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Math Help - reduction of order...

  1. #1
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    reduction of order...

    the indicated y1 is a solution of the given differential equation. find a second solution y2(x).

    so my problem is: y'' -4y' + 4y = 0 , y1 = e^(2x)

    if y=u(x)y1(x) then y= u(x)e^(2x).

    after finding y'' and y' i substituted back into the original equation and got,

    u''e^(2x) = 0. and this is where i am stuck. I am not sure what to do.

    i know that u'' must be 0.

    my book goes on to say that, u= c1x + c2? and that c1= 1 and c2=0 and that
    y2= xe^(2x)?

    a little help please....

    thanks in advance.
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  2. #2
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    If \displaystyle \frac{d^2u}{dx^2} = 0, integrate once to get \displaystyle \frac{du}{dx}, integrate again to get \displaystyle u.
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  3. #3
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    ok so with that being said u'=c1 and u would be c1x + c2

    ok i see that now thanks...ok but how did the book know to make c1 = 1 and c2 = 0?
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  4. #4
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    Were you given any initial conditions or boundary conditions?
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  5. #5
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    no i wasnt
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  6. #6
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    In that case you will need to substitute \displaystyle y = e^{2x} + (c_1x + c_2)e^{2x} into your DE, simplify, then equate the coefficients of like powers of \displaystyle x...
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  7. #7
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    What you're looking for is the second independent solution. If you keep your two constants as they are you have

    y = \left(c_1 + c_2x\right)e^{2x} = c_1  e^{2x} + c_2 x e^{2x}

    The first is your first solution ( c_1 = 1, c_2 = 0) and the second the second independent solution ( c_1 = 0, c_2 = 1)
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