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
.