Suppose $$f\left( x \right) = {e^{ax}} + {e^{bx}},$$    where $$a \ne b,$$  and that $$f''\left( x \right) - 2f'\left( x \right) - 15f\left( x \right) = 0$$       for all $$x.$$ Then the product $$ab$$   is :

Solution :
$$\eqalign{ & \left( {{a^2} - 2a - 15} \right){e^{ax}} + \left( {{b^2} - 2b - 15} \right){e^{bx}} = 0 \cr & {\text{or, }}\left( {{a^2} - 2a - 15} \right) = 0{\text{ and }}\left( {{b^2} - 2b - 15} \right) = 0 \cr & {\text{or, }}\left( {a - 5} \right)\left( {a + 3} \right) = 0{\text{ and }}\left( {b - 5} \right)\left( {b + 3} \right) = 0 \cr & {\text{i}}{\text{.e}}{\text{., }}a = 5{\text{ or }} - 3{\text{ and }}b = 5{\text{ or }} - 3 \cr & \therefore \,a \ne b \cr & {\text{Hence, }}a = 5{\text{ and }}b = - 3{\text{ or }}a = - 3{\text{ and }}b = 5 \cr & {\text{or }}ab = - 15 \cr} $$