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Solve differential equation

  • Thread starter syj
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  • #1
syj
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Homework Statement



[itex]\frac{dr}{dt}=r(u-r)[/itex]

Homework Equations



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

i think this is like bernoulli's equation



The Attempt at a Solution


so i let [itex]w=\frac{1}{r}[/itex] so that [itex]r=\frac{1}{w}[/itex]
this gives me

[itex]\frac{dr}{dt}=-1w^{-2}\frac{dw}{dt}[/itex]

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

if i divide by [itex]-w^{-2}[/itex]

then i get a linear differential equation

[itex]\frac{dw}{dt}+uw=1[/itex]

the integrating factor is [itex]e^{ut}[/itex]

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

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

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

is this right?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
check by differentiating and comparing to the original DE

note this equation is separable, so you could do it by direct integration
 
  • #3
syj
55
0
please explain how i would solve by separation.
thanks
 
  • #4
lanedance
Homework Helper
3,304
2
[tex]
\frac{dr}{dt}=r(u-r)
[/tex]

gives
[tex]
\int \frac{dr}{r(u-r)}=\int dt
[/tex]

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)
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,804
931
[tex]\frac{dr}{dt}= r(u-r)[/tex]
with u constant.

[tex]\frac{dr}{r(u-r)}= dt[/tex]

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!)
 
  • #6
syj
55
0
ok so i get
[itex]\frac{ln(r)-ln(r-u)}{u}=t[/itex]

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

if this is correct then i think i can do:
[itex]
\frac{r}{r-u}=e^{ut}
[/itex]

i know something is wrong cos i dont have [itex] r_0 [/itex]

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

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