Hi, I need some help with a question involving differential equation.

The first one is:

Find the solution to the differential equation.

y"-9y'+8y = 0, y(0) = 8, y'(0) = 8

Express your answer as a function of the variable t.

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Here's what I did for the first question:

I used y = e^(λt) and found the first and second derivatives for y.

y' = λ*e^(λt)

y" = (λ^2)*e^(λt)

And I subbed it into y"-9y'+8y = 0, which is:

(λ^2)*e^(λt) - 9λe^(λt) + 8e^(λt) = 0

factored out the e^(λt) and got (λ^2 - 9λ +8) = 0.

Which results - (λ - 8)(λ - 1) = 0.

λ = 8 and λ= 1. I plugged the numbers into e^(λt) and got roots

A*e^(1(t)) + B*e^(8t) where A and B are constants.

After that I not sure what to do with it or how to plug in y(0) = 8 and

y'(0) = 8 into the equation.

I would be grateful for any advice offered

.