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Math Help - 2nd Order Differential Equations

  1. #1
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    2nd Order Differential Equations

    Hello,

    Second order differential equation:

    y'' - y = exp[x]

    Solving for the complementary function, y_c:

    m^2 - 1 = 0
    -> m = 1 and m = 0

    :. y_c = A*exp[-1] + B*exp[0] = A*exp[-1] + B

    Solving for the particular integral, y_p:
    Try y = a*exp[x] -> this is a solution of the homogeneous eqn, so multiply the function by x:

    y = ax*exp[x]
    y' = a*exp[x] + ax*exp[x]
    y'' = a*exp[x] + a*exp[x] + ax*exp[x]

    Substituting into y'' - y' = x*exp[x], the equation reduces to:

    a + a*x = x -> How to work the constant (alpha) out?

    Thank you
    Last edited by algorithm; February 16th 2009 at 12:35 PM.
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  2. #2
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    Also, for the differential equation:

    y'' - y = Sin(2x)sin(x)

    How would one find the particular integral?

    Thank you
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  3. #3
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    Quote Originally Posted by algorithm View Post
    Hello,

    Second order differential equation:

    y'' - y = exp[x]

    Solving for the complementary function, y_c:

    m^2 - 1 = 0
    -> m = 1 and m = 0 (*)

    :. y_c = A*exp[-1] + B*exp[0] = A*exp[-1] + B (**)

    Solving for the particular integral, y_p:
    Try y = a*exp[x] -> this is a solution of the homogeneous eqn, so multiply the function by x:

    y = ax*exp[x]
    y' = a*exp[x] + ax*exp[x]
    y'' = a*exp[x] + a*exp[x] + ax*exp[x]

    Substituting into y'' - y' = x*exp[x], (***) the equation reduces to:

    a + a*x = x -> How to work the constant (alpha) out?

    Thank you
    In the step above (*), [tex]m = \pm 1[tex] and in this step (**), it should read

    y_c = A e^x + B e^{-x}.

    And finally in this step (***), you should get from your original ODE

    y'' - y = 2a e^x + ax e^x - ax e^x = 2ae^x = e^x giving a = \frac{1}{2} so your solution is

    y = A e^x + B e^{-x} + \frac{1}{2} x e^x
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  4. #4
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    Quote Originally Posted by algorithm View Post
    Also, for the differential equation:

    y'' - y = Sin(2x)sin(x)

    How would one find the particular integral?

    Thank you
    Use the sum/difference formula for \sin x \sin 2x.
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  5. #5
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    Hello,

    Thank you for the help.

    This is another ODE:

     y'' + y' = x^{2},
    conditions: y(0) = 0, y'(0) = 0

    Complementary function, y_c:
    m^{2} + m = 0
    m = 0, -1

    y_c = (C1 + C2x)e^{-x}

    Particular integral y_p:
    Try y_p = Ax^{2} + Bx + C
    y_p' = 2Ax + B
    y_p'' = 2A

    => 2A + 2Ax + B = x^{2}
    2A + B = 0, and 2A = 0, giving A = B = 0.

    y = C1 + C2e^{-x}
    y' = -C2e^{-x}
    0 = C1 + C2
    0 = -C2
    C1 = C2 = 0, giving the final solution:
    y = 0

    The correct answer has more x's.
    Thank you
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  6. #6
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    Quote Originally Posted by algorithm View Post
    Hello,

    Thank you for the help.

    This is another ODE:

     y'' + y' = x^{2},
    conditions: y(0) = 0, y'(0) = 0

    Complementary function, y_c:
    m^{2} + m = 0
    m = 0, -1

    y_c = (C1 + C2x)e^{-x}

    Particular integral y_p:
    Try y_p = Ax^{2} + Bx + C this C is part of your complimentary solution
    y_p' = 2Ax + B
    y_p'' = 2A

    => 2A + 2Ax + B = x^{2}
    2A + B = 0, and 2A = 0, giving A = B = 0.

    y = C1 + C2e^{-x}
    y' = -C2e^{-x}
    0 = C1 + C2
    0 = -C2
    C1 = C2 = 0, giving the final solution:
    y = 0

    The correct answer has more x's.
    Thank you
    Your complimentary solution is

    y = c_1 + c_2 e^{-x}

    Since part of this complimentary is x^0, for your guess you'll need t obump up your particualr to

    y_p = Ax^3 +Bx^2 + Cx
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