Question

Write the given second order equation as its equivalent system of first order equations. u′′−5u′−4u=1.5sin(3t),u(1)=1,u′(1)=2.5 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.)

Answer 1

Answer:

hi your question options is not available but attached to the answer is a complete question with the question options that you seek answer to

Answer:  v = 5v + 4u + 1.5sin(3t),

• 0

• 1

•  4

• 5

• 0

• 1.5sin(3t)

• 1

•  2.5

Step-by-step explanation:

u" - 5u' - 4u = 1.5sin(3t)        where u'(1) = 2.5   u(1) = 1

v represents the "velocity function"   i.e   v = u'(t)

As v = u'(t)

u' = v

since u' = v

v' = u"

v'  = 5u' + 4u + 1.5sin(3t)   ( given that u" - 5u' - 4u = 1.5sin(3t) )

    = 5v + 4u + 1.5sin(3t)  ( noting that v = u' )

so v' = 5v + 4u + 1.5sin(3t)

d/dt $\left[\begin{array}{ccc}u&\\v&\\\end{array}\right]$= $\left[\begin{array}{ccc}0&1&\\4&5&\\\end{array}\right]$  $\left[\begin{array}{ccc}u&\\v&\\\end{array}\right]$ + $\left[\begin{array}{ccc}0&\\1.5sin(3t)&\\\end{array}\right]$

Given that u(1) = 1 and u'(1) = 2.5

since v = u'

v(1) = 2.5

note: the initial value for the vector valued function is given as

$\left[\begin{array}{ccc}u(1)&\\v(1)\\\end{array}\right]$  = $\left[\begin{array}{ccc}1\\2.5\\\end{array}\right]$