y = e". b) Show that by assuming y = e" we are led to the equation (3) an' + ba+ c= 0. (10 points) Our solution then becomes y = Cell + Czer where , and 2 are solutions to equation (3). So solving linear second-order differential equations really just involves solving a related quadratic equation. c) Use this technique to solve the equation y" - 5y' + 6y = 0 (10 points)