# reduction of order with homogeneous equations...

• Feb 28th 2011, 01:54 AM
slapmaxwell1
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....
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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.

• Feb 28th 2011, 03:02 AM
HallsofIvy
Quote:

Originally Posted by slapmaxwell1
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....
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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.

Quote:

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.

Quote:

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.