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Math Help - nonhomogeneous 2nd order linear eqn

  1. #1
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    nonhomogeneous 2nd order linear eqn

    y''+y' -2y=2-2e^-2x ...... find general soln

    ay''+by'+cy= G(x)

    y''+y'-2y=0

    r^2+r-2=(r-1)(r+2)=0

    roots = 1 & -2

    Yc = c1 e ^x + c2 e^-2x

    Yp???? stuck!
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  2. #2
    Member kjchauhan's Avatar
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    Use method of variation of parameters,
    You get,
    y=\left( \frac{2}{9}\left(e^{-3x}-3e^{-x}\right)+c_1\right)e^x-\left(\frac{1}{3}\left(e^{2x}-x\right)+c_2\right)e^{-2x}
    Last edited by kjchauhan; December 18th 2009 at 10:18 AM.
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  3. #3
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    "Variation of parameters" always works but is typically much harder than "undetermined coefficients" when that works- which it will as long as the right hand side consists of the kind of functions we get for homogeneous linear equations with constant coefficients, polynomials, exponentials, sine and cosine, or products of those.

    Here, your right hand side is 2- 2e^{-2x}. Because of the constant "2", try some constant A. Because of the " -2e^{-2x}", your first thought should be " Be^{-2x}". However, e^{-2x} is already a solution to the associated homogeneous equation so try " Bxe^{-2x}" instead.

    In other words, try Y_p(x)= A+ Bxe^{-2x}.
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  4. #4
    Member kjchauhan's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    "Variation of parameters" always works but is typically much harder than "undetermined coefficients" when that works- which it will as long as the right hand side consists of the kind of functions we get for homogeneous linear equations with constant coefficients, polynomials, exponentials, sine and cosine, or products of those.

    Here, your right hand side is 2- 2e^{-2x}. Because of the constant "2", try some constant A. Because of the " -2e^{-2x}", your first thought should be " Be^{-2x}". However, e^{-2x} is already a solution to the associated homogeneous equation so try " Bxe^{-2x}" instead.

    In other words, try Y_p(x)= A+ Bxe^{-2x}.
    Ok..That also works..
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  5. #5
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    could somebody show me the next line combining yp to yc
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  6. #6
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    i am now at 10 posts.
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  7. #7
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    yp'= -2(A+Bxe^-2x)
    yp''= 4(A+Bxe^-2x)

    therefore from orginal equation
    y''+y' -2y

    4(A+Bxe^-2x) + -2(A+Bxe^-2x) - 2*(A+Bxe^-2x) = 2-2e-2x

    ????
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  8. #8
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    Quote Originally Posted by harveyo View Post
    yp'= -2(A+Bxe^-2x)
    What are you taking as yp? I suggested y_p= A+ Bxe^{-2x} but that certainly is not its derivative.

    If y_p= A+ Bxe^{-2x}, then y_p'= 0+ Be^{-2x}- 2Bxe^{-2x}

    yp''= 4(A+Bxe^-2x)
    Even with " y_p'= -2(A+ Bxe^{-2x})" this would not be the second derivative. You are aware that the derivative of -2A is 0, aren't you?

    You should have y_p"= -4Be^{-2x}+ 4Bxe^{-2x}

    therefore from orginal equation
    y''+y' -2y

    4(A+Bxe^-2x) + -2(A+Bxe^-2x) - 2*(A+Bxe^-2x) = 2-2e-2x

    ????
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  9. #9
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    makes complete sense now.
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