Question

Problem 1 consider the ODE dy (B): 2y+2t-- 1. Show that y(t)2 is a solution to (E) with initial value y(0)-0 and that Show that yi(t)-t2 + Lis a solution to (E) with initial value y(0) 1 2, if y(t) is a solution of (E) with initial value y(0) = 0.4, what can t?

Answer 1

(E): dy/dt = -y^2 + y + 2yt^2 + 2t - t^2 - t^4 y_1 = t^2 implies dy_1/dt = 2t substitute y_1 = t^2 in(E) 2t = -(t^2)^2 + t^2 + 2t^2 t^2 + 2t - t^2 - t^4 2t = -t^4 + t^2 + 2 t^4 + 2t - t^2 - t^4 2t = 2t, true. Hence y_1(t) = t^2, y_1(0) = 0 is solution now consider y_1 = t^2 + 1 implies dy_1/dt = 2t from(E) 2t = -(t^2 + 1)^2 + (t^2 + 1) + 2(t^2 + 1) t^2 + 2t - t^2 - t^4 2t = -t^4 - 1 - 2t^2 + t^2 + 1 + 2t^4 + 2t^2 + 2t - t^2 - t^4 2t = 2t , true hence y_1 = t^2 + 1 is solution from (1) y_1 = t^2.y_1(0) = 0 y_2 = t^2 + 1,y_1(0) = 1 y_1 = t^2 + 0.4, y(0) = 0.4, forall t