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unclear how to finish this equation

1/P*dP/dt=b+aP

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- Thread starter zoldman
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- #1

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unclear how to finish this equation

1/P*dP/dt=b+aP

- #2

HallsofIvy

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[tex]\frac{1}{P}\frac{dP}{dt}= b+aP[/tex]

is separable:

[tex]\frac{dP}{P(b+aP)}= dt[/tex]

Which I'm sure you knew since you ask about partial fractions.

We can write

[tex]\frac{1}{P(b+ aP)}= \frac{A}{P}+ \frac{B}{b+ aP}[/tex]

for some A and B. Multiply both sides by P(b+ aP):

[tex]1= A(b+ aP)+ BP[/tex]

What do you get if P= 0? What do you get if P= -a/b? Put those values for A and B into the fractions and integrate.

- #3

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I appreciate the confirmation of the partial Fraction step. I arrive at the same situation:

A=1/b and B=-a/b. And I know both a and b from a linear regression.

So substituting back I get

((1/b)/P+((-a/b)/(b+aP)=dt then integrate both sides

(1/b)lnP +(-1/b)ln(b+aP)=t +C

Now my question is here how do I solve for P= f(t).

- #4

HallsofIvy

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The rest is algebra:zoldman said:I appreciate the confirmation of the partial Fraction step. I arrive at the same situation:

A=1/b and B=-a/b. And I know both a and b from a linear regression.

So substituting back I get

((1/b)/P+((-a/b)/(b+aP)=dt then integrate both sides

(1/b)lnP +(-1/b)ln(b+aP)=t +C

Now my question is here how do I solve for P= f(t).

[tex]\frac{1}{b}ln P- \frac{1}{b}ln(b+aP)= ln\left(\frac{P}{b+aP}\right)^\frac{1}{b}= t+ c[/tex]

Take exponential of both sides:

[tex]\left(\frac{P}{b+aP}\right)^\frac{1}{b}= e^{t+ C}= C'e^t[/tex]

(C= e

Take a

[tex]\frac{P}{b+aP}=C"e^{at}[/tex]

(C"= C'

multiply both sides by b+ aP and expand:

[tex] P= C"e^{at}(b+ aP)= C"be^{at}+ aC"e^{at}P[/tex]

[tex] P- aC"e^{at}P= P(1- aC"e^{at})= C"be^{at}[/tex]

[tex] P= \frac{C"be^{at}}{1- aC"e^{at}[/tex]

- #5

mathwonk

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= aP(1 - P/N).

of course it is separable and solving it, shows that the populaion approaches N as time goes on.

- #6

saltydog

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Starting with:

[tex]\frac{1}{b}ln(P)-\frac{1}{b}ln(b+aP)=t+c[/tex]

multiplying by b and collecting logarithms:

[tex]ln\left[\frac{P}{b+aP}\right]=b(t+c)[/tex]

Taking exponentials:

[tex]\frac{P}{b+aP}=e^{b(t+c)}[/tex]

multiplying both sides by b+aP and collecting the P's:

[tex]P\left[1-ae^{b(t+c)}\right]=be^{b(t+c)}[/tex]

Isolating the P and then multiplying the top and bottom of the rational expression by [itex]e^{b(t+c)}[/itex] to make it cleaner leaves:

[tex]P(t)=\frac{b}{e^{-b(t+c)}-a}[/tex]

Now, how about completely characterizing the solutions in terms of a and b assuming some initial condition like P(0)=1. What would it look like whatever it was? (just a suggestion )

[tex]\frac{1}{b}ln(P)-\frac{1}{b}ln(b+aP)=t+c[/tex]

multiplying by b and collecting logarithms:

[tex]ln\left[\frac{P}{b+aP}\right]=b(t+c)[/tex]

Taking exponentials:

[tex]\frac{P}{b+aP}=e^{b(t+c)}[/tex]

multiplying both sides by b+aP and collecting the P's:

[tex]P\left[1-ae^{b(t+c)}\right]=be^{b(t+c)}[/tex]

Isolating the P and then multiplying the top and bottom of the rational expression by [itex]e^{b(t+c)}[/itex] to make it cleaner leaves:

[tex]P(t)=\frac{b}{e^{-b(t+c)}-a}[/tex]

Now, how about completely characterizing the solutions in terms of a and b assuming some initial condition like P(0)=1. What would it look like whatever it was? (just a suggestion )

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Am I going in the right direction?

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