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.
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 them 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 .
Yea I did solve but I got stuck when I plugged y(0) = 8 and y'(0) = 8 into the equation y = A*e^(t) + B*e^(8t).
For y(0) = 8, I got 8 = Ae^(0) + Be^(8(0)), which results in 8 = A + B
and for y'(0) = 8 I got 8 = Ae^(0) + 8*Be^(8(0)), which is 8 = A + 8B.
e^(0) = 1.
I am stranded with what to do with the two equations; I tried to solve for one equation for A or B and subbed it in to the other equation.
For example, solving for B in 8 = A + B. I got 8 - B = A and plugged it into 8 = A + 8B, which results in 8 = (8 - B) + 8B, which I got B = 0.
I'm confused .