Can These Second Order ODEs Model Planetary Trajectories?

[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

Could anyone of you please help me?

Thanks in advance :)

 

[HallsofIvy]

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 <b>not</b> 0 to get an equation for c (the &quot;indicial&quot; equation). They try to get a recurrance relation for a_n.

 

[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, because 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 learned how to solve DEs with power series :( but I'll try the ansatz and see what will come out :)

 

[Marin]

[EDIT]: it should be: mr''[t] = - GMm/r^2[t] and k = -GM , but 'k' is a constant nevertheless :)

 

[HallsofIvy]

" 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?

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
u= dr/dt= \sqrt{C- 2k/r}

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, because 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 learned how to solve DEs with power series :( but I'll try the ansatz and see what will come out :)

 

[Marin]

dr/dt= sqrt{C- 2k/r}

dr/ sqrt{C- 2k/r} = dt

http://integrals.wolfram.com/index.jsp?expr=1/+sqrt{C-+2k/x}&random=false

is this solution, providing us with the planet trajectories?

and 2nd question: How am I supposed to find r(t) after the integration?

best regards, Marin