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High order linear differential equation

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data
    (x^2)(y'') + xy' - 4y = 0

    a) show that x^2 and x^-2 are linear independent solutions of this equation on the interval 0 <x<infinity
    b) write the general solution of the given equation
    c) find the solution that satisfies the conditions y(2) = 3, y'(2) = -1. Explain why htis solution is unique. Over what interval is this solution defined?


    2. Relevant equations



    3. The attempt at a solution
    ok im pretty sure i got part a). i just found y' and y'' for y = x^2 and plugged it in. then did the same for y = x^-2. both give me an answer of 0. it doesnt seem to matter what x is so i guess 0<x<infinity doesnt contain any false statement so i said it was true.

    for part b), im having trouble because either my professor didnt explain well or i took bad notes. the general equation is supposed to be:
    y(x) = Yc(x) + yp(x)

    i beleive Yc(x) = c_1(x^2) + c_2(x^-2)
    in the example in class yp(x) = 1/2 but there's no explanation on how we got that. and nothing in the book on "general equation" so any help on this would be appreciated

    for part c). i replaced x with 2, y with 3, and y' with -1 then i got y'' = 7/2. was that what i was supposed to do? then i was going to integrate back to find the equation of y that satisfies all the conditions. there's no clear explanation on this step either, so do i have the right idea for this?
     
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  3. Feb 8, 2009 #2

    Hootenanny

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    For part (b)
    yp(x) is called the particular integral and is only non-zero when the ODE is inhomogeneous, i.e. when the ODE has a non zero right hand side. Note that in general a nth order ODE has n linearly independent solutions. Since your ODE is second order and you already have two linearly independent solutions you can safely set yp(x) = 0.

    Note that in this case even if you were to find an expression for yp(x), it would be some linear combination of the two solutions that you already have and therefore wouldn't make any difference to the solution.

    For part (c)
    Yes, you are indeed meant to use the conditions that you have been given. However, you are meant to find the particular solution for this ODE. The solution you have already found yc(x) is the general solution where you have two undetermined coefficients (c1 and c2). To find the particular solution you need to use the conditions that you have been given to determine that values of c1 and c2.

    Do you follow?
     
  4. Feb 8, 2009 #3

    HallsofIvy

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    One of the things you should have learned before dealing with equation like this (called "Euler-type" or "equipotent" equations since the power of x is the same as the order of the derivative) You should have learned that the set of solutions to any linear homogeneous nth order differential equation forms a vector space of dimension n. That means that any n indepenedent solutions form a basis and that any solution can be written as a linear combination of n independent solutions. Since this is a second order linear homogeneous equation you only need two independent solutions- which you are given. Any solution to this equation is a linear combination of [itex]x^2[/itex] and [/itex]x^{-2}[/itex]- that is, [itex]y(x)= Ax^2+ Bx^{-2}[/itex]. Because this is a "homogeneous" equation, the 'particular solution', yp is 0 as Hootenanny said.

    By the way, that means that you haven't finished part (a). You have shown that the two given functions satisfy the equation but you haven't shown that they are "independent". "Independent", here, means that the only linear combination equal to 0 for all x is the "trivial" solution, with coefficients equal to 0. What you need to do is show that the only solution to [itex]Ax^2+ Bx^{-2}= 0[/itex], for all x> 0, is A= B= 0. You can do that, for example, by taking x= 1 and 2 to get two equations for A and B or by taking x= 1 in that and the derivative [itex]2Ax- 2Bx^{-2}= 0[/itex] to get two equations.

    Also, you are wrong that "it doesnt seem to matter what x is". Obviously x-2 is not defined at x= 0. That's why the interval 0< x< infinity does not include 0.

    For part (c), no you do not put the initial conditions into the equation, you put them into the general solution to evaluate the constants.
     
    Last edited: Feb 8, 2009
  5. Feb 8, 2009 #4
    ok so basically the definition of "independent" states that Ax^2 + Bx^-2 cannot = 0 unless both A and B = 0 right? and otherwise it is considered "dependent"?
     
  6. Feb 8, 2009 #5

    HallsofIvy

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    This one reason I recommend students take Linear Algebra before Differential Equations! The theory behind "linear differential equations" (which are what a first course in differential equations is basically about) is linear algebra.

    Yes, in linear algebra, a set of "vectors", {v1, v2, ..., vn} is said to be independent if the only solution to the equation a1v1+ a2v2+ ...+ anvn= 0 is the "trivial" solution a1= a2= ...= an= 0. Notice that if there were a solution to that equation in which one coefficient, say ai, we NOT 0, we could solve for that vector: aivi= -a1v1- a2v2- ...-anvn (moving everything except aivi to the right side) and since ai is no 0, vi= -(a1/ai)v1- (a2/ai)v2- ...- (an/ai)vn so we could replace vi with that expression and don't need vi at all.
     
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