Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

DGL 2. Order

  1. Jul 22, 2008 #1
    Hello everybody!

    Here are two ODE 2nd order I tried to solve, but I failed :(

    r''[t] - k/(r[t])^2 = 0

    xy''[x] = ay[x] + b

    Could anyone of you please help me?

    Thanks in advance :)
     
  2. jcsd
  3. Jul 22, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since the independent variable, t, does not appear in the first problem, it can be handled by "quadrature". Let u= r'. r"= u' and, by the chain rule, u'= du/dt= (du/dr)(dr/dt)= ru'. The equation for u is then ru'+ k/r^2= 0 which can be solved by a direct integration: ru'= -k/r^2 so u'= -k/r^3= -kr^(-3) and, finally, du= -kr^(-3)dr. Integrating, r'= u= (1/2)kr^(-2)+ C. That is a separable equation for r:
    [tex]\frac{dr}{(1/2)kr^{-2}+ C}= \frac{r^2 dr}{(1/2)k+ Cr^2}= dt[/tex]
    You may find the left side of that to be a very difficult integration.

    As for the second, xy"= ay+ b, that is a linear differential equation with constant coefficients. It probably would be simplest to do this by taking y to be a Taylor's series solution. Since the leading coefficient is x, you will have to use Frobenius' method: let
    [tex]y= \sum_{n=0}^\infty a_n x^{n+c}[/itex] where c is an unknown number, not necessarily positive or integer. Do the differentiations term by term, put into the equation and assume that a0 is not 0 to get an equation for c (the "indicial" equation). They try to get a recurrance relation for an.
     
  4. Jul 22, 2008 #3
    " Let u= r'. r"= u' and, by the chain rule, u'= du/dt= (du/dr)(dr/dt)= ru'. "

    u= r' - they are both functions of t, aren't they?
    r"= u' is ok
    u'= du/dt= (du/dr)(dr/dt) - chain rule is ok, but how do you get to this:
    u'= ru' from the chain rule?

    As to the integral: I got:

    t = r/C - Sqrt(k/(2C^3))*arctan(r*Sqrt(2C/k)) + C_1

    now it's ok, but I cannot find r(t) :( and I need it, cuz the DE is Newton's 2nd Law applied to the gravitational force, considered k = GM ( M- mass, and G the grav. constant)

    mr''[t] = GMm/r^2[t] <=> r''[t] - k/r^2[t] = 0

    I was expecting ellipses or the conical intersections


    2. DE: unfortunately I haven't learnt how to solve DEs with power series :( but I'll try the ansatz and see what will come out :)
     
  5. Jul 23, 2008 #4
    [EDIT]: it should be: mr''[t] = - GMm/r^2[t] and k = -GM , but 'k' is a constant nevertheless :)
     
  6. Jul 23, 2008 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Sorry, I miswrote: u' (du/dr)(dr/dt)= u du/dr The equation becomes u du/dr= k/r2 so u du= k r-2 dr which gives (1/2)u^2= -k/r+ C or
    [tex]u= dr/dt= \sqrt{C- 2k/r}[/tex]

     
    Last edited: Jul 23, 2008
  7. Jul 23, 2008 #6
    Last edited by a moderator: Jul 23, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?