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Math Help - differential and proof

  1. #1
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    Post differential and proof

    Hey guys!

    How would you solve the following:


    1) If  y = e^{-kt}(Acoshqt +Bsinhqt) [A, B, k, q are constants]

    show that  \frac{d^2y}{dt^2} + 2k\frac{dy}{dt} + (k^2 - q^2)y=0




    2) Find \frac{dy}{dx} when  y = tanh^{-1}\left(\frac{2x}{1+x^2}\right)


    Thanks.
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  2. #2
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    Hello, dadon!

    2) Find \frac{dy}{dx} when  y \,= \,\text{tanh}^{-1}\left(\frac{2x}{1+x^2}\right)

    We have: . \text{tanh }y \:=\:\frac{2x}{1+x^2}

    From the defintion of \text{tanh }y, we have: . \frac{e^y - e^{-y}}{e^y + e^{-y}} \:=\:\frac{2x}{1+x^2}

    Multiply the left fraction by \frac{e^y}{e^y}\!:\;\;\frac{e^{2y} - 1}{e^{2y} + 1} \:=\:\frac{2x}{1 + x^2}

    . . (1+x^2)(e^{2y} - 1) \:=\:2x(e^{2y} + 1)\quad\Rightarrow\quad (1+x^2)e^{2y} - (1 + x^2) \:=\:2xe^{2y} + 2x

    . . (1 + x^2)e^{2y} - 2xe^{2y} \:=\:2x + (1 + x^2)

    . . (x^2 - 2x + 1)e^{2y}\:=\:x^2 + 2x + 1

    . . e^{2y} \:=\:\frac{x^2+2x+1}{x^2-2x+1} \:=\:\frac{(x+1)^2}{(x-1)^2} \:=\:\left(\frac{x+1}{x-1}\right)^2

    . . 2y \:=\:\ln\left(\frac{x+1}{x-1}\right)^2 \:=\:2\ln\left(\frac{x+1}{x-1}\right)

    . . y \;=\;\ln\left(\frac{x+1}{x-1}\right) \;= \;\ln(x+1) - \ln(x-1)


    Therefore: . \frac{dy}{dx} \;=\;\frac{1}{x+1} - \frac{1}{x - 1}\;=\;\frac{-2}{x^2-1} \;=\;\frac{2}{1-x^2}

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  3. #3
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    thanks!

    can any one help with question 1?
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  4. #4
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    Quote Originally Posted by dadon View Post
    Hey guys!

    How would you solve the following:


    1) If  y = e^{-kt}(Acoshqt +Bsinhqt) [A, B, k, q are constants]

    show that  \frac{d^2y}{dt^2} + 2k\frac{dy}{dt} + (k^2 - q^2)y=0
     y = e^{-kt}(A \cosh(qt) +B\sinh(qt))

     y' = -k e^{-kt}(A \cosh(qt) +B\sinh(qt)) +e^{-kt}q (A \sinh(qt) +B\cosh(qt))

    .....  = -k y +e^{-kt}q (A \sinh(qt) +B\cosh(qt))

    So:

    y'' = -k y' -k e^{-kt}q (A \sinh(qt)  +B\cosh(qt))+e^{-kt}q^2 (A \cosh(qt) +B\sinh(qt))

    hence:

    y'' = -k y' -k e^{-kt}q (A \sinh(qt)  +B\cosh(qt))+q^2y

    Now rework the middle term:

    y'' = -k y' -k (y'+ky) +q^2y

    Which rearranges into the required differential equation.

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
     y' = -k e^{-kt}(A \cosh(qt) +B\sinh(qt)) +e^{-kt}q (A \sinh(qt) +B\cosh(qt))

    .....  = -k y +e^{-kt}q (A \sinh(qt) +B\cosh(qt))

    So:

    y'' = -k y' -k e^{-kt}q (A \sinh(qt)  +B\cosh(qt))+e^{-kt}q^2 (A \cosh(qt) +B\sinh(qt))

    hence:

    y'' = -k y' -k e^{-kt}q (A \sinh(qt)  +B\cosh(qt))+q^2y

    Now rework the middle term:

    y'' = -k y' -k (y'+ky) +q^2y

    Which rearranges into the required differential equation.

    RonL
    hey captain!

    what do you do on the second line starting y' and also line y''.

    thanks

    dadon
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  6. #6
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    Quote Originally Posted by CaptainBlack View Post
     y = e^{-kt}(A \cosh(qt) +B\sinh(qt))

     y' = -k e^{-kt}(A \cosh(qt) +B\sinh(qt)) +e^{-kt}q (A \sinh(qt) +B\cosh(qt))

    .....  = -k y +e^{-kt}q (A \sinh(qt) +B\cosh(qt))

    So:

    y'' = -k y' -k e^{-kt}q (A \sinh(qt)  +B\cosh(qt))+e^{-kt}q^2 (A \cosh(qt) +B\sinh(qt))

    hence:

    y'' = -k y' -k e^{-kt}q (A \sinh(qt)  +B\cosh(qt))+q^2y

    Now rework the middle term:

    y'' = -k y' -k (y'+ky) +q^2y

    Which rearranges into the required differential equation.

    RonL
    Quote Originally Posted by dadon View Post
    hey captain!

    what do you do on the second line starting y' and also line y''.

    thanks

    dadon

    The first line where y' appears is using the product rule on the previous line (the expression for y). Where the e^{-kt} is one term and A \cosh(qt)+B \sinh(qt) is the other term. Then you need to know that (d/dx) \cosh(x)=\sinh(x) and (d/dx) \sinh(x)=\cosh(x).

    Similarly where y'' appears this is the derivative of y', the first term is the derivative of the first term in y', and the remaining terms come from applying the product rule to the second term in the expression for y'.

    RonL
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post

    y'' = -k y' -k (y'+ky) +q^2y

    Which rearranges into the required differential equation.

    RonL
    thank you
    Last edited by dadon; March 11th 2007 at 08:15 AM.
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  8. #8
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    Quote Originally Posted by CaptainBlack View Post

    Now rework the middle term:

    y'' = -k y' -k (y'+ky) +q^2y
    How do you get -k(y' +ky) in the quoted text?

    Thank you
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by dadon View Post
    How do you get -k(y' +ky) in the quoted text?

    Thank you
    From the top of that post we have:

    y' = -k y +e^{-kt}q (A sinh(qt) +B cosh(qt))

    y' + ky = e^{-kt}q (A sinh(qt) +B cosh(qt))

    which should do the job.

    RonL
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