Solving a system in polar coordinates

Hey Everybody.

for the system:

[tex] r' = r(1-r) [/tex]
[tex] \theta' = 1 [/tex]

with

[tex]r(0) = x; \theta(0) = 0 [/tex];

the answer is

[tex]r(t) = \frac{xe^{t}}{1-x+xe^{t}} [/tex]

[tex]\theta(t) = t [/tex]

This answer was given in class as part of a process, and I can't remember how that answer is calculated. Can someone help me?
 

HallsofIvy

Science Advisor
Homework Helper
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Hey Everybody.

for the system:

[tex] r' = r(1-r) [/tex]
[tex] \theta' = 1 [/tex]

with

[tex]r(0) = x; \theta(0) = 0 [/tex];

the answer is

[tex]r(t) = \frac{xe^{t}}{1-x+xe^{t}} [/tex]

[tex]\theta(t) = t [/tex]

This answer was given in class as part of a process, and I can't remember how that answer is calculated. Can someone help me?
Pretty basic. You have two completely separate equations- just solve each one separately. [itex]\theta'= 1[/itex] so, integrating, [itex]\theta= t+ C[/itex]. Since [itex]\theta(0)= 0+ C= 0[/itex], C= 0 and [itex]\theta(t)= t[/itex]. [itex]r'= r(1- x)[/itex] is only slightly harder:[itex]dr/dt= r(1-r)[/itex] can be rewritten as
[tex]\frac{dr}{r(1- r)}= dt[/tex]
and that can be integrated by "partial fractions".
 
Pretty basic. You have two completely separate equations- just solve each one separately. [itex]\theta'= 1[/itex] so, integrating, [itex]\theta= t+ C[/itex]. Since [itex]\theta(0)= 0+ C= 0[/itex], C= 0 and [itex]\theta(t)= t[/itex]. [itex]r'= r(1- x)[/itex] is only slightly harder:[itex]dr/dt= r(1-r)[/itex] can be rewritten as
[tex]\frac{dr}{r(1- r)}= dt[/tex]
and that can be integrated by "partial fractions".
[tex] \int\frac{dr}{r(1-r)} \ = \ \int dt [/tex]

where [tex] \int\frac{dr}{r(1-r)} \ = \ \int\left(\frac{1}{r} \ + \ \frac{1}{1-r} \right)dr \ = \ ln(r) \ - \ ln(1-r) + C_{1} [/tex]

so [tex] ln(r) \ - \ ln(1-r) + C_{1} = t + C_{2} [/tex]

[tex] 2r -1 + C_{3} = e^{t} + C_{4} [/tex]

[tex] r = \frac{e^{t} + C + 1}{2} [/tex]

[tex] r(0) = \frac{e^{0} + C + 1}{2} = x [/tex]

[tex] C = x-1 [/tex];

[tex] r = \frac{e^{t} + x}{2} [/tex];

I'm still not getting the right answer.
 
Last edited:

HallsofIvy

Science Advisor
Homework Helper
41,682
864
[tex] \int\frac{dr}{r(1-r)} \ = \ \int dt [/tex]

where [tex] \int\frac{dr}{r(1-r)} \ = \ \int\left(\frac{1}{r} \ + \ \frac{1}{1-r} \right)dr \ = \ ln(r) \ - \ ln(1-r) + C_{1} [/tex]

so [tex] ln(r) \ - \ ln(1-r) + C_{1} = t + C_{2} [/tex]

[tex] 2r -1 + C_{3} = e^{t} + C_{4} [/tex]
No. ln(r/(1-r))= t+ C' (I've combined your C1 and C2).

Now r/(1-r)= Cet. (C is eC')
So r= Cet- rCet, (1- Cet)r= Cet, and
[tex]r(t)= \frac{Ce^{t}}{1- Ce^{t}}[/tex]
Since r(0)= x,
[tex]r(0)= \frac{C}{1- C}= x[/tex]
so C= (1- C)x, C= x- Cx, C+ Cx= C(1+ x)= x so
[tex]C= \frac{x}{1+x}[/tex]

[tex] r = \frac{e^{t} + C + 1}{2} [/tex]

[tex] r(0) = \frac{e^{0} + C + 1}{2} = x [/tex]

[tex] C = x-1 [/tex];

[tex] r = \frac{e^{t} + x}{2} [/tex];

I'm still not getting the right answer.
 
Thanks so much, I get it now.
 

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