The volume Vcm3 of fluid in a container at time t seconds can be modelled using the differential system dVdt=2V−15. If V=15cm3 at time t=0, calculate the volume of fluid in the container after 2 seconds. Give your answer to the nearest cubic centimetre. Do not include any other symbols or units in your answer.

The ODE is separable, as

dV/dt = 2V - 15   →   dV/(2V - 15) = dt

Integrate both sides to get

1/2 ln|2V - 15| = t + C

(To compute integral, consider substituting U = 2V - 15.)

The volume is V = 15 cm³ when t = 0, so that

1/2 ln(15) = C

and so the volume at any time t is such that

1/2 ln|2V - 15| = t + 1/2 ln(15)

Solve for V :

ln|2V - 15| = 2t + ln(15)

2V - 15 = exp(2t + ln(15))

2V - 15 = exp(2t) exp(ln(15))

2V - 15 = 15 exp(2t)

2V = 15 + 15 exp(2t)

V = 15/2 (1 + exp(2t))

(where exp(x) = eˣ )

Then the volume of fluid after t = 2 s is

V = 15/2 (1 + exp(4)) ≈ 417 cm³

true

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