Solutions 2. 2: we have p (t) = 3t, so the general solution is y = c exp( 3t). The initial condition leads to c = 3: we have p(t) = 4/t, so p (t) = 4 ln t, and the general solution is y = ct4. The initial condition yields c = 1: the integrating factor is e 2t. Hence we can put the equation in the form e 2t(y 2y) = d dt (e 2ty) = et. Integration leads to i. e. e 2ty = et + c, y = e3t + ce2t. Hence p(t) = 3(t2 + 1), p (t) = t3 3t, and the general solution is y = C exp(t3 + 3t): the integrating factor is e2t. Integration leads to i. e. e2t(y + 2y) = d dt (e2ty) = et. e2ty = et + c, y = e t + ce 2t: the integrating factor is e2t.
