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