A tank holds 20 litres of pure water. Brine with 3 g of salt per litre flows in at 2 litres a minute, and the well-mixed solution flows out at 1 litre a minute, so after $t$ minutes the tank holds $20 + t$ litres and the salt $Q$ (in grams) satisfies $\frac{dQ}{dt} = 6 - \frac{Q}{20 + t}$ . How much salt is in the tank after 20 minutes?

3 gCorrect answerYour answer

120 - \frac{120}{e}

gCorrect answerYour answer90 gCorrect answerYour answer120 gCorrect answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

$\frac{dQ}{dt} + \frac{1}{20 + t}Q = 6$ has the factor $e^{\log(20 + t)} = 20 + t$ .
$Q(20 + t) = \int 6(20 + t)\,dt = 120t + 3t^{2} + C$ ; $Q(0) = 0$ gives $C = 0$ .
At $t = 20$ : $Q = \frac{2400 + 1200}{40} = 90$ g.

Worth noting

$120 - \frac{120}{e}$ keeps the volume at 20 litres, solving $\frac{dQ}{dt} = 6 - \frac{Q}{20}$ ; the tank fills by a litre each minute.

Ten questions from Homogeneous & Linear Differential Equations, at your own pace.

Explanation