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.)

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

 v = 5v + 4u + 1.5sin(3t),

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 =   +

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

 =

it's both, now can you answer my most recent question?

the answer is 13 cm