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Differential Equations reduction of order

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Find y2 given
    t^2y'' + 3ty' + y = 0 y1(t) = 1/t


    2. Relevant equations



    3. The attempt at a solution
    i found y, y' and y'' and plugged it back in to the Differential equation,
    after doing some work, i'm basicly down to
    tv''+v' = 0
    where y2 = v/t

    so u = v' and u' = v'' Plugging that in...
    tu' + u = 0
    how would i solve this for u?
    i think its separable equations but that would give me
    du/-u = dt/t
    which would be ln (u) = -ln(t) + c .... u = 1/t +c so v' = 1/t +c which means v = lnt
    pluging v into y2 would give me lnt/t
    Sorry its kind of confusing but the answer's not in the back of the book and i need to know if i did this right

    Thank you!
     
  2. jcsd
  3. Oct 19, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, the problem reduces to tu'+ u= 0 which is separable: du/u= -dt/t. Integrating ln(u)= -ln(t)+ c and, taking the exponential of both sides u= C/t where C= ec. (Notice that the C is multiplied, not added: ea+c[/b]= eaec, not "ea+ c".)

    The integral of u= v'= C/t is v(t)= Cln(t)+ D and then, since you made the reduction by assuming a solution of the form y= v(t)y1= v(t)(1/t), your general solution to the original differential equation is y(t)= Cln(t)/t+ D/t. How does that compare with the solution in the back of your book?
     
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