# Word Problem - ODE?

#### Beetle37

Hi

I have an equation dr/dt = 1/5R(2-R); R(0) = 0.5

I can see it is in terms of R and t however need to solve exactly for R(t)..Do I:

Re-arrange so all R's together:

1/dt = (1/5R(2-R))dr

then integrate both sides.

If yes I can't automatically see how to integrate RHS, any pointers would be welcome.

Kind regards
Beetle

#### Archie

You haven't put the R terms together correctly. Attempting to interpret your notation, you should get $$\displaystyle {5 \over R(2-R)}\,\mathrm d R$$ or $$\displaystyle {5R \over 2-R}\,\mathrm d R$$ or $$\displaystyle 5R(2-R)\,\mathrm d R$$.

For the first, you use partial fractions. For the last, expand the brackets. And for the second write $$\displaystyle R=2-(2-R)$$.

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#### Beetle37

Hi, I can see how I went wrong putting the R terms together however I can't see where to go. I can see integrating the expression gives me a new expression but it is still all in terms of R.

The only t expression I have comes from the dr/dt at the beginning of the equation. I simply don't understand how to find the t terms - am I meant to be getting dt by itself? It is a word equation about the evolution of a population and the first step is to solve the system exactly for R(t) - which I can't seem to do.

Kind regards
Beetle

#### Beetle37

Seperable ODE - going in circles

In order to solve dr/dt = 1/5R(2-R) "exactly for R(t)"

I tried to rearrange the expression to get 1dt on its own - figuring I could integrate that to get a t term.

So ended up with:

dt=5/R(2-R)dr

This clearly a separable DE and all x terms were on LHS and R terms on LHS. So looking at the RHS I used partial fractions:

5/R(2-R) = 5/R + 5/(2-R)

Substituting A and B for the numerator and solving the simultaneous equation led to

2A-R(A+b), so
A+B=0; and
2A=5

Therefore: A=5/2 and B= -5/2

Substitute A into equation led to:

2/5R so taking the integral led to:

2/5(R^-1) = 2/5*0 R^0

leaving a total of 0 - can this be correct?

The other fraction led to the integral of 2/5 [(2-R)] which I think equals

2R-1/2R^(2)

This process essentially takes me back to the original equation just replacing dx with x??

So confused (and tired)

Cheers
Beetle

#### chiro

MHF Helper
Re: Seperable ODE - going in circles

Hey Beetle37.

When you integrate a 1/R form you get a logarithm. Take a look at the results for logarithms and 1/x with differentiation and integration (it is a special case the power rule doesn't apply to).

Also 1/R + 1/(2-R) = (2-R+R)/[R(2-R)] = 2/denominator and if you adjust this 5/2 you get the correct partial fraction decomposition.

This means A = B = 5/2.

• 2 people

#### Archie

There's not enough information here for anyone to advise you. Why don't you post the question?

#### Beetle37

Hi, I had re-posted this as it not strictly a word problem.
The question is: The evolution of a population of rhinoceros, R(t), in hundreds, times in years is given by the equation

dr/dt=1/5 R(2-R); R(0) = 0.5

(a) Solve the system exactly for R(t).
(b) What happens as the time t to infinity ie what is the population a long time in the future
(c) Write an Euler scheme and compute until the population levels off
(d) Plot both of your solutions on the same set of axis and comment

Kind regards
Beetle

#### Beetle37

Re: Seperable ODE - going in circles

Thank you, I have no idea how I didn't see that, however I can only see it for the first partial fraction, not the second is that correct or is it back to the drawing board?

#### topsquark

Forum Staff
I merged two threads so the order of the posts might be a little off.

-Dan

• 1 person

#### skeeter

MHF Helper
• 1 person