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Homework Help: ODE question - Is it linear?

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data
    Is the following ODE linear? If so, is it homogeneous?

    xy'' + siny = 0, where y = y(x)

    2. Relevant equations
    Linear = coefficients of unknown function y(x) and its derivatives only depend on x, not the unknown

    Homogeneous: can be written in form y'' + p(x)y' + q(x)y = 0

    3. The attempt at a solution

    I'm confused in that the only way to get rid of the sin from the y is to put an arcsin in front of the y'' term. Can one just 'ignore' the sin and say it is linear and also homogeneous by the definitions above?

    Thanks for any help.
  2. jcsd
  3. Feb 3, 2010 #2
    Write [tex]L[y] = x y'' + \sin(y)[/tex]. [tex]L[/tex] is called a differential operator (a function of a function), and the given ODE is the same as finding the y so that
    [tex]L[y] = 0[/tex]

    An ODE is linear if its differential operator ([tex]L[/tex] above) is linear: in other words, for any two functions [tex]y_1,y_2[/tex] and two numbers [tex]a,b[/tex] we have
    [tex]L[a y_1 + b y_2] = a L[y_1] + b L[y_2] [/tex]
    So the given ODE cannot be linear, since
    [tex]L[a y_1 + b y_2] = axy_1'' + bxy_2'' + \sin(a y_1 + b y_2)[/tex]
    [tex]aL[y_1] + bL[y_2] = axy_1'' + bxy_2'' + a\sin(y_1) + b\sin(y_2)[/tex]
    which are not equal.

    Notice that if [tex]L[y] = a_0(x) y(x) + a_1(x) y' + \ldots + a_n(x) y^{(n)}(x) [/tex] then
    [tex]L[/tex] is linear by the above definition.
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