$y = c_1 e^{-3x} + c_2 e^{2x}$ → equation (1)
$y' = -3c_1 e^{-3x} + 2c_2 e^{2x}$ → equation (2)
$y'' = 9c_1 e^{-3x} + 4c_2 e^{2x}$ → equation (3)
3 × equation (1) + equation (2)
$3y + y' = 5c_2 e^{2x}$ → equation (4)
3 × equation (2) + equation (3)
$3y' + y'' = 10c_2 e^{2x}$ → equation (5)
2 × equation (4) - equation (5)
$2(3y + y') - (3y' + y'') = 0$
$6y + 2y' - 3y' - y'' = 0$
$6y - y' - y'' = 0$ answer
Note: The methods of elimination vary with the way in which the constants enter the given relation.
