Question

Please read and help?! thanks

Answer 1

See below

Explanation:

It is given that `y=e^(rx)`

In the differential equation the first derivative of y and the second derivative should be substituted

So, differentiating for the first time you get ,
`y^'=re^(rx)` --------(i)
(Using chain rule here , `d(e^x)/dx=e^x` and `d(rx)/dx=r`)

**Differentiating (i) (Second derivative)you get ,
`y^('')=r*re^(rx)`
which is equal to -> `y^('')=r^2e^(rx)` -----(ii)

Substituting (i) and (ii) into the given differential equation you get ,

`r^2e^(rx)-6re^(rx)+8e^(rx)=0`

Taking e^(rx) common out and then factorising the equation you get

`r=2 and r=4`

Therefore , the values of r that can be used are

`r=2 and r=4`