
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by dwsmith View Post
    \displaystyle\frac{\cos{y}}{1+e^{-x}}=Ae^x

    \displaystyle\frac{\cos{\frac{\pi}{4}}}{1+1}=A

    \displaystyle\frac{\sqrt{2}}{4}=A
    but the book answer is (1+e^x)Sec y=2\sqrt{2}
    Must be a typo in the book.
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  9. #9
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    Quote Originally Posted by bugatti79 View Post
    but the book answer is (1+e^x)Sec y=2\sqrt{2}
    Must be a typo in the book.
    \displaystyle\cos{y}=\frac{\sqrt{2}}{4}(1+e^x)\Rig  htarrow\frac{1}{\cos{y}}=\frac{1}{\frac{\sqrt{2}}{  4}(1+e^x)}\Rightarrow\sec{y}=\frac{2\sqrt{2}}{1+e^  x}

    See it?
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  10. #10
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by dwsmith View Post
    \displaystyle\cos{y}=\frac{\sqrt{2}}{4}(1+e^x)\Rig  htarrow\frac{1}{\cos{y}}=\frac{1}{\frac{\sqrt{2}}{  4}(1+e^x)}\Rightarrow\sec{y}=\frac{2\sqrt{2}}{1+e^  x}

    See it?
    I do now, thanks. Just testing the TeXForm function in mathematica to convert the code to latex....

    \sec (y)=\frac{2 \sqrt{2}}{e^x+1}

    THat is so handy! :-)

    Cheers
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