Question

Lab 2 -Page 8 of 9 Math 225)4 er the the following initial vahve problem dr dt (a) 3 points Solve the iaitial valoe problem using the separable DE method. (b) 4 points Solve the same IVP (initial value problem) using the integrating factor method. Hint: Uniqueness of solutions means your answers should be the same in parts (a) and (b).

Answer 1

dx/dt = 8 t + 2 t x x (0) = 1 (1) (a) dx/dt = 2 t (4 + x) = > dx/4 + x = 2 t dt Integrating both sides, integral dx/4 + x = integral 2 t dt + c, where c is arbitrary constant log (4 + x) = t^2 + c log (4 + x) = t^2 + log c_1 where (log c_1 = c) log (4 + x/c_1) = t^2 4 + x/c_1 = e^t^2 4 + x = c_1 e^t^2 x = c_1 e^t^2 - 4 (2) Applying initial condition, 1 = c_1 - 4 = > c_1 = 5 Putting in (2), we get x = 5 e^t^2 - 4, which is Required solution. \r\n (b) dx/dt = 8 t + 2 t x x (0) = 1 dx/dt - 2 t x = 8 t Integrating factor is e^integral - 2 t dt I.F. = e^- t^2 then x (t) e^- t^2