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Homogeneous Differential Equation

  1. Mar 22, 2006 #1


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    Solve the initial value problem
    y''' - 5y'' + 100y' - 500y = 0; y(0) = 0, y'(0) = 10, y''(0) = 250
    given that y_1(x) = e^5x is one particular solution of the differential equation.
    r^3 - 5r^2 + 100r - 500 = 0
    r = +/- 10i or 5
    complementary solution y_c = e^5x(c1cos10x + c2sin10x)
    general solution y_g = y_c + y_p = e^5x(c1cos10x + c2sin10x + 1)
    y(0) = c1 + 0 + 1 = 0
    c1 = -1
    y'(0) = 5e^5x((-1)cos10x + c2sin10x) + e^5x(-10(-1)sin10x + 10c2cos10x) = 10

    5*(-1) + 10(0) + 10c2 = 10
    10c2 = 15
    c2 = 3/2
    y(x) = e^5x(-cos10x + (3/2)sin10x + 1)

    The book's answer is 2e^5x - 2cos10x. How did they get that? And do I need y''(0) = 250? I have one real root and 2 complex roots, so
    e^ax(c1cosbx + c2sinbx) should work. And I don't need y'' because I don't have a c3 to solve for, yes? What am I doing wrong?
  2. jcsd
  3. Mar 22, 2006 #2


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    You are given three initial conditions: one to determine each parameter c1, c2, and c3; the last of these you have left out somewhere: look for it in your y_g.
  4. Mar 22, 2006 #3


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    I have one real root and 2 complex roots. Euler's formula e^ax(c1cosbx + c2sinbx) should suffice. e^ax does not get a c_n coefficient. There is also no c coefficient in y_1(x) = e^5x.

    If I multiply e^5x by c3 to make y_g = e^5x(c1cos10x + c2sin10x) + c3e^5x
    I get:
    [c1 + 0 + c3 = 0]
    [5c1 + 10c2 + 5c3 = 0]
    [125c1 + 100c2 + 25c3 = 250]

    After row reducing, I get c1 = 3/2, c2 = 1, c3 = -3/2
    These coefficients bring me nowhere near the book's answer. Are you sure there is a c3 in this differential equation? I wouldn't know in which term it would be.
  5. Mar 22, 2006 #4


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    You've misunderstood.
    You have one solution for each of the roots:
    [tex]C_{0}e^{ax}, C_{1}\cos(bx), C_{2}\sin(bx), a=5, b=10[/tex]
  6. Mar 22, 2006 #5
    Do you really need a particular solution to find the general solution of a _homogeneous_ diff. equation? I think it's useful only for non-homogeneous ones. You should review that part.

    If the roots of the characteristic equation are +10i, -10i and 5, then the general solution for your homogeneous diff. equation has the form
    [itex] y = C_1 e^{5x} + C_2 e^{10ix} + C_3 e^{-10ix} [/itex]
    Differentiating and applying the initial conditions I get that [itex] C_1 + C_2 + C_3 = 0 [/itex], [itex] 5 C_1 + 10i C_2 - 10i C_3 = 10 [/itex] - and this is enough to see that [itex] C_2 = C_3 [/itex] and thus [itex] C_1 = 2, C_2 = C_3 = -1 [/itex]
    Now using the the (Euler?) identity [itex] e^{ix} = \cos x + i \sin x [/itex], you can simplify [itex] y = 2e^{5x} - e^{10ix} - e^{-10ix} = 2e^{5x} - \cos 10x - i \sin 10x - \cos 10x + i \sin 10x = 2e^{5x} - 2\cos10x [/itex], which agrees with your book's solution.

    Last edited: Mar 22, 2006
  7. Mar 22, 2006 #6


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    You do not multiply the solution e5t by the other solutions, you form a linear combination!

    The general solution is, as vladb said,
    [itex] y = C_1 e^{5x} + C_2 e^{10ix} + C_3 e^{-10ix} [/itex]
    or, equivalently
    [tex]y= C_1 e^{5x}+ C_2 cos(10x)+ C_3 sin(10x)[/tex]

    [tex]e^{5x}(C_1 cos(10x)+ C_2 sin(10x))[/tex]
    would be appropriate for characteristic solutions 5+ 10i and 5- 10i.

    vladb, "particular solution" does not necessarily apply to non-homogeneous equations, it simply means one (not the general) solution to an equation. Here, of course, it tells us one solution to the cubic characteristic equation, allowing us to reduce to a simple quadratic.
  8. Mar 22, 2006 #7
    I know that, that's why I said "useful". I only wanted to emphasize that one doesn't necessarily need "particular solutions" to find the general solution of a homogen. diff. equation. (And this is where OP's confusion appeared first, imho). And finding a particular solution y = e^5x is a problem of the same difficulty as guessing a root r=5 of the characteristic polynomial.
  9. Mar 22, 2006 #8


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    Thank You!
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