Given

`y = 5ln(x)`

`y' = 5/x`

`y''=-5/x^2`

`y'''=10/x^3`

`y''''=-30/x^4`

so we have to check whether` y^((4)) -16y=0`

`y'''' -16y=-30/x^4 -16(5ln(x)) != 0 `

So it's not a solution to the differential equation.

Given

`y = 5ln(x)`

`y' = 5/x`

`y''=-5/x^2`

`y'''=10/x^3`

`y''''=-30/x^4`

so we have to check whether` y^((4)) -16y=0`

`y'''' -16y=-30/x^4 -16(5ln(x)) != 0 `

So it's not a solution to the differential equation.