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2nd order differential equations with constant coeff. The Particular integrals.

  1. Jul 3, 2008 #1

    rock.freak667

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    For the differential equation

    [tex]\frac{d^2y}{dx^2}+4 \frac{dy}{dx}=sinx[/tex]


    One root of the auxiliary equation is '0' meaning the particular integral for the right hand side is x(Asinx+Bcosx). But is there any formal proof for making this claim that for 0 as one root is it is x(Asinx+Bcosx) or 0 were the two roots, the PI would be x^2(Asinx+Bcosx)?
     
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  3. Jul 3, 2008 #2

    Ben Niehoff

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    It seems to me that [itex]A \sin x + B \cos x[/itex] will work just fine. You only need the additional factor of [itex]x[/itex] if you have a double root.

    To see why it works, simply substitute [itex]y = A \sin x + B \cos x[/itex] into your equation. You should end up with

    [tex]f(A,B) \sin x + g(A,B) \cos x = \sin x[/tex]

    To get both sides to be equal for all x, you need to solve two equations in the two unknowns, A and B.
     
  4. Jul 3, 2008 #3

    rock.freak667

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    Ah dumb me I was thinking of the wrong example and made the wrong statement.

    But what I really wanted to know is if there is any proof for why a PI should be

    x(Acosax+Bsinax) when 'a' is one or both roots of the auxiliary equation (RHS=sine or cosine)
    or xe^ax for 'a' as one root and x^2e^ax for 'a' as both roots (RHS=some exp. function)

    New example:

    [tex]\frac{d^2y}{dx^2}+4 \frac{dy}{dx}+4y=6e^{-2x}[/tex]


    For this example: The PI is of the form [itex]Ax^2e^{-2x}[/itex], but how did we know that we needed to multiply by [itex]x^2[/itex]?
     
  5. Jul 3, 2008 #4

    Defennder

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    There's a rule for the choice of polynomial used in the method of undetermined coefficients. I quote this from my notes:

    Suppose [itex]r(x)=P_m(x)e^{\mu x}[/itex] is the RHS of the 2nd order linear ODE of polynomial degree m, then the DE has a particular solution of the following form:
    [tex]y= x^k Q_m (x) e^{\mu x}[/tex]
    where Qm(x) is an undetermined polynomail with degree m, k is the multiplicity of the root [itex]\mu[/itex] in the characteristic/auxiliary equation [itex]\lambda^2 + a\lambda + b = 0[/itex]. If [itex]\mu[/itex] is not a root of the equation, then k=0.

    Unfortunately I don't know how to prove that the above rule would always work.
     
  6. Jul 4, 2008 #5

    Ben Niehoff

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    Defennder:

    To prove these things, all you need to do is plug the formula in and take the derivatives. Then use induction to prove it for all orders of linear, constant-coefficient ODEs.
     
  7. Jul 4, 2008 #6

    Defennder

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    Differentiating the above twice gives me a very complicated and tedious expression to work with. Ouch.
     
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