dx/dt = -5x/2, t>=0. Given that x=60 when t=0, solve the differential equation, giving x in terms of t.

dx/dt = -5x/2 to solve this we must firstly separate the variables ∫2/x dx = -∫5 dt then we solve the integrals using basic integration formulae 2lnx = -5t+c. When it comes to the exam, many students forget the +c and lose an easy mark so always remember to add this when integrating. We know x=60 when t=0, so we can substitutes these in to solve for c and complete the equation 2ln60 = -5(0)+c > c = 2ln60 it is often easier to leave c in log form since it can sometimes make later calculations easier. We can now sub our c into the original equation we solved and simplify to find 2lnx = -5t + 2ln60 > lnx = -5t/2 + ln60 > lnx - ln60 = -5t/2 > ln(x/60) = -5t/2 (Using basic log rules) > x/60 = e^(-5t/2) since the question asks us to find x in terms of t, we can find x = 60e^(-5t/2).