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Math Help - Another head scratcher

  1. #1
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    Another head scratcher

    I am trying to set up the appropriate particular solution for this differential equation but don't know where to start. It doesn't even seem like a ode since I only have y terms on the LHS and no derivatives of y.

    Any help would be appreciated just to set this up. Thanks

    (D-1)^3(D^2-4)y=xe^x +e^2x +e^-2x

    Last edited by doke; November 1st 2006 at 01:43 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by doke View Post
    I am trying to set up the appropriate particular solution for this differential equation but don't know where to start. It doesn't even seem like a ode since I only have y terms on the LHS and no derivatives of y.

    Any help would be appreciated just to set this up. Thanks

    (D-1)^3(D^2-4)=xe^x +e^2x +e^-2x

    If D is supposed to be a differential operator, we need something like
    (D-1)^3(D^2-4)y

    If not, what is D?

    -Dan
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  3. #3
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    Clarification

    Yes, I'm sorry. There is a y at the end of the parenthesis. I just edited my post.
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  4. #4
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    Hello, doke!

    Are you in self-study?
    This notation should have been explained to you in class.


    (D-1)^3(D^2-4)y \:=\:xe^x + e^{2x} +e^{-2x}

    When using the "Method of Operators", they often use this form of notation.

    Dy means \frac{dy}{dx} . . . D^2y means \frac{d^2y}{dx^2} . . . D^3y means \frac{d^3y}{dx^3}


    Example

    An equation like \frac{dy}{dx} - 2y \;= \;x can be written: . Dy - 2y \;=\;x

    Then we can factor out the y (ha!): . (D - 2)y \;=\;x

    To solve for y, we would "divide by (D - 2)"

    A really silly thing to say because \frac{x}{D - 2} has no meaning . . . or does it?

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Suppose we have the equation: . \frac{dy}{dx} - ay \:=\:P(x)

    We can solve it with an integrating factor: . I \:=\:e^{\int \text{-}a\,dx} \:=\:e^{\text{-}ax}

    Multiply through by I:\;\;e^{\text{-}ax}\frac{dy}{dx} + ae^{\text{-}ax}y \;=\;e^{\text{-}ax}P(x)

    The left side is the derivative of a product: . \frac{d}{dx}\left(e^{\text{-}ax}y\right) \;=\;e^{\text{-}ax}P(x)

    Integrate: . e^{\text{-}ax}y \;=\;\int e^{\text{-}ax}P(x)\,dx

    Therefore: . \boxed{y \;= \;e^{ax}\int e^{\text{-}ax}P(x)\,dx}


    The same equation can be written: . Dy - ay \;=\;P(x)

    . . which gives us: . (D - a)y \;=\;P(x)\quad \Rightarrow \quad y \;=\;\boxed{\frac{P(x)}{D-a}}


    We have now defined that 'rather silly' statement:

    . . . . . \boxed{\boxed{\frac{P(x)}{D - a} \;= \;e^{ax}\int e^{-ax}P(x)\,dx}}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Back to our example . . .

    We had: . (D - 2)y\:=\:x\quad\Rightarrow\quad y \:=\:\frac{x}{D - 2}

    By definition: . y \;\;= \;\;\frac{x}{D - 2} \;\;= \;\;e^{2x}\!\!\int xe^{-2x}dx

    Integrate by parts and get: . y \;= \;e^{2x}\left[-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + C\right]

    Therefore: . y \;= \;-\frac{1}{2}x - \frac{1}{4} + Ce^{2x}\quad\Rightarrow\quad \boxed{y \;= \;-\frac{1}{4}\left(2x + 1 + Ce^{2x}\right)}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Now, your problem is inhumanely long and difficult!

    You have: . (D-1)^3(D-2)(D+2)y \:=\:xe^x + e^{2x} +e^{-2x}

    You have: P(x) \:=\:xe^x + e^{2x} + e^{-2x}

    And you are expected to "divide by D - 1" three times
    . . then "divide by D - 2", then "divide by D + 2".

    . . . I'll wait in the car.

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  5. #5
    Forum Admin topsquark's Avatar
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    (D-1)^3(D^2-4)y = xe^x + e^{2x} + e^{-2x}

    We need a particular solution to this. Note that:
    (D-1)^3(D^2-4) = D^5 + \alpha D^4 + ... + \beta D + 4

    To get a homogeneous solution for this differential equation, note that all of the coefficients of the D operator are constants. This means that the homogeneous solutions are going to be of the form:
    y_h(x) = Ae^{ax} + Be^{bx} + ...
    where the a, b, ... solve the characteristic polynomial.

    As to the solution of the characteristic polynomial, we know them, since the problem gives us a factored version of the operator. The characteristic polynomial is:
    (m - 1)^3(m^2 - 4) = 0.
    This has roots at the three cube roots of 1 and at m = 2, -2.

    This suggests the following form for the particular solution:
    y_p(x) = Cxe^x + Dxe^{2x} + Exe^{-2x}

    We need the first term to produce the xe^x on the RHS of the differential equation. We need the second and third terms the way they are in order to produce the e^{2x} + e^{-2x}. We need to multiply by the extra "x" in this case because both e^{2x} and e^{-2x} are already in the homogeneous solution.

    I hope I explained that well enough!

    -Dan
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