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Inhomogeneous differential equation, boundary

  1. Oct 18, 2009 #1
    y"+by'+c=0
    y(0)=y(1)=0
     
  2. jcsd
  3. Oct 19, 2009 #2
    Actually, the equation only implies y', so the equation for z=y' is first order and can be directly integrated by variation of constants (assume the same solution as for the homogeneous equation, z=C*exp(-b/a *x), and the assume C is a function of x). solve for C, and then you will get z=y'. integrate this and you will get y
     
  4. Oct 19, 2009 #3

    HallsofIvy

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    Do you have a question about this? It looks like homework to me and so belongs in the homework section. What have YOU done on it. It is a relatively simple linear equation with constant coefficients. What is the associated homogeneous equation? What is its characteristic equation?
     
  5. Oct 19, 2009 #4
    Thanks a million Gato!
    ==========================================
    No, Hall. It's not homework. Probably the terse manner in which I plastered on the board gave that impression.
     
  6. Oct 19, 2009 #5
    This is what I had done previously:

    Take the homogeneous side:y"+by'=0
    Condition: y is never zero.
    General solution for homogeneous: y = e^mx
    Substituting: (m^2+bm)e^mx = 0
    e^mx is never zero.
    Then: m^2 + bm = 0 (characteristic equation)
    b never zero (condition above).
    Then: m(m+b)=0; m=0; m=-b
    y=c1 + c2e^(-bx)
    y(0)=0; then, c1 + c2 = 0
    y(1)=0; then, c2e^(-b)= 0

    I have a non-trivial solution only if the det: [1 1]
    [0 e^(-b)]
    {which is equal to e^(-b)} is zero.
    If not, then y=0 (trivial solution)

    But this is a Reynold's equation (fluid dynamics) and the function cannot be zero throughout
     
    Last edited: Oct 19, 2009
  7. Oct 20, 2009 #6

    HallsofIvy

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    the theorem you state, that there is a non-trivial solution only if the Jacobian is zero is only for homogeneous equations. The trivial solution is always a solution to homogeneous equations so there is a non-trival solution only if the solution is not unique. That is not the case for non-homogeneous equations as you have here.

    Okay, the characteristic equation is [itex]r^2+ br= r(r+b)= 0[/itex] and so the characteristic roots are 0 and -b. The general solution to the associated homogeneous equation is [itex]y= A+ Be^{-bt}[/itex].

    Since a constant already satisfies the homogeneous equations, try a particular solution of the form Pt. Then y'= P, y"= 0 so y"+ y'= bP= c. P= c/b and the particular solution is (c/b)t.

    The general solution to the entire equation, then, is y(t)= A+ Be^{-bt}+ (c/b)t.

    Setting t= 0, y(0)= A= 0.

    Setting t= 1, [itex]y(1)= Be^{-b}+ (c/b)= 0[/itex] so [itex]Be^{-b}= -c/b[/itex] and [itex]B= (-c/b)e^{b}[/itex].

    [itex]y(t)= (-c/b)e^{b}e^{-bt}+ (c/b)t= (c/b)(t- e^{b(1- t)})[/itex] is the only solution to that problem.
     
    Last edited: Oct 20, 2009
  8. Oct 20, 2009 #7
    Thanks Hall. I had stopped at the homogeneous because I didn't think of the Pt type solution of the inhomogeneous.
    However, for t=0, y(0)=0= A+B , instead of A (as you wrote). (1/(e^0)=1)
    Right?
    Or have I lost all my faculties?
     
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