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Math Help - Finding the value of the term named 'K'

  1. #1
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    Red face Finding the value of the term named 'K'

    Question : if <br />
y = e^t cos t , x = e^t sin t <br />
and y'' (x + y)^2 = K (xy' - y). Find K
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  2. #2
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    What have you tried? Since x= e^tcos(t) and y= e^tsin(t), y'= e^t (cos(t)+ sin(t) and y"= e^t(cos(t)+ sin(t))+ e^t(-sin(t)+ cos(t)) = 2e^tcos(t)

    Also (x+y)^2= (e^t(cos(t)+ sin(t))^2= e^{2t}(cos^2(t)+ 2sin(t)cos(t)+ sin^2(t)) = e^{2t}(1+ 2sin(t)cos(t)).

    Put those into the equations and see what you get.
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  3. #3
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    Quote Originally Posted by zorro View Post
    Question : if <br />
y = e^t cos t , x = e^t sin t <br />
and y'' (x + y)^2 = K (xy' - y). Find K
    x = e^t\sin{t}.


    y = e^t\cos{t}.


    y' = e^t\cos{t} - e^t\sin{t}

     = e^t(\cos{t} - \sin{t}).


    y'' = e^t(\cos{t} - \sin{t}) + e^t(-\sin{t} - \cos{t})

     = -2e^t\sin{t}.



    Substitute this all into the DE.

    y'' (x + y)^2 = K (xy' - y)

    -2e^t\sin{t}(e^t\sin{t} + e^t\cos{t})^2 = K[e^t\sin{t}e^t(\cos{t} - \sin{t}) - e^t\cos{t}]

    Now try to solve for K.
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  4. #4
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    Red face After going to the steps i am stuck at this eq

    Quote Originally Posted by Prove It View Post
    x = e^t\sin{t}.


    y = e^t\cos{t}.


    y' = e^t\cos{t} - e^t\sin{t}

     = e^t(\cos{t} - \sin{t}).


    y'' = e^t(\cos{t} - \sin{t}) + e^t(-\sin{t} - \cos{t})

     = -2e^t\sin{t}.



    Substitute this all into the DE.

    y'' (x + y)^2 = K (xy' - y)

    -2e^t\sin{t}(e^t\sin{t} + e^t\cos{t})^2 = K[e^t\sin{t}e^t(\cos{t} - \sin{t}) - e^t\cos{t}]

    Now try to solve for K.

    After going thrght the steps and procedure i am stuck at this steps

    -2e^t sint (sint + cost)^2 = K   e^t (sint cost) - sint^2 - cost
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    What have you tried? Since x= e^tcos(t) and y= e^tsin(t), y'= e^t (cos(t)+ sin(t) and y"= e^t(cos(t)+ sin(t))+ e^t(-sin(t)+ cos(t)) = 2e^tcos(t)

    Also (x+y)^2= (e^t(cos(t)+ sin(t))^2= e^{2t}(cos^2(t)+ 2sin(t)cos(t)+ sin^2(t)) = e^{2t}(1+ 2sin(t)cos(t)).

    Put those into the equations and see what you get.
    I think this is not correct.

    y'=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt  }}

    so, we have now:

    \frac{dy}{dt}=e^t\cdot \left(cos(t)-sin(t)\right)
    \frac{dx}{dt}=e^t\cdot \left(cos(t)+sin(t)\right)
    \frac{dy}{dx}=\frac{cos(t)-sin(t)}{cos(t)+sin(t)}

    from here you can calculate:

    y''=\frac{d^2y}{dx^2}=\frac{dt}{dx}\frac{d}{dt} \left(\frac{dy}{dt}\right)= \frac{\frac{d}{dt}\left(\frac{cos(t)-sin(t)}{cos(t)+sin(t)}\right) }{e^t\cdot (sin(t)+cos(t))}=...

