A-Level Differential equations: worked solution

Question

With v = dP/dt = 20/(t + C) and P = 0, v = 10 at t = 0, find the particular solution for P in terms of t.

Step 1: use the initial value of v to pin down C. v = 20/(t + C), and we are told v = 10 when t = 0: 10 = 20/(0 + C) 10C = 20 C = 2 So dP/dt = 20/(t + 2). Step 2: integrate with respect to t to recover P. P = ∫ 20/(t + 2) dt The numerator 20 is 20 times the derivative of (t + 2), which is 1, so this is a standard logarithmic integral: P = 20 ln|t + 2| + D, where D is the constant of integration. For t ≥ 0 we have t + 2 > 0, so the modulus can be dropped: P = 20 ln(t + 2) + D Step 3: use the second condition, P = 0 when t = 0, to find D. 0 = 20 ln(0 + 2) + D 0 = 20 ln 2 + D D = -20 ln 2 Step 4: substitute D back and combine the two logarithms using ln A - ln B = ln(A/B). P = 20 ln(t + 2) - 20 ln 2 P = 20[ln(t + 2) - ln 2] P = 20 ln((t + 2)/2) Check: at t = 0 this gives 20 ln 1 = 0, as required.