TS EAMCET 5 May 2018 Shift 1 Solved Paper

The equation is given as ,
y=a+be2x+ce−3x …… (1)
Differentiate the above equation with respect to x .
y1=

dy

dx

=2be2x−3ce−3x …… (2)
Differentiate the above equation with respect to x .
y2=

d2y

dx2

=4be2x+9ce−3x…… (3)
Differentiate the above equation with respect to x .
y3=

d3y

dx3

=8be2x−27ce−3x …… (4)
Add equation (3) and equation (4), and subtract 6 × equation (2).
y2+y3−6y1=(12be2x−18ce−3x) −(12be2x−18ce−3x)
=0