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Thread: DIFFERENTIAL equation

  1. #1
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    Cool DIFFERENTIAL equation

    hi
    can someone please help me with the following equation

    y''-y=$\displaystyle x^n$

    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
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    If we write the DE in terms of LT we have...

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

    ... and from it we obtain...

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    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

    $\displaystyle integralx^n*e^x dx$
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  4. #4
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    Variation of parameters:
    The general solution to the associated homogeneous equation, y"- y= 0, is, as you say, $\displaystyle Ce^x+ De^{-x}$.

    In "variation of parameters" we look for a solution of the form $\displaystyle y(x)= u(x)e^x+ v(x)e^{-x}$. 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 $\displaystyle y(x)= u(x)e^x+ v(x)e^{-x}$ then $\displaystyle y'= u'e^x+ ue^x+ v'e^{-x}- ve^{-x}$. Since there are an infinite number of "u" and "v" that will work, we can "narrow the search" and simplify by requiring, also, that $\displaystyle u'e^x+ v'e^{-x}= 0$.

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

    Our requirement that "$\displaystyle u'e^x+ v'e^{-x}= 0$ means that we will have no u" or v" and the fact that $\displaystyle e^x$ and $\displaystyle e^{-x}$ 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 $\displaystyle u'e^x+ v'e^{-x}= 0$ and $\displaystyle u'e^x- v'e^{-x}= x^n$ "algebraically" for u' and v'. Adding the two equations gives $\displaystyle 2u'e^x= x^n$ so $\displaystyle u'= \frac{1}{2}x^n e^{-x}$. Subtracting the second equation from the first gives $\displaystyle 2v'e^{-x}= -x^n$ so $\displaystyle v'= -\frac{1}{2}x^ne^x$.

    Integrate to find u(x) and v(x). You can do those integrals by integration by parts, taking $\displaystyle u= x^n$ and $\displaystyle dv= e^xdx$ in the one case and $\displaystyle dv= e^{-x}dx$ in the other. That will reduce to an integral in $\displaystyle x^{n-1}$. Repeat until you have completely reduced that to $\displaystyle x^0$. 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 $\displaystyle y(x)= A_nx^n+ A_{n-1}x^{n-1}+ \cdot \cdot \cdot + A_1 x+ A_0$, 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
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    Post Reduction formula

    wow!!!

    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 $\displaystyle x^0$ is n!
    so t sum up all the series to find a general solution ....
    will upload the result
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