Question

The differential equation of the family of curves y=c₁e^x+c₂e^-x is......

(a)`(d^2y)/dx^2+y=0`

(b)`(d^2y)/dx^2-y=0`

(c)`(d^2y)/dx^2+1=0`

(d)`(d^2y)/dx^2-1=0`

Answer 1

y=c₁e^x+c₂e^-x 

^{differentiate w.r.t 'x'}

`dy/dx=c_1e^x-c_2e^(-x) ..............(1)`

differentiate equation (1) w.r.t 'x'

`(d^2y)/(dx^2)=c_1e^x+c_2e^(-x)`

`(d^2y)/(dx^2)-y=0`

=L.H.S

`=c_1e^x+c_2e^(-x)-(c_1e^x+c_2e^(-x))`

`=c_1e^x+c_2e^(-x)-c_1e^x-c_2e^(-x)`

`=0`

Hence

`(b) (d^2y)/(dx^2)-y=0`