# Solving a system in polar coordinates

#### Somefantastik

Hey Everybody.

for the system:

$$r' = r(1-r)$$
$$\theta' = 1$$

with

$$r(0) = x; \theta(0) = 0$$;

$$r(t) = \frac{xe^{t}}{1-x+xe^{t}}$$

$$\theta(t) = t$$

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?

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#### HallsofIvy

Homework Helper
Hey Everybody.

for the system:

$$r' = r(1-r)$$
$$\theta' = 1$$

with

$$r(0) = x; \theta(0) = 0$$;

$$r(t) = \frac{xe^{t}}{1-x+xe^{t}}$$

$$\theta(t) = t$$

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. $\theta'= 1$ so, integrating, $\theta= t+ C$. Since $\theta(0)= 0+ C= 0$, C= 0 and $\theta(t)= t$. $r'= r(1- x)$ is only slightly harder:$dr/dt= r(1-r)$ can be rewritten as
$$\frac{dr}{r(1- r)}= dt$$
and that can be integrated by "partial fractions".

#### Somefantastik

Pretty basic. You have two completely separate equations- just solve each one separately. $\theta'= 1$ so, integrating, $\theta= t+ C$. Since $\theta(0)= 0+ C= 0$, C= 0 and $\theta(t)= t$. $r'= r(1- x)$ is only slightly harder:$dr/dt= r(1-r)$ can be rewritten as
$$\frac{dr}{r(1- r)}= dt$$
and that can be integrated by "partial fractions".
$$\int\frac{dr}{r(1-r)} \ = \ \int dt$$

where $$\int\frac{dr}{r(1-r)} \ = \ \int\left(\frac{1}{r} \ + \ \frac{1}{1-r} \right)dr \ = \ ln(r) \ - \ ln(1-r) + C_{1}$$

so $$ln(r) \ - \ ln(1-r) + C_{1} = t + C_{2}$$

$$2r -1 + C_{3} = e^{t} + C_{4}$$

$$r = \frac{e^{t} + C + 1}{2}$$

$$r(0) = \frac{e^{0} + C + 1}{2} = x$$

$$C = x-1$$;

$$r = \frac{e^{t} + x}{2}$$;

I'm still not getting the right answer.

Last edited:

#### HallsofIvy

Homework Helper
$$\int\frac{dr}{r(1-r)} \ = \ \int dt$$

where $$\int\frac{dr}{r(1-r)} \ = \ \int\left(\frac{1}{r} \ + \ \frac{1}{1-r} \right)dr \ = \ ln(r) \ - \ ln(1-r) + C_{1}$$

so $$ln(r) \ - \ ln(1-r) + C_{1} = t + C_{2}$$

$$2r -1 + C_{3} = e^{t} + C_{4}$$
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
$$r(t)= \frac{Ce^{t}}{1- Ce^{t}}$$
Since r(0)= x,
$$r(0)= \frac{C}{1- C}= x$$
so C= (1- C)x, C= x- Cx, C+ Cx= C(1+ x)= x so
$$C= \frac{x}{1+x}$$

$$r = \frac{e^{t} + C + 1}{2}$$

$$r(0) = \frac{e^{0} + C + 1}{2} = x$$

$$C = x-1$$;

$$r = \frac{e^{t} + x}{2}$$;

I'm still not getting the right answer.

#### Somefantastik

Thanks so much, I get it now.

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