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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.)

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batolisis

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

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

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

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