    Putting all this in the given differential equation, you can obtain K.
    (I got 2)

    Coomast
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  6. #6
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    Is y'' correct

    Quote Originally Posted by Coomast View Post
    I think this is not correct.

    y'=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt  }}

    so, we have now:

    \frac{dy}{dt}=e^t\cdot \left(cos(t)-sin(t)\right)
    \frac{dx}{dt}=e^t\cdot \left(cos(t)+sin(t)\right)
    \frac{dy}{dx}=\frac{cos(t)-sin(t)}{cos(t)+sin(t)}

    from here you can calculate:

    y''=\frac{d^2y}{dx^2}=\frac{dt}{dx}\frac{d}{dt} \left(\frac{dy}{dt}\right)= \frac{\frac{d}{dt}\left(\frac{cos(t)-sin(t)}{cos(t)+sin(t)}\right) }{e^t\cdot (sin(t)+cos(t))}=...

    Putting all this in the given differential equation, you can obtain K.
    (I got 2)

    Coomast


    I am getting y'' as
    y'' = \frac{\frac{-2(cos^2 t + sin^2 t)}{(cos t + sin t)^2}}{e^t(cos t + sin t)}


    is that correct mate...
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  7. #7
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    Quote Originally Posted by zorro View Post
    I am getting y'' as
    y'' = \frac{\frac{-2(cos^2 t + sin^2 t)}{(cos t + sin t)^2}}{e^t(cos t + sin t)}


    is that correct mate...
    "Mate" that's correct. Written more simply as:

    y''=\frac{-2}{e^t \cdot \left(cos(t)+sin(t)\right)^3}

    And putting all in the original DE gives for K....

    Coomast
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  8. #8
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    Need ur suggestion

    Quote Originally Posted by Coomast View Post
    "Mate" that's correct. Written more simply as:

    y''=\frac{-2}{e^t \cdot \left(cos(t)+sin(t)\right)^3}

    And putting all in the original DE gives for K....

    Coomast

    mate this might be silly question but need to make sure
    In the above equation where did (cos^2 t + sin^2 t) go how did u get rid of it, did u take it as '1'
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  9. #9
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    Quote Originally Posted by zorro View Post
    mate this might be silly question but need to make sure
    In the above equation where did (cos^2 t + sin^2 t) go how did u get rid of it, did u take it as '1'
    Yes, Zorro, indeed, it is equal to 1. It is one of the most basic goniometric relations. Why would you assume this is not so?

    coomast
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  10. #10
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    Stuck again !!!

    Quote Originally Posted by Coomast View Post
    Yes, Zorro, indeed, it is equal to 1. It is one of the most basic goniometric relations. Why would you assume this is not so?

    coomast

    I guess mate i am doubting myself now days....
    1 more question ....
    (xy' - y) = (e^t sint) (e^t (cos t - sin t)) - (e^t cos t)

     = (e^t)^2 sin t cos t - (e^t)^2  sin^2 t - (e^t cos t)

    I am stuck her now
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  11. #11
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    Quote Originally Posted by zorro View Post
    I guess mate i am doubting myself now days....
    1 more question ....
    (xy' - y) = (e^t sint) (e^t (cos t - sin t)) - (e^t cos t)

     = (e^t)^2 sin t cos t - (e^t)^2  sin^2 t - (e^t cos t)

    I am stuck her now
    No, this is wrong. Expand the given factor of the DE as:

    xy'-y=x\cdot \frac{dy}{dx}-y=e^tsin(t) \cdot \left( \frac{cost(t)-sin(t)}{cost(t)+sin(t)} \right) -e^tcos(t)

    Can you proceed from here?

    Coomast

    edit: a "+" sign had to be a "-" sign....corrected
    Last edited by Coomast; December 1st 2009 at 12:27 PM.
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  12. #12
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    Thumbs up Thanks mate ........

    Quote Originally Posted by Coomast View Post
    No, this is wrong. Expand the given factor of the DE as:

    xy'-y=x\cdot \frac{dy}{dx}-y=e^tsin(t) \cdot \left( \frac{cost(t)+sin(t)}{cost(t)+sin(t)} \right) -e^tcos(t)

    Can you proceed from here?

    Coomast


    Yes got the aanswer 2


    Thanks mate ........God bless u ........and thanks again for helping me .......Cheers mate
    Last edited by mr fantastic; December 2nd 2009 at 02:48 AM.
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  13. #13
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    Zorro,

    You're welcome

    Coomast
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