Quiz 2b solutions: consider the equation y 3y = 0. (a) show that y1(x) = 1 and y2(x) = e3x are independent solutions. Evidently y1 is a solution, as y and y . 2 = 9e3x 3 3e3x = 0. These two solutions are linearly independent because one is not a scalar multiple of the other. We seek constants c1, c2 such that y = c1 + c2e3x satis es the given initial conditions. Now y = 3c2e3x, so y (0) = 2 gives. 3 , so c1 = 14 c2 = 2/3, so y = c1 2. Then y(0) = 4 gives 4 = c1 2 y = 2. (a) find the general solution of the equation (x + 5)y + y = ex. First put the equation in standard form: y + This is a linear rst-order de, so we compute the integrating factor:
