Question

3. Solve the initial value problems. a) z' + (5/t)x = 1 + t, x(1) = 1, d) N, = N-W", N(0) = No e) cos θν'tu-3, u(n/2)-1 1+t2
Just b and d

Answer 1

x^1 = (a + b/t) x, x (1) = 1 i.e dx/dt = (a + b/t) x implies dx/dx = (a + b/t) dt Therefore integral dx/x = integral (a + b/t) dt implies log x = at + log t + log c implies log x - log t - log c = at implies log (x/ct) = at implies x/ct = e^at implies x = ct e^at since x (1) = 1, 1 = c times 1 times e^a times 1 is 1 = c times e^a implies c = 1/e^a Therefore x = 1/e^a t e^at = t e^at = -a is, x = t e^a (t - 1) (d) N' = N - 9 e^- t, N (0) = N_0 is N^1 - N = -9 e^- t, which is in linear form with p = -1 and Q = -9 e^- t integral p dt = integral - 1 dt = -t Therefore integrating factor IF = e^integral pdt = e^- t \r\n Hence the solution is N (IF) = integral (IF)