The equation can be rewritten as y + 1 + t t y = 1, y(1) = 2. so the domain is (0, ). (b) find the solution to this initial value problem. This is a linear di erential equation so we use the method of integrating factor. (t) = er 1+t t dt = er 1 t +1dt = eln t+t = tet then y(t) = 1 tet (cid:2)tet et + c(cid:3) = 1 . C tet with the initial value y(1) = 2 we get so c = 2e and. 1: consider the initial value problem x = tx(1 x) , x(0) = x0. (a) find the solution to this initial value problem. We use seperation of variables for this one. 1 x(1 x) dx = z t. Nding the partial fractions on the left and substitution u = 1 + t on the right we have integrating.
