Solving a Nonlinear Differential Equation Using Bernoulli's Method

[syj]

Homework Statement

\frac{dr}{dt}=r(u-r)

Homework Equations

i rewrote this as \frac{dr}{dt}-ur=-r^2

i think this is like bernoulli's equation

The Attempt at a Solution

so i let w=\frac{1}{r} so that r=\frac{1}{w}
this gives me

\frac{dr}{dt}=-1w^{-2}\frac{dw}{dt}

so now i have
-w^{-2}\frac{dw}{dt}-uw^{-1}=-w^{-2}

if i divide by -w^{-2}

then i get a linear differential equation

\frac{dw}{dt}+uw=1

the integrating factor is e^{ut}

so i get w=\frac{1}{u}+w_0e^{-ut}

from my definition of w i have w_0=\frac{1}{r_0} and w=\frac{1}{r}

so i get r=\frac{ur_0}{r_0+ue^{-ut}}

is this right?

 

[lanedance]

check by differentiating and comparing to the original DE

note this equation is separable, so you could do it by direct integration

 

[syj]

please explain how i would solve by separation.
thanks

 

[lanedance]

<br /> \frac{dr}{dt}=r(u-r)<br />

gives
<br /> \int \frac{dr}{r(u-r)}=\int dt<br />

though even if you can solve the integral (partial fractions may be a good start) it may be difficult to get into an analytic function for r(t)

 

[HallsofIvy]

\frac{dr}{dt}= r(u-r)
with u constant.

\frac{dr}{r(u-r)}= dt

Integrate the left side with respect to r and the right side with respect to t.

( I really hate it when lanedance gets in one or two minutes ahead of me but to have him first with exactly the same timestand- arrrgh!)

 

[syj]

ok so i get
\frac{ln(r)-ln(r-u)}{u}=t

which i then manipulate (or try to at least ) to get:
<br /> ln\frac{r}{r-u}=ut<br />

if this is correct then i think i can do:
<br /> \frac{r}{r-u}=e^{ut}<br />

i know something is wrong cos i don't have r_0

im working through an example in jordan and smith (nonlinear ordinary differential equations)
so i know what the answer should be, I am trying to get to it tho.
:(
please please correc this.