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Homework Help: Linear Equation with Constant Coefficients Problem

  1. Sep 20, 2005 #1
    Find all solutions [tex] \phi [/tex] of [tex] y''+y=0 [/tex] satisfying:
    1) [tex] \phi(0)=1, \phi(\pi/2)=2 [/tex]
    2) [tex] \phi(0)=0, \phi(\pi)=0 [/tex]
    3) [tex] \phi(0)=0, \phi'(\pi/2)=0 [/tex]
    4) [tex] \phi(0)=0, \phi(\pi/2)=0 [/tex]

    I cannot seem to solve parts 2-4 in a way that would result in the coefficients being any constant other than zero. Whenever I solve for one of the constants in the equation [tex] \phi(0)=0 [/tex] , I find that [tex] C_1=-C_2 [/tex] ; which isn't alarming yet since the magnitude of either constant is not yet known. When I try to solve the second property in each of the subsequent equations, I find that [tex] C_2=0 [/tex] , for example, which implies that [tex] C_1=0 [/tex] since [tex] C_1=-C_2 [/tex] . Below, I have included a portion of the work that I completed to get to the current situation that I am in. Any assistance offered would be much appreciated. Thank you for your time.

    [tex] y''+y=0 [/tex] .
    Characteristic Polynomial: [tex] p(r)=r^2+1=(r-i)(r+i) [/tex]
    [tex] \implies [/tex] Roots [tex] =\pm i [/tex] .
    [tex] \therefore \phi(x)=C_1e^ {ix} +C_2e^ {-ix} [/tex] where [tex] C_1 [/tex] & [tex] C_2 [/tex] are constants.
    [tex] \phi(0)=C_1e^0+C_2e^0=C_1+C_2=1 \implies C_1=1-C_2 [/tex] .
    [tex] \phi(\pi/2)=C_1e^ {\pi i/2} +C_2e^{ -\pi i/2} =(1-C_2)e^ {\pi i/2} +C_2e^ {-\pi i/2} =\cdots=(1-2C_2)i=2 \implies C_2=(\frac 1 2 +i) \implies C_1=(\frac 1 2 -i ) [/tex].
    [tex] \therefore \phi(x)=(\frac 1 2 -i)e^ {ix} +(\frac 1 2 +i)e^ {-ix} =\cdots=cos(x)+2sin(x) [/tex] .
     
  2. jcsd
  3. Sep 20, 2005 #2

    George Jones

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    The first condition of 2) gives, as you say, [itex]C_1 = -C_2[/itex]. Using this and the second condition of 2), I get that [itex]C_2[/itex] can be anything.

    Regards,
    George
     
  4. Sep 20, 2005 #3
    b) y=(c1)sinx
    c) y=(c1)sinx
    d) y=o

    (c1)= any constant

    For b and c, I get (c2)=0, so (c1) can be any constant and there is no cosx term
    For d, I get both (c1) and (c2) are 0, so the only soln. is y=0
     
  5. Sep 20, 2005 #4

    George Jones

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    Careful - I think your C's are different JM00404's C's. This doesn't affect the final answers, but it does affect the values of the C's.

    JM00404 has chose C's such that

    [tex]\phi(x) = C_1 e^{ix} + C_2 e^{-ix}[/tex].

    I think your C's are such that

    [tex]\phi(x) = C_1 sinx + C_2 cosx[/tex].

    Regards,
    George
     
  6. Sep 20, 2005 #5
    You're right

    You're right. The general solution to the equation that I am using is

    y(x) = (c1)sinx + (c2)cosx
     
  7. Sep 20, 2005 #6
    Lets look @ (2). we have a general solution y=(c1)sinx+(c2)cosx, the first restriction y(0)=0, inplies that c2=0. The second restriction y(Pi)=0 does nothing because for any value of c1, sinx will be zero at Pi. so the solution is a bunch of sine curves with any amplitude. This is a good example of a second order d.e. that has 2 boundary conditions that don't determine a unique solution.
     
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