Find r(t) that is the solution to the following initial value problem.

Question

Find r(t) that is the solution to the following initial value problem.

r''(t) = 2i + cos(t)j + e^2tk

r'(0) = i + 3j

r(0) = 3i + k

Answer 1

r'(t) = ∫(2i + cos(t)j + e^2tk)dt = 2ti + sintj +1/2e^2tk + C_1. To get constant C₁ we have initial condition: r'(0) = i + 3j. So, r'(0) = 2·0i + sin0j + 1/2e^2·0k + C₁ = i + 3j. From here C₁ = i + 3j - 1/2k. Because r'(t) = (2t + 1)i +(sint + 3)j + (1/2e^2t - 1/2)k. Now r(t) = ∫[(2t + 1)i +(sint + 3)j + (1/2e^2t - 1/2)k]dt = (t² + t)i + (3t - cost)j + (1/4e^2t - 1/2t)k + C₂. To get Cconstant C₂ we use second initial condition: r(0) = 3i + k.

So, r(0) = - j + 1/4k + C₂ = 3i + k. From this C₂ = 3i + j + 3/4k.

Now r(t) = (t² + t + 3)i + (3t - cost + 1)j + 1/4(e^2t - 2t + 3)k