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Linear Algebra

Exercise 2.7.18

Answers

1.

The auxiliary polynomial is

m t^{2} + rt + k

. The polynomial has two zeroes

α = \frac{- r + \sqrt{r^{2} - 4 mk}}{2 m}

and

β = \frac{- r - \sqrt{r^{2} - 4 mk}}{2 m} .

So the general solution to the equation is

y (t) = C_{1} e^{αt} + C_{2} e^{βt} .

2.

By the previous argument assume the solution is

y (t) = C_{1} e^{αt} + C_{2} e^{βt} .

Consider the two initial conditions

y (0) = C_{1} + C_{2} = 0

and

y^{'} (0) = α C_{1} + β C_{2} = v_{0} .

Solve that

C_{1} = {(α - β)}^{- 1} v_{0}

and

C_{2} = {(β - α)}^{- 1} v_{0} .

3.

The limit tends to zero since the real parts of

α

and

β

is both

- \frac{r}{2 m}

, a negative value, by assuming the

r^{2} - 4 mk \leq 0

. Even if

r^{2} - 4 mk > 0

, we still know that

α

and

β

are negative real number.

jephianlin

2011-06-27 00:00

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