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Thread: DIFFERENTIAL equation

  1. #1
    Nov 2009

    Cool DIFFERENTIAL equation

    can someone please help me with the following equation

    y''-y= x^n

    using method of separation of variables...

    the roots come out to be 1,-1

    so when i find variable parameters A(X) and B(x)
    i get stuck

    please guide me.
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  2. #2
    MHF Contributor chisigma's Avatar
    Mar 2009
    near Piacenza (Italy)
    If we write the DE in terms of LT we have...

    s^{2}\cdot Y(s) -s\cdot y(0) -y^{'}(0) - Y(s) = \frac{n!}{s^{n+1}} (1)

    ... and from it we obtain...

    Y(s)= \frac{s\cdot y(0)+y^{'}(0)}{s^{2}-1} + \frac{n!}{s^{n+1}\cdot (s^{2}-1)} (2)

    The solution of the DE can be [formally...] obtained computing the inverse LT of (2) and is...

    y(t)= y(0)\cdot \cosh x + y^{'}(0)\cdot \sinh x + \int_{0}^{x} (x-u)^{n}\cdot \sinh u\cdot du (3)

    Kind regards

    \chi \sigma
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  3. #3
    Nov 2009

    Post about the solution above

    Many thanks for the reply ...
    but instead of using the Laplace Method...we are supposed to do it via variation of parameters...
    and when i solve for the parameters i get the integral

    integralx^n*e^x dx
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  4. #4
    MHF Contributor

    Apr 2005
    Variation of parameters:
    The general solution to the associated homogeneous equation, y"- y= 0, is, as you say, Ce^x+ De^{-x}.

    In "variation of parameters" we look for a solution of the form y(x)= u(x)e^x+ v(x)e^{-x}. Note that there are an infinite number of such solutions. In fact, given any solution, Y(x), we could just take u(x) identically 0 and then solve for v(x) or, conversely, take v(x) identically 0 and solve for u(x).

    In any case, if y(x)= u(x)e^x+ v(x)e^{-x} then y'= u'e^x+ ue^x+ v'e^{-x}- ve^{-x}. Since there are an infinite number of "u" and "v" that will work, we can "narrow the search" and simplify by requiring, also, that u'e^x+ v'e^{-x}= 0.

    With that we have y'= ue^x- ve^{-x} and, differentiating again, y"= u'e^x+ ue^x- v'e^{-x}+ ve^{-x}. Putting that into the differential equation, y"- y= u'e^x+ ue^x- v'e^{-x}+ ve^{-x}- (ue^x+ ve^{-x})= u'e^x- v'e^{-x}= x^n.

    Our requirement that " u'e^x+ v'e^{-x}= 0 means that we will have no u" or v" and the fact that e^x and e^{-x} satisfy the homogeneous equation means there is no u or v. We have only u' and v' in the equation.

    Now we can solve the two equations u'e^x+ v'e^{-x}= 0 and u'e^x- v'e^{-x}= x^n "algebraically" for u' and v'. Adding the two equations gives 2u'e^x= x^n so u'= \frac{1}{2}x^n e^{-x}. Subtracting the second equation from the first gives 2v'e^{-x}= -x^n so v'= -\frac{1}{2}x^ne^x.

    Integrate to find u(x) and v(x). You can do those integrals by integration by parts, taking u= x^n and dv= e^xdx in the one case and dv= e^{-x}dx in the other. That will reduce to an integral in x^{n-1}. Repeat until you have completely reduced that to x^0. After a few example, you should be able to find a formula for the general case.

    I will confess that I would have done this a third way, using "undetermined coefficients". Try a specific solution of the form y(x)= A_nx^n+ A_{n-1}x^{n-1}+ \cdot \cdot \cdot + A_1 x+ A_0, put that into the differential equation and solve for the coefficients.
    Last edited by mr fantastic; Nov 11th 2009 at 03:29 AM. Reason: Fixed some latex.
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  5. #5
    Nov 2009

    Post Reduction formula


    thanks a lot HallsofIvy for the post...and certainly the part solved is almost same as i did except the last part...and yes i am trying to do the integral...its like series of n n-1 n-2 n-3 and in the end for the power of x^0 is n!
    so t sum up all the series to find a general solution ....
    will upload the result
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