Integration by separation of variables: algebraic rearrangement?

Question

A lake contains 5,000,000 million litres of unpolluted water. A river flows into the lake at 100,000 litres per day. Due to polluters, the river flowing in contains 5 grams per litre of pollutant. A river flows out of the lake at 100,000 litres per day. Find an expression for the amount of pollutant in the lake.

I have:
$\frac{d p}{d t} = 500, 000 - \frac{p}{50}$ $(dp)/dt = 500,000-p/50$
$p = 25, 000, 000 + e^{- \frac{t}{50} + c}$ $p=25,000,000+e^(-t/50+c)$

Answer says $p = 25, 000, 000 (1 - e^{- \frac{t}{50}})$ $p=25,000,000(1-e^(-t/50))$

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Answer 1 · 2016-05-13T16:50:49

your answer is almost correct. Needs to get rid of c only, as explained below.

Your derivation is ok. Only thing left is to determine the constant of integration c.

For this apply the initial condition that at t=0, p=0 (there was no pollution initially)

Thus $0 = 25, 000, 000 + e^{c}$ $0= 25,000,000+e^c$

Thus $e^{c} = - 25, 000, 000$ $e^c= -25,000,000$ . Your answer would then become

$p = 25, 000, 000 + e^{- \frac{t}{50}} . e^{c}$ $p= 25,000,000+e^(-t/50) . e^c$

$p = 25, 000, 000 - 25, 000, 000 e^{- \frac{t}{50}}$ $p= 25,000,000-25,000,000 e^(-t/50)$

$= 25, 000, 000 (1 - e^{- \frac{t}{50}})$ $=25,000,000(1-e^(-t/50))$