EAMCET Engineering 2004 Solved Paper

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