Question

Solve the initial-value problem. dr/dt + 2tr = r, r(0) = 5

Answer 1

$\displaystyle\frac{dr}{dt} + 2tr = r\ \Rightarrow\ \frac{dr}{dt} = r - 2tr\ \Rightarrow\ \frac{dr}{dt} = r(1 - 2t) \ \Rightarrow \\ \\ \frac{dr}{r} = (1 - 2t)\, dt\ \Rightarrow \int \frac{dr}{r} = \int(1 - 2t) \, dt\ \Rightarrow\ \ln|r| = t - t^2 + C. \\ \\ |r| = e^{t - t^2 + C} = ke^{t - t^2}.\ \text{Since } r(0) = 5, 5 = ke^0 = k. \\ \\ \text{Thus, } r(t) = 5e^{t - t^2}.$