Results 1 to 5 of 5

Math Help - DIFFERENTIAL equation

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    3

    Cool DIFFERENTIAL equation

    hi
    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2009
    Posts
    3

    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,583
    Thanks
    1418
    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; November 11th 2009 at 03:29 AM. Reason: Fixed some latex.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2009
    Posts
    3

    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 x^0 is n!
    so t sum up all the series to find a general solution ....
    will upload the result
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: May 8th 2011, 12:27 PM
  2. Replies: 1
    Last Post: April 11th 2011, 01:17 AM
  3. [SOLVED] Solve Differential equation for the original equation
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: February 21st 2011, 01:24 PM
  4. Partial differential equation-wave equation - dimensional analysis
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: August 28th 2009, 11:39 AM
  5. Replies: 13
    Last Post: May 19th 2008, 08:56 AM

Search Tags


/mathhelpforum @mathhelpforum