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Reduction of Order to find 2nd solution of DE

  1. Feb 25, 2013 #1
    1. The problem statement, all variables and given/known data
    t^2y''-4ty'+6y=0 t>0, y1(t)=t^2 find the 2nd solution using method of reduction of order.


    2. Relevant equations



    3. The attempt at a solution

    Let y=v(t)t^2
    y'=2tv+v't^2
    y''=2v+2tv'+v''t^2+2tv'

    put in back into eq and solving reduces to (checked it twice)
    t^4*v''=0
    v''=0
    Then I integrated with respect to T
    v'=c (c is some constant)
    v(t)=ct where c is a constant.

    I substituted v(t) back into y=v(t)*t^2 and got y2=ct^3 where c is some constant so y2=t^3? I know thats the answer but did I correctly do all these steps? Also do I need to show the wronskian in order to prove that the solution is acceptable as y1 and y2 need to form a fundamental set of solutions? Thanks I just want to make sure I am doing this correctly.

    --------Problem 2: same question different equation

    (x-1)y''-xy'+y=0 x>1 y1=e^x

    let y=v(t)e^x
    y'=v'e^x +ve^x
    y''=v''e^x +2v'e^x +ve^x

    into the eq:
    e^x[(x-1)(v''+2v'+v)-x(v'+v)+v]=0
    (x-1)v''+(x-2)v'=0 then I get confused
     
    Last edited: Feb 25, 2013
  2. jcsd
  3. Feb 25, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You have done problem 1 correctly.

    For problem 2, after you have
    e^x[(x-1)(v''+2v'+v)-x(v'+v)+v]=0
    (x-1)v''+(x-2)v'=0
    let u= v' so your differential equation is (x- 1)u'+ (x- 2)u= 0, a differential equation of order 1 (this was, after all, a "reduction of order" method).

    That is, in fact, a separable equation- it can be written as (x- 1)u'= (2- x)u and then [tex]\frac{du}{u}= \frac{2- x}{x- 1}dx[/tex]. Integrate both sides.
     
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