# DGL 2. Order

1. Jul 22, 2008

### Marin

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

2. Jul 22, 2008

### HallsofIvy

Staff Emeritus
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:
$$\frac{dr}{(1/2)kr^{-2}+ C}= \frac{r^2 dr}{(1/2)k+ Cr^2}= dt$$
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
$$y= \sum_{n=0}^\infty a_n x^{n+c}[/itex] where c is an unknown number, not necessarily positive or integer. Do the differentiations term by term, put into the equation and assume that a0 is not 0 to get an equation for c (the "indicial" equation). They try to get a recurrance relation for an. 3. Jul 22, 2008 ### Marin " Let u= r'. r"= u' and, by the chain rule, u'= du/dt= (du/dr)(dr/dt)= ru'. " u= r' - they are both functions of t, aren't they? r"= u' is ok u'= du/dt= (du/dr)(dr/dt) - chain rule is ok, but how do you get to this: u'= ru' from the chain rule? As to the integral: I got: t = r/C - Sqrt(k/(2C^3))*arctan(r*Sqrt(2C/k)) + C_1 now it's ok, but I cannot find r(t) :( and I need it, cuz the DE is Newton's 2nd Law applied to the gravitational force, considered k = GM ( M- mass, and G the grav. constant) mr''[t] = GMm/r^2[t] <=> r''[t] - k/r^2[t] = 0 I was expecting ellipses or the conical intersections 2. DE: unfortunately I haven't learnt how to solve DEs with power series :( but I'll try the ansatz and see what will come out :) 4. Jul 23, 2008 ### Marin [EDIT]: it should be: mr''[t] = - GMm/r^2[t] and k = -GM , but 'k' is a constant nevertheless :) 5. Jul 23, 2008 ### HallsofIvy Staff Emeritus Sorry, I miswrote: u' (du/dr)(dr/dt)= u du/dr The equation becomes u du/dr= k/r2 so u du= k r-2 dr which gives (1/2)u^2= -k/r+ C or [tex]u= dr/dt= \sqrt{C- 2k/r}$$

Last edited: Jul 23, 2008
6. Jul 23, 2008

### Marin

Last edited by a moderator: Jul 23, 2008