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1 Answer
Feb 13, 2018

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