# Thread: Second Order Differential Equation

1. ## Second Order Differential Equation

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 .

2. Originally Posted by Kenny12345
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 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 .
you have $\displaystyle y(t) = Ae^t + Be^{8t}$

so that $\displaystyle y'(t) = Ae^t + 8Be^{8t}$

Now, $\displaystyle y(0) = 8$ means that when you plug in $\displaystyle t = 0$ into $\displaystyle y(t)$ the result is 8. $\displaystyle y'(0) = 8$ means something similar.

can you finish?

3. 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 .

4. Originally Posted by Kenny12345
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 .
that's a long way to solve it, but that's fine. given these initial conditions, A = 8 and B = 0

don't see why you were confused...

5. What's an easier way to solve it then? I would like to know .

6. Thanks for the advice. I found the solution while I was typing my post.