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  1. #1
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    second order linear nonhomogenous

    3>Find the general solution for y” + y’=e^ x –sin(x)

    5.Given that y=e ^(–x) is a solution for the differential equation xy” +(x-1) y’ – y=0,
    find its general solution
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by harry View Post
    3>Find the general solution for y + y=e^ x sin(x)
    First find the homogeneous solution:
    yh'' + yh' = 0

    The characteristic equation is:
    m^2 + m = 0
    m(m + 1) = 0

    m = 0 or m = -1

    Thus
    yh(x) = A + Be^{-x}

    Now for the particular solution. Since this is a linear equation we may propose a form:
    yp = yp1 + yp2
    where yp1 produces the -sin(x) term and yp2 produces the e^{x} term.

    I would suggest trying:
    yp(x) = Csin(x) + Dcos(x) + Ee^{x}

    Thus
    yp'(x) = Ccos(x) - Dsin(x) + Ee^{x}

    yp''(x) = -Csin(x) - Dcos(x) + Ee^{x}

    So
    yp1'' + yp1' = (C - D)cos(x) + (-C - D)sin(x) + 2Ee^{x} = -sin(x) + e^{x}

    Thus
    C - D = 0
    -C - D = -1
    2E = 1

    The top equation says D = C, so the middle equation says:
    -2C = -1
    Thus D = C = 1/2 so

    yp(x) = 1/2 * (sin(x) + cos(x)) + e^{x}

    Thus the full solution will be:
    y(x) = A + Be^{-x} + 1/2 * (sin(x) + cos(x)) + 1/2 * e^{x}

    Check:
    y'(x) = -Be^{-x} + 1/2 *(cos(x) - sin(x)) + 1/2 * e^{x}

    y''(x) = Be^{x} + 1/2 *(-sin(x) - cos(x)) + 1/2 * e^{x}

    Adding these up:
    y''(x) + y'(x) = -sin(x) + e^{x} (Check!)

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by harry View Post
    5.Given that y=e ^(x) is a solution for the differential equation xy +(x-1) y y=0,
    find its general solution
    If I've got this right (I had to look it up. It's been a while!) This is the method.

    We have one solution y1(x) = e^{-x} to this differential equation. We posit that there is another solution y2(x) = v(x)*y1(x) = e^{-x} * v(x) where v(x) is unknown at this point. So put this into the differential equation:
    y2 = e^{-x}*v
    y2' = -e^{-x}*v + e^{-x}*v'
    y2'' = e^{-x}*v - 2e^{-x}*v' + e^{-x}*v''

    So
    x*[e^{-x}*v - 2e^{-x}*v' + e^{-x}*v''] + (x - 1)*[-e^{-x}*v + e^{-x}*v'] - [e^{-x}*v] = 0

    Or
    x*e^{-x}*v'' - (x + 1)*e^{-x}*v' = 0

    x*v'' - (x + 1)v' = 0

    Now reduce the order. Let z(x) = v'(x). Then:
    x*z' - (x + 1)*z = 0

    z' -[(x + 1)/x]z = 0

    This is a first order homogeneous linear differential equation. To put it mildly, it's been studied. The general solution of z' + p(x)*z = q(x) is:
    z(x) = {Int[u(x)*q(x)dx] + C}/u(x)

    with u(x) = exp{Int[u(x)dx]}

    Here we have p(x) = -(x + 1)/x and q(x) = 0.

    u(x) = exp{-Int[(x + 1)/x dx]} = exp{-x - ln(|x|)}

    u(x) = (1/|x|)*e^{-x}

    So
    z(x) =C*|x|*e^{x}

    I'm going to do something bad so I can finish this problem. I'm going to drop the absolute value bars on x and let
    z(x) = C*x*e^{x}

    Then
    v'(x) = z(x) = C*x*e^{x}

    v(x) = C*Int[x*e^{x} dx] = C*(x - 1)*e^{x}

    So
    y2(x) = v(x)*e^{-x} = C*(x - 1)

    You can verify that this indeed solves the given differential equation.

    So the general solution of x*y'' + (x - 1)y' - y = 0 will be arbitrary constants times our two solutions:
    y(x) = A*e^{-x} + B*(x - 1)

    (The justification for dropping the absolute value bars is that this does generate an acceptable solution. I have no other intelligent comment about it, except that it was necessary to generate a solution.)

    -Dan
    Last edited by topsquark; April 7th 2007 at 02:18 PM.
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