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Math Help - reduction of order with homogeneous equations...

  1. #1
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    reduction of order with homogeneous equations...

    i got to the end of this problem and then me and the book went in two different directions. once i found my u i wasnt sure how to finish the problem, because what i had was so different from what my book had. next i didnt see how the book found yp to be -1/2.(it says be inspection?)

    y'' -4y = 2 ; y1 = e^(-2x)

    so y = ue^(-2x), y' = -2ue^(-2x), y''= 4ue^(-2x)-4u'e^(-2x)+u''e^(-2x)

    so e^(-2x)(u'' -4u')= 0. so i let v = u'

    v'-4v= 0. the integrating factor is e^(-4x)

    e^(-4x)v'-e^(-4x)(4v)=0

    (ve^(-4x))' = integral (0)dx

    so v = c/e^(-4x)= ce^(4x)=u'

    so u =c1e^(4x) +c2<----- im thinking this is where i went off on tangent....
    ------
    the book says u = c1e^(4x). the second solution is y2 = e^(-2x)e^(4x)=e^2x and yp = -1/2

    so that y = c1e^(-2x) + c2e^(2x) - 1/2.

    i know with a homogeneous equation that the general solution is y = yc + yp

    jus a lil lost on this one..any help would be appreciated.

    thanks in advance
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  2. #2
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    Quote Originally Posted by slapmaxwell1 View Post
    i got to the end of this problem and then me and the book went in two different directions. once i found my u i wasnt sure how to finish the problem, because what i had was so different from what my book had. next i didnt see how the book found yp to be -1/2.(it says be inspection?)

    y'' -4y = 2 ; y1 = e^(-2x)

    so y = ue^(-2x), y' = -2ue^(-2x), y''= 4ue^(-2x)-4u'e^(-2x)+u''e^(-2x)

    so e^(-2x)(u'' -4u')= 0. so i let v = u'

    v'-4v= 0. the integrating factor is e^(-4x)

    e^(-4x)v'-e^(-4x)(4v)=0

    (ve^(-4x))' = integral (0)dx

    so v = c/e^(-4x)= ce^(4x)=u'

    so u =c1e^(4x) +c2<----- im thinking this is where i went off on tangent....
    ------
    Recall that you started with " y(x)= u(x)e^{-2x}". With u(x)= c_1e^{4x}+ c_2, that will give y(x)= (c_1e^{4x}+ c_2)e^{-2x}= c_1e^{2x}+ c_2e^{-2x}. That second constant of integration, c_2 gives you the solution you already had.

    the book says u = c1e^(4x).
    Yes, because then y= c_1e^{4x}e^{-2x}= c_1e^{2x} and you already have [tex]e^{-2x} as a solution so your book is just not bothering to write that again. Both methods are correct.

    the second solution is y2 = e^(-2x)e^(4x)=e^2x and yp = -1/2

    so that y = c1e^(-2x) + c2e^(2x) - 1/2.

    i know with a homogeneous equation that the general solution is y = yc + yp

    jus a lil lost on this one..any help would be appreciated.

    thanks in advance
    A little lost where? The general theorem for linear non-homogeneous equation is that if Y(x) is the general solution to the associated homogeneous equation and y(x) is any single solution to the entire equation, then Y(x)+ y(x) is the general solution to the entire equation. Having found the general solution to the associated homogeneous equation and the a specific solution to the entire equation, add them.
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