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Thread: 2nd order differentiation query

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

    Hi All,

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

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

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

    $\displaystyle u(x)=Ae^x$ therefore

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

    Therefore $\displaystyle A^2u(x)=k^2u(x)$ implies

    $\displaystyle A_1=k and A_2=-k$

    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 $\displaystyle u'(x)=Au(x)$ and $\displaystyle u(x)=Ae^x$ ?? 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,

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

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

    $\displaystyle m=+ - n$

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

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

    $\displaystyle u(x)=Ae^x$ therefore

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

    $\displaystyle =Au'(x)=A^2u(x)$ ????

    Therefore $\displaystyle A^2u(x)=k^2u(x)$ implies

    $\displaystyle A_1=k and A_2=-k$

    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 $\displaystyle e^{Ax}$ instead?
    If $\displaystyle u(x)= e^{Ax}$, then $\displaystyle u'= A e^{Ax}= Au$ so $\displaystyle u''= Au'= A(Ae^{Ax})= A^2 e^{Ax}$.
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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,

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

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

    $\displaystyle m=+ - n$

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

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

    By the way, using two consecutive single quotes in LaTex rather than a double quote will give $\displaystyle u''$ rather than $\displaystyle u"$.
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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 $\displaystyle u''= k^2u$, then the characteristic equation is $\displaystyle m^2= k^2$ so that $\displaystyle m= \pm k$ and the general solution to the differential equation is $\displaystyle u(x)= Ae^{kx}+ Be^{-kx}$. Note that this could also be written as $\displaystyle u(x)= Ccosh(kx)+ Dsinh(kx)$.

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