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Alternate Method for solving two Second Order DEs.

  1. Jun 22, 2007 #1
    Hello everyone

    The following two differential equations appear on page 93 of George Simmons' book on Differential Equations. While I have been able to solve them, I have some questions. [Not HW]

    I. [tex](x^2-1)y'' - 2xy' + 2y = (x^2-1)^2[/tex]
    II. [tex](x^2 + x)y'' + (2-x^2)y' - (2+x)y = x(x+1)^2[/tex]

    How I solved them:

    For the first equation, [itex]y = x[/itex] is a solution of the homogeneous part as can be verified by inspection. From this we can construct a second solution [itex]y = xv(x)[/itex] with [itex]v(x)[/itex] given by

    [tex]v(x) = \int \frac{1}{x^2}e^{\int \frac{2x}{x^2-1}dx}dx = \left(x + \frac{1}{3x^3}\right)[/tex]

    The homogeneous equation thus has the solution

    [tex]y_{h}(x) = c_{1}x + c_{2}\left(x^2 + \frac{1}{3x^2}\right)[/tex]

    For the general solution, add a polynomial and compare coefficients.

    For the second equation, [itex]y = 1/x[/itex] is a solution of the homogenous part. The second solution is [itex]y = v(x)/x[/itex] where [itex]v(x)[/itex] is given by

    [tex]v(x) = \int x^2 e^{-\int \frac{2-x^2}{x^2 + x} dx}dx = \int x^2 \left(\frac{1+x}{x^2}\right) e^{x}dx = \int (x+1)e^{x}dx = xe^{x}[/tex]

    The general solution to the homogenous part is thus

    [tex]y_{h}(x) = \frac{c_{1}}{x} + c_{2}e^{x}[/tex]

    The general solution for the nonhomogeneous equation can be constructed as for the first part.

    Now, I have two questions:

    1. I could "guess" the simple solutions to the homogeneous equations in both cases, but how do I get to this solution rigorously? (I tried some substitutions but none of them worked.)

    2. Is there some totally different way out to solve these equations too?

    Thanks.
     
    Last edited: Jun 22, 2007
  2. jcsd
  3. Jun 22, 2007 #2
    from my experience, DE is mostly guess and check
     
  4. Jun 22, 2007 #3
    Ice, I'm sure there's some other way out...the question doesn't say guess and check...
     
  5. Jun 22, 2007 #4
    Does your textbook have a method for solving equations of the form:

    DQy" + My' + Ny = DQ^2

    where D, Q, M, and N are all functions of x.

    Because both your equations are of that form. The homogenous equation for the first one is not too far off an Euler equation, so maybe that might help?

    The first one obviously has a particular solution of [tex]y = \frac{x^4}{6} + \frac{1}{2}\ [/tex], from assuming a quartic solution of [tex]y = ax^4 + bx^3 + cx^2 + dx + e[/tex]. But solving the homogenous is tricky. I'd say try a series solution.
     
    Last edited: Jun 22, 2007
  6. Jun 23, 2007 #5
    Hi morson, thanks for your reply. Yes, it does but only for a restricted class of such equations which are reducible to constant coefficient d.e.s (such as Euler's equidimensional equation). The general result is proved as a problem in the book:

    This is the only method I am aware of. In this particular case, the guessed solutions worked and so did the usual method thereafter. But no concrete methods for solving such equations have been developed so far. Of course I can solve them with the power series method, but I was wondering if there are other methods I do not know yet...

    PS (To Moderator): I think I posted this in the wrong forums, can you please shift it to the Differential Equations forum if you think thats the right place for it.
     
  7. Jun 23, 2007 #6

    siddharth

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    Hi maverick280857,

    One way of solving a second order non homogenous linear diff eqn would be by the method of variation of parameters.

    Let's say that the diff eqn you need to solve is of the form,
    [tex] \ddot{y} + a(x)\dot{y} + b(x)y = c(x)[/tex]

    First, consider the corresponding homogenous equation,
    [tex] \ddot{y} + a(x)\dot{y} + b(x)y = 0[/tex]

    First, the solution to the homogenous equation will be of the form [tex]y_h(x) = c1y_1(x) + c2y_2(x)[/tex]

    Next, the trick in the variation of parameters method is to look for a particular solution of the form,
    [tex] y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x) [/tex]
    such that,
    [tex] \begin{cases} u_1^\prime y_1 + u_2^\prime y_2 = 0 \\ u_1^\prime y_1^\prime + u_2^\prime y_2^\prime = c(x) \end{cases }[/tex]

    (Why this? Substitute [tex]y_p[/tex] back into the original eqn and check that it satisfies)

    This seems to be what you've done in your solution you posted.

    To find the solution to the homogenous eqn, you might have to "guess" it, or use the power series method. I can't think of any other way.
     
    Last edited: Jun 23, 2007
  8. Jun 23, 2007 #7
    Yup I already have :smile:

    Fine, so there is no other reduction of form for these two equations I guess...
     
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