Getting the general solution by using partial fractions [Differential Equation]

**DVS** · September 30th 2012, 10:38 PM

Hi all,

I'm not really confident with partial fractions, so this question really has me stumped at the moment.

For (factorized form of Constant Harvest/Constant Effort models):

$\frac{dP}{dt} = \frac{r}{K} (P-Pu)(Ps-P)$

How would I use variable separation and partial fractions to find the general solution?

So far I've made it to:

$\frac{dP}{(P-Pu)(Ps-P)} = \frac{r}{K} dt$

Am I on the right track?

P does not have to be the subject.

Thank you!

**Prove It** · September 30th 2012, 10:52 PM

Originally Posted by DVS

Hi all,

I'm not really confident with partial fractions, so this question really has me stumped at the moment.

For (factorized form of Constant Harvest/Constant Effort models):

$\frac{dP}{dt} = \frac{r}{K} (P-Pu)(Ps-P)$

How would I use variable separation and partial fractions to find the general solution?

So far I've made it to:

$\frac{dP}{(P-Pu)(Ps-P)} = \frac{r}{K} dt$

Am I on the right track?

P does not have to be the subject.

Thank you!

Are you using $\displaystyle \begin{align*} Pu \end{align*}$ to represent $\displaystyle \begin{align*} P\cdot u \end{align*}$ or $\displaystyle \begin{align*} P_u \end{align*}$ ? Same for Ps?

**chiro** · September 30th 2012, 10:55 PM

Hey DVS.

The first thing I recommend is to take P outside of both terms. Doing this will give you:

[1/P^2]*[1-u)^(-1)*[s-1]^(-1) which means you don't have to do a thing since s and u are independent variables of the one that is being integrating (i.e. you treat this as [1/P^2]*d where d is a constant.

So the integral of d*[1/P^2] is just -d/P + C for an indefinite integral but you can carry this to the RHS and put the C there.

**DVS** · September 30th 2012, 10:56 PM

Originally Posted by Prove It

Are you using $\displaystyle \begin{align*} Pu \end{align*}$ to represent $\displaystyle \begin{align*} P\cdot u \end{align*}$ or $\displaystyle \begin{align*} P_u \end{align*}$ ? Same for Ps?

Apologies for not mentioning that.

$P_{u}$ is the unstable equilibrium
$P_{s}$ is the stable equilibrium

So it would be $P_{u}$ , not $P\cdot u$

**JJacquelin** · September 30th 2012, 11:29 PM

You just have to rewrite 1/((P-Pu)(Ps-P)) on the form = (A/(P-Pu))+(B/(P-Ps)) with A, B to be expressed as functions on Pu and Ps.

**DVS** · October 1st 2012, 04:43 AM

Had another look at it earlier,

I now have:

$\frac{1}{(P-P_{u})(P_{s}-P)}$ = $\frac{A}{(P-P_{u})}$ + $\frac{B}{(P_{s}-P)}$

Then as I believe, I am supposed to multiply both sides by the denominator of the left side, which gives me:

$1 = \frac{A}{(P-P_{u})} (P - P_{u})(P_{s} - P) + \frac {B}{(P_{s} - P)} (P-P_{u})(P_{s} - P)$

But I don't see what that gets me or where I go to find the general solution from here.

**JJacquelin** · October 1st 2012, 05:09 AM

First, simplify. Then expand.

**DVS** · October 1st 2012, 05:33 AM

Now:

$1 = A(Ps-P) + B(P-Pu)$

But I still don't understand - how does expanding that help?

**JJacquelin** · October 1st 2012, 05:51 AM

1 = B*P - A*P +A*Ps -B*Pu
What is the relationship between B and A so that the expression be constant for any value of P ?

**hollywood** · October 1st 2012, 06:52 AM

Originally Posted by DVS

Now:

$1 = A(Ps-P) + B(P-Pu)$

But I still don't understand - how does expanding that help?

You want the two polynomials $1$ and $A(P_s-P) + B(P-P_u)$ to be equal. That means that all the coefficients of $P$ have to be equal. To figure that out, you need to expand $A(P_s-P) + B(P-P_u)$ :
$A(P_s-P) + B(P-P_u)$
$=AP_s-AP + BP-BP_u$
$=(B-A)P+(AP_s-BP_u)$
Which has to equal $1$ . That means that $B-A=0$ and $AP_s-BP_u=1$ . This is two linear equations in the two unknowns $A$ and $B$ , so you can solve for them. The answer ends up being pretty intuitive - not always the case with partial fractions. I usually like to double-check my final partial fraction result before moving on.

There's another way to figure out what $A$ and $B$ are, which is based on the principle that if two polynomials in P are equal, then they have to be equal for all values of P.

- Hollywood

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