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Write the given second order equation as its equivalent system of first order equations. u′′+2u′+2u=0 Use v to represent the "velocity function", i.e. v=u′(t). Use v and u for the two functions, rather than u(t) and v(t). (The latter confuses webwork. Functions like sin(t) are ok.)

Respuesta :

Answer:

u′ = v                  eq. 1

v′ = -2v -2u        eq. 2

Step-by-step explanation:

We are given a second order equation of the form

u′′ + 2u′ + 2u = 0

We have to convert it into a system of first order equations.

Let

u′ = v      eq. 1

Then

u′′ = v′

so

(v′) + 2(v) + 2u = 0

v′ + 2v + 2u = 0

v′ = -2v -2u    eq. 2

Hence the two first order equations are

u′ = v                  eq. 1

v′ = -2v -2u        eq. 2

We can also write these equations in matrix form as

d/dt(X) = AX

[tex]\frac{d}{dt} \left[\begin{array}{ccc}u\\v\end{array}\right] = \left[\begin{array}{ccc}0&1\\-2&-2\end{array}\right] \left[\begin{array}{ccc}u\\v\end{array}\right][/tex]

Where X is the column vector and A is the square matrix (2x2)

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