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Math Help - Second order DEs and Self Adjoint Form

  1. #1
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    Second order DEs and Self Adjoint Form

    How do you go about proving that all second order DEs

    a(x)y" + b(x)y' + c(x)y = 0

    can be written in self adjoint form

    [p(x)y']' + q(x)y = 0?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by kiwijoey View Post
    How do you go about proving that all second order DEs

    a(x)y" + b(x)y' + c(x)y = 0

    can be written in self adjoint form

    [p(x)y']' + q(x)y = 0?
    Well, if you let b(x)=a^{\prime}(x), it follows that a(x)y^{\prime\prime}+a^{\prime}(x)y^{\prime}=\left[a(x)y^{\prime}\right]^{\prime} (def. of product rule)

    So it follows that a(x)y^{\prime\prime}+b(x)y^{\prime}+c(x)y=0 can be written as \left[a(x)y^{\prime}\right]^{\prime}+c(x)y=0\sim\left[p(x)y^{\prime}\right]^{\prime}+q(x)y=0

    Does this make sense?
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  3. #3
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    Yeah it does, thanks mate, I knew it wouldn't be very complicated.
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    Well, if you let b(x)=a^{\prime}(x), it follows that a(x)y^{\prime\prime}+a^{\prime}(x)y^{\prime}=\left[a(x)y^{\prime}\right]^{\prime} (def. of product rule)

    So it follows that a(x)y^{\prime\prime}+b(x)y^{\prime}+c(x)y=0 can be written as \left[a(x)y^{\prime}\right]^{\prime}+c(x)y=0\sim\left[p(x)y^{\prime}\right]^{\prime}+q(x)y=0

    Does this make sense?
    That does not show what was asked. The original questions was to show that any second order differential equation could be put in that form. Chris L has only showed that you can write an equation of the form a(x)y"+ a'(x)y'+ c(x)y= 0 in that form. What if b(x) is not equal to a'(x)?

    Just as with first order equations, try to find an "integrating factor", u(x). Look at (a(x)u(x)y')'= a(x)u(x)y"+ (a'(x)u(x)+ a(x)u'(x))y' and compare that with a(x)u(x)y"+ b(x)u(x)y'. In order that those be the same we must have a'(x)u(x)+ a(x)u'(x)= b(x)u(x) which is the same as the first order, separable, equation a(x)u'(x)= (b(x)- a'(x))u(x). Integrating \frac{du}{u}= \frac{b(x)- a'(x)}{a(x)} gives the integrating factor, u(x). Multiplying both sides of the equation by that u(x) puts it in the form required.
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