Differential equation - Reduction of order

fineTuner
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Hi,

i have to reduce the order of a 2nd order differential equation, to solve it with a numerical method.
The equation is:

\ddot{r}+a\dot{r}+\frac{b}{r^{2}}=0

with a,b\geq0

I tried to reduce it substituting \dot{r}=v, but i don't know what to do with the term \frac{b}{r^{2}} .

Can someone explain me what i should do?

Thank you,

J.
 
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In general, you can't just reduce the order of a differential equation. What you can do is convert a second order equation to two first order differential equations or, equivalently to a two dimensional vector equation. Here, if you let v= r', then r''= v' so your first equation becomes, as you saw, v'+ av+ b/r^2= 0 and your second equation is r'= v. Equivalently, if we think of vector, V, as being \begin{bmatrix}r(t) \\ v(t)\end{bmatrix} we have the vector equation
V'= \begin{bmatrix}r(t) \\ v(t) \end{bmatrix}'= \begin{bmatrix}-v- b/r^2 \\ v\end{bmatrix}
 
Hi !
Another manner to present it :
Let dr/dt = Y(r)
d²r/dt² = d(Y(r))/dt = (dY/dr)*(dr/dt) = (dY/dr)*Y
r''+a*r'+b/r² = (dY/dr)*Y+a*Y+b/r²
Y' Y + a Y = -b/r² is an first order ODE, where Y(r) is the unknown function.
Solving it leads to Y(r)
Then dt = dr/Y(r)
But the analytic integration of 1/Y(r) will probably be too arduous. That is the raison why numerical solving of the first order ODE is more realisic in practice.
 
Hi!
Thank you JJacquelin and HallsofIvy for your answers. The most useful form is the first one, in the matrix form. I will put it into a f90 program to simulate a body orbiting around a planet with air drag... now it's time to find the right units to avoid problems with my pc, let's see what will happen!
 
HallsofIvy said:
In general, you can't just reduce the order of a differential equation. What you can do is convert a second order equation to two first order differential equations or, equivalently to a two dimensional vector equation. Here, if you let v= r', then r''= v' so your first equation becomes, as you saw, v'+ av+ b/r^2= 0 and your second equation is r'= v. Equivalently, if we think of vector, V, as being \begin{bmatrix}r(t) \\ v(t)\end{bmatrix} we have the vector equation
V'= \begin{bmatrix}r(t) \\ v(t) \end{bmatrix}'= \begin{bmatrix}-v- b/r^2 \\ v\end{bmatrix}
But shouldn't it read
V&#039;=\begin{bmatrix} r&#039; \\ v&#039; \end{bmatrix}<br /> =\begin{bmatrix} v \\ r&#039;&#039; \end{bmatrix}<br /> =\begin{bmatrix} v \\ -a v-b/r^2 \end{bmatrix}
 
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