Show that `y=a e^(2x)+b e^(-x)` is a solution of the differential equation `(d^2y)/(dx^2)-2y=0.`

Explanation

`y=ae^(2x)+be^(-x)`……….`(1)`
Differentiate with respect to `x`
`(dy)/(dx)=a*e^(2x)*(2)+b*e^(-x)(-1)`
`implies (dy)/(dx)=2a*e^(2x)-b*e^(-x)`………`(2)`
Again differentiate with respect to `x`
`(d^(2)y)/(dx^(2))=2a*e^(2x)*(2)-b*e^(-x)(-1)`
`implies (d^(2)y)/(dx^(2))=4a*e^(2x)+b*e^(-x)`..............`(3)`
Now `L.H.S=(d^(2)y)/(dx^(2))-(dy)/(dx)-2y`
`=(4a*e^(2x)+be^(-x))`
`-(2ae^(2x)-b*e^(-x))-2(ae^(2x)+be^(-x))`
[Putting the values from `(1)`, `(2)` and `(3)`]
`=4ae^(2x)+be^(-x)-2ae^(2x)+be^(-x)-2ae^(2x)-2be^(-x)`
`=0=R.H.S`.
`:. y=ae^(2x)+be^(-x)` is a solution of the given differential equation.