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Diff EQ reduction of Order problem need help!

  1. Aug 18, 2008 #1
    1. The problem statement, all variables and given/known data
    One solution y1(t) of the differential equation is given
    a) Use the method of reductions of order to obtain a second solution y2(t)

    b) Compute the wronskian formed by y1(t) and y2(t)


    2. Relevant equations

    (t + 1)^2y'' - 4(t+1)y' + 6y = 0

    y1(t) = (t+1)^2

    3. The attempt at a solution

    i assume that the second solutions is y1(t)*u(t) where u(t) is some function of t that statisfies the diff eq

    y2(t) = y1(t)*u(t) = (t+1)^2*u(t)

    y2(t)' = 2(t+1)*u(t) + (t+1)^2*u(t)'

    y2(t)'' = 2u(t) + 2(t+1)*u(t)' + 2(t+1)*u(t)' + (t+1)^2*u(t)''
    or
    y2(t)'' = 2u(t) + 4(t+1)*u(t)' + (t+1)^2*u(t)''

    I plug these into the eqaution and reduce and I wind up with

    (t+1)^4*u(t)'' = 0

    I do a reduction of order through v(t) = u(t)' and v(t)' = u(t)''

    and the equations becomes (t+1)^4*v(t)' = 0

    Here is where I may be going wrong, I divide the (t+1)^4 and just have
    v(t)' = 0 which then becomes u(t) = C1t + C2

    so y2(t) = y1(t)* C1t + C2]
    or
    y2(t) = C1*y1(t)*t + C2y1(t)

    C1 can be folded into complimentary solution constant, so I assign 1 to that. C2 can be zero since its a multiple of the first solution.

    My problem is the answer is wrong. The book has (t+1)^3, I checked both y1 and the books answer and they both work. So I am lost as to what I did wrong. I showed it to a couple of post docs and they got the answer I got. Any help is much appreciated.
     
  2. jcsd
  3. Aug 18, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    Everybody is correct. The book writes the general answer as D1*(t+1)^2+D2*(t+1)^3. That's the same as (t+1)^2*(D1+D2+D2*t). If you pick C1=D1+D2 and C2=D2, that's exactly your answer. They are both the same set of functions. There are many different choices for the 'other linearly independent function' y2. E.g. they also could have written y2=t*(t+1)^2.
     
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