y=ae2x+be−3x+cex..........(1)
Differentiating above equation w.r.t. x, we have
y′=2ae2x−3be−3x+cex..........(2)
On subtracting eqn(1) from (2), we have
y′−y=2ae2x−3be−3x+cex−(ae2x+be−3x+cex)
y′−y=2ae2x−3be−3x+cex−ae2x−be−3x−cex
y'−y=ae2x−4be−3x..........(3)
Differentiating above equation w.r.t. x, we have
y′′−y′=2ae2x+12be−3x..........(4)
Multiplying eqn(3) by 2, we have
2y′−2y=2ae2x−8be3x..........(5)
On subtracting eqn(5) from (4), we have
(y′′−y′)−(2y′−2y)=2ae2x+12be−3x−(2ae2x−8be−3x)
y′′−y′−2y′+2y=2ae2x+12be−3x−2ae2x+8be−3x
y′′−3y'+2y=20be−3x..........(6)
Differentiating above equation w.r.t. x, we have
y'''−3y''+2y'=−60be−3x..........(7)
Multiplying eqn(6) by 3, we have
3y′′−9y′+6y=60be−3x..........(8)
On adding eqn(7)&(8), we have
(y'''−3y''+2y')+(3y′′−9y′+6y)=−60be−3x+(60be−3x)
y'''−3y''+2y'+3y′′−9y′+6y=−60be−3x+60be−3x
y'''−7y'+6y=0
Hence, the correct answer is y′′′−7y′+6y=0.