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Math Help - 2nd order differentiation query

  1. #1
    Senior Member bugatti79's Avatar
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    2nd order differentiation query

    Hi All,

    Given  u (x) = Ae^x and u"(x)=k^2 u (x)

    then u"(x) is calculated as u"(x) = A^2 u (x)

    How was this derived?

    I calculate both u'(x) and u"(x) to be equal to Ae^x. I dont know how the A^2 appears in front of u(x)!!!.....

    Thanks
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  2. #2
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    Why do you say that u''(x) "is calculated as u''(x)= A^2u(x)? If u(x)= Ae^x then u'= Ae^x and u''= Ae^x. That can be written as u''= k^2u(x) if and only if k^2= A so that k= \sqrt{A}.
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  3. #3
    Senior Member bugatti79's Avatar
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    My lecture notes reads
     u"(x)=k^2(x)

    u(x)=Ae^x therefore

    u"(x)=(u'(x))'=(Au(x))'=Au'(x)=A^2u(x) ????

    Therefore A^2u(x)=k^2u(x) implies

    A_1=k and A_2=-k

    I dont understand the third line!!...I dont think its the standard way of getting solution for a nd order equation?
    Thanks
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  4. #4
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    so you have u'(x)=Au(x) and u(x)=Ae^x ?? In the third line, that seems strange?
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  5. #5
    Senior Member bugatti79's Avatar
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    I dont know what his method is. I think I will just stick with the standard way of getting the solution ie,

    u"(x)=k^2u(x) is of the form

    u"(x)-n^2u(x)=0 giving roots
    m^2-n^2=0 therefore

     m=+ - n

    therefore solution is u(x)=Ae^{nx} +Be^{-nx}

    This ok?
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  6. #6
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    [QUOTE=bugatti79;591430]My lecture notes reads
     u"(x)=k^2(x)

    u(x)=Ae^x therefore

    u"(x)=(u'(x))'=(Au(x))'[/quote
    This is wrong. If u= Ae^x then u'= A(e^x) not Au.

    =Au'(x)=A^2u(x) ????

    Therefore A^2u(x)=k^2u(x) implies

    A_1=k and A_2=-k

    I dont understand the third line!!...I dont think its the standard way of getting solution for a nd order equation?
    Thanks
    Are you sure you don't have something like e^{Ax} instead?
    If u(x)= e^{Ax}, then u'= A e^{Ax}= Au so u''= Au'= A(Ae^{Ax})= A^2 e^{Ax}.
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  7. #7
    Senior Member bugatti79's Avatar
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    aha, thats what it was. It mustve been a typo on behalf or the lecturers :-)
    But the standard derivation in post 5is correct? Thanks
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  8. #8
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    Quote Originally Posted by bugatti79 View Post
    I dont know what his method is. I think I will just stick with the standard way of getting the solution ie,

    u"(x)=k^2u(x) is of the form

    u"(x)-n^2u(x)=0 giving roots
    m^2-n^2=0 therefore

     m=+ - n

    therefore solution is u(x)=Ae^{nx} +Be^{-nx}

    This ok?
    Why did you change from k to n? If the differential equation is u''= k^2u, then the characteristic equation is m^2= k^2 so that m= \pm k and the general solution to the differential equation is u(x)= Ae^{kx}+ Be^{-kx}. Note that this could also be written as u(x)= Ccosh(kx)+ Dsinh(kx).

    By the way, using two consecutive single quotes in LaTex rather than a double quote will give u'' rather than u".
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  9. #9
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Why did you change from k to n? If the differential equation is u''= k^2u, then the characteristic equation is m^2= k^2 so that m= \pm k and the general solution to the differential equation is u(x)= Ae^{kx}+ Be^{-kx}. Note that this could also be written as u(x)= Ccosh(kx)+ Dsinh(kx).

    By the way, using two consecutive single quotes in LaTex rather than a double quote will give u'' rather than u".
    The k to n, just a typo. Thanks.
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