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.)

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

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

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation: