Given that, y=Aex+Be2x+Ce3x . . . (i) On differentiating both sides w.r.t. x, we get y′=Aex+2Be2x+3Ce3x From Eq. (i) ‌Aex‌=y−Be2x−Ce3x ∴‌y′‌=y+Be2x+2Ce3x . . . (ii) Again differentiating w.r.t. x, we get y′′=y′+2Be2x+6Ce3x From Eq. (ii) ∴‌‌Be2x‌=y′−y−2Ce3x y′′‌=y′+2y′−2y−4Ce3x+6Ce3x y′′‌=3y′−2y+2Ce3x . . . (iii) Again differentiating w.r.t. x, we get y′′′=3y′′−2y′+6Ce3x From Eq. (iii) ‌2Ce3x=y′′−3y′+2y ∴‌y′′′=3y′′−2y′+3(y′′−3y′+2y) ⇒y′′′=6y′′−11y′+6y ⇒y′′′−6y′′+11y′−6y=0
