Question

An object which is released from a hot-air balloon at a height experiences a force of air resistance which is directly proportional to the velocity, v , after t seconds satisfies the differential equation dv/dt + 0.25v = 10, where v is in meter per second and t is seconds. Assuming that the initial velocity of the object is 0, find v in terms of t.

Answer 1

$\displaystyle \frac{\mathrm{d}v}{\mathrm{d}t} + 0.25v = 10\\ \textsf{This can be solved by multiplying}\\\textsf{both sides of the differential equation}\\\textsf{by an integrating factor.}\\ \frac{\mathrm{d}v}{\mathrm{d}t} + 0.25v = 10\\ I.F = \textsf{Integrating factor}\\ I.F = e^{\int 0.25 \, \mathrm{d}t} = e^{\frac{t}{4}}\\ \textsf{The solution to the first order linear}\\\textsf{differential equation is thus;}\\ v\, I.F = \int 10 I.F \, \mathrm{d}t\\ \begin{aligned} ve^{\frac{t}{4}} &= \int 10\,e^{\frac{t}{4}} \, \mathrm{d}t\\ ve^{\frac{t}{4}} &= 10 \times 4 e^{\frac{t}{4}} + C\\ ve^{\frac{t}{4}} &= 40e^{\frac{t}{4}} + C\\ v&= 40 + Ce^{-\frac{t}{4}}\\ v(t) &= 40 + Ce^{-\frac{t}{4}} \end{aligned}\\ \textsf{It is given that} \,v(0) = 0\\ v(0)= 40 + C = 0 \implies C = -40\\ \therefore v(t)= 40 - 40e^{-\frac{t}{4}} = 40\left(1 - e^{-\frac{t}{4}}\right)$