Simple ODE need to double check if correct

**chogo** · September 18th 2011, 01:03 PM

Can some one tell me if i am correct or completely off

$\frac{dP}{dA} = h(1-P)-rP$ with initial condition $P(0)=1$

here $r$ and $h$ are constants.

$\frac{dP}{dA} = h(1-P)-rP$

by substitution

$\frac{log(h-ph-pr)}{h+r}=-A+C$

therefore

$h-ph-pr=Ce^{-A(h+r)}$

$p=\frac{h}{h+r}-Ce^{-A(h+r)}$

using condition $p(0)=1$

$C=\frac{h}{h+r}-1}$

the solution is

$p=\frac{h}{h+r} - \frac{h}{h+r}e^{-A(h+r)} + e^{-A(h+r)}$

which can be factored as

$p=\frac{h+re^{-A(h+r)}}{h+r}$

is this correct?

this is in reference to the paper in Malaria Journal Smith et al 2007 'Standardising estimates of the Plasmodium Falciparum parasire rate' this equation is listed on page 3 and i cant recapture their solution

their solution is

$p=\frac{h}{h+r} (1-e^{-A(h+r)})$

I must be doing something stupid

**JJacquelin** · September 19th 2011, 12:38 AM

In the referenced paper, the solution corresponds to the condition p(0)=0
while your solution corresponds to p(0)=1.

**chogo** · September 19th 2011, 02:34 AM

Thank you JJacquelin. They have miss specified the initial condition in the paper then.

Thanks for your time and input

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