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Second order homogenous with variable coeffecients

  1. Mar 15, 2007 #1
    In general how do we deal with linear second order differential equations with variable coeffecients?
     
  2. jcsd
  3. Mar 16, 2007 #2
    In general, you can't. You have to specify the form of those coefficients and hope to match it to an equation for which there is a known solution, probably involving a substitution or two along the way. (Unless it lends itself to direct integration, or something.)

    Which equation are you concerned with?
     
    Last edited: Mar 16, 2007
  4. Mar 16, 2007 #3
    why can't you just plug in in e^rt like you do with the constant coefficient ones and just let r be variable(guessing)? For example when you plug e^rt into the homogenous equation you can find r by solving a quadratic equation involving the coffecient. You need to solve for r. however since in this case the coeeficients depend on time, can't we just have variable solutions (r). note that I haven't thought about this much.

    I was just curious as the notes I found didn't discuss them much.
     
    Last edited: Mar 16, 2007
  5. Mar 16, 2007 #4
    Let's say your equation is

    [tex]y^{\prime \prime} + \alpha(t) y^{\prime} + \beta(t) y = 0[/tex]

    where alpha and beta are arbitrary functions. Now, try your method of subbing

    [tex]y(t) = e^{r(t)} [/tex]

    bung this into the first equation, and you get

    [tex]r^{\prime \prime} e^{r(t)} + r^{\prime 2} e^{r(t)} + \alpha(t) r^{\prime} e^{r(t)} + \beta(t) e^{r(t)} = 0[/tex]

    Now, divide out the expoenents, and your equation becomes

    [tex]r^{\prime \prime} + r^{\prime 2}+\alpha(t) r^{\prime} +\beta(t) =0[/tex]

    So, you now have a different equation, but, more than likely it's not one which is any easier to solve-- unless you get lucky.

    Note however, this new one is effectively a first order equation -- i.e. by letting [tex]q(t) =r^{\prime}(t)[/tex].
     
    Last edited: Mar 16, 2007
  6. Mar 16, 2007 #5
    Err... if you want ot find out what kind of second-order equations are soluble, you can look here. They also have some solutions of PDEs on other pages.
     
  7. Mar 17, 2007 #6

    HallsofIvy

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    Typically, series solutions are used for linear differential equations with variable coefficients.
     
  8. Mar 17, 2007 #7
    decomposition in series...

    Is general recipe ,but sometimes such eqs can be solved explicitely and in finite ,closed form.
    It depends on [itex]\alpha(t),\beta(t)[/itex] functions coefficients involved.
     
  9. Mar 17, 2007 #8
    The brute force method, usually a method of last resort is the method of frobenius. The problem is you'll generate infinite series solutions which rarely have a closed form. The method is necessary for laplaces equation in cylindrical and spherical coordinates.
     
  10. Mar 18, 2007 #9

    HallsofIvy

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    Yes, particularly "Euler type" or "equi-potential" equations.

    Strictly speaking, "Frobenius" method only applies to series expansion about regular singular points, not general series expansions.
     
  11. Mar 18, 2007 #10
    There is the y(x)=u(x)*v(x)...
    solve for v(x), and end up with something like this u'' +(something)u=0
    if your lucky enough it's an easy equation...
    Edit:If the Diff Eq was this: y'' +a(x)y'+b(x)y=0
    then replacing y=v(x)*u(x)
    we have this :u''*v+u'(2v'+a(x)v)+u(v"+a(x)v'+b(x)v)=0
    Take (2v'+a(x)v)=0 solve for v , then replace in the Eq above.
    Then this is what I meant if you were lucky (v"+a(x)v'+b(x)v) should simplify into something related to v in order to solve...
     
    Last edited: Mar 18, 2007
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