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Differential equation - 2nd order diff equation

  1. Mar 11, 2007 #1
    Consider the equation 2t^2y'' + 3ty' - y = 0
    (a) Show that y1(t) = sqrt(t) and y2(t) = 1/t are soltuions of the equation on the interval 0<t<infinity
    (b) Compute W[y1,y2](t). What happens as t approaches zero?
    (c) Show that y1(t) and y2(t) form a fundamental set of solutions of the equation on the interval 0<t<infinity.

    For (a), i just did lhs rhs and proved they equaled, i did not do anything with the interval, am i suppose to?
    For (b) I computed it to be W=-3/(2t^(3/2)) which is the correct answer. and i got as t->0, W-> - infinity, is this correct? can someone confirm.
    For (c) i'm not sure how to show that it is a fundamental set of solutions on the interval. (a) and (b) confirms that it is a fundamental set of solutions, but on the interval, do i substitute in the ends of the interval? or what? i am unsure, someone please help.
  2. jcsd
  3. Mar 12, 2007 #2
    For part (a) you ought to do what the question asks, so yes you are supposed to find what happens on the interval. What is W[y1,y2], the wronskian of the two? The last part is asking whether or not the two solutions form a basis for the solution space.
  4. Mar 12, 2007 #3
    so can i just sub in the endpoints? im not sure how i show this.
  5. Mar 12, 2007 #4
    and yes it is the wronskian of the two
    im pretty sure i got (a) and (b) , it is just (c) that is giving me a lot of trouble. i was never taught how to show fundamental set of solutions on an interval.
  6. Mar 12, 2007 #5


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    The reason for mentioning the interval 0< t< infinity: sqrt(t) is only defined on that interval (plus 0) while 1/t is not defined at t= 0 so a solution cannot cross it- it is the largest interval on which both of those functions are defined. You don't have to do any more than show that the two functions satisfy the equation on that interval which is what you did.

    Do you know the definition of "fundamental set of solutions"? That is a set of solutions such that any solution can be written as a linear combination of them in a unique way. More technically, they form a basis for the vector space of all solutions to the linear differential equation. Since there are two functions here, and the differential equation is of second order, it is sufficient to show that they are independent.

    You probably have a theorem that says solutions to a linear differential equation are independent if and only if their Wronskian is never 0 but you can also show it directly from the definition of "independent".

    Suppose [itex]A\sqrt{t}+ B/t= 0[/itex] for all positive t, for some numbers A and B. What must A and B be?
  7. Jul 23, 2007 #6
    It seems you got (a) and (b) correct. For (c), to put it simply, if

    [tex]W[y1,y2][/tex] is not equal to 0, then [tex]y1[/tex] and [tex]y2[/tex] form a fundamental set of solutions. So since you got

    [tex]W[y1,y2] = -\frac{3}{2t^{3/2}}[/tex]
    [tex]y1[/tex] and [tex]y2[/tex] form a fundamental set of solutions.
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