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Math Help - Nonhomogeneous Linear Differential Equation

  1. #1
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    Nonhomogeneous Linear Differential Equation

    I have a problem similar to one already posted here which has been bothering me as well.

    (x-1)y'' - xy' + y = (x-1)^2; y1=x, y2=e^x

    I need some help figuring out the general solution via variation of parameters, I am finding it quite troublesome matching it up with the answer the book gives. Some insight would be helpful.
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  2. #2
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    Quote Originally Posted by Jukodan View Post
    I have a problem similar to one already posted here which has been bothering me as well.

    (x-1)y'' - xy' + y = (x-1)^2; y1=x, y2=e^x

    I need some help figuring out the general solution via variation of parameters, I am finding it quite troublesome matching it up with the answer the book gives. Some insight would be helpful.
    Okay, it is easy to check that y1 and y2 really are independent solutions to the homogeneous equation (and you should always check).

    We look for a solution to the entire equation of the form y(x)= u(x)x+ v(x)e^x. Then y'= u'x+ u+ v'e^x+ ve^x. There are, in fact, an infinite number of such functions, u(x) and v(x), (given a solution y, pick any function to be u(x) and solve y(x)= u(x)x+ v(x)e^x for v), we can simplify, and narrow the search, by requiring that u' x+ v' e^x= 0. That reduces y' to y'= u+ ve^x so y"= u'+ v'e^x+ ve^x and the equation becomes
    (x- 1)(u'+ v'e^x+ ve^x)- x(u+ ve^x)+ ux+ ve^x= (x- 1)^2
    xu'+ xv'e^x+ xve^x- u'- v'e^x- ve^x- xu- xve^x+ xu+ ve^x = xu'+ xv'e^- u'- v'e^x= (x-1)^2
    Our clever requirement that u'x+ v'e^x= 0 means that there are no u" or v" terms in the equation and the fact that y1 and y2 satisfy the homogeneous equation means that there are no u or v terms! We have, now, two equations, (x-1)u'+ (x-1)e^xv'= (x-1)^2 and xu'+ e^xv'= 0 to solve for u' and v'. For the first, we can divide through by x-1 to get u'+ e^xv'= x-1. Solve those equations, algebraically, for u' and v', then integrate to find u and v.
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  3. #3
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    for a2 y" + a1 y ' +a0 y = f(x)

    then with variation of parameters the general solution is

    1) y= y1 z1 + y2 z2

    where y1 and y2 are solutions to the homog eqn and W(y1,y2) is the wronskian.
    In your pblm W(x,e^x) = e^x (x-1)

    and z1 ' = - y2f(x)/(a2W(y1,y2))

    z2 ' = y1 f(x)/(a2W(y1,y2))

    In your case z1 ' = - 1 yields z1 = -x + C1

    z2 ' = xe^(-x) yields z2 = -x*e^(-x)- e^(-x) + C2

    plug into 1) to obtain general solution
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