Write the given second order equation as its equivalent system of first order equations. u″+5u′+4u=0 u″+5u′+4u=0 use vv to repre sent the "velocity function", i.e. v=u′(t)v=u′(t). use vv and uu for the two functions, rather than u(t)u(t) and v(t)v(t).
1 answer:
v = u', so v' = u'' and the original equation can be written
... v' + 5v + 4u = 0
Then the system of first-order equations is
... u' = v
... v' = -5v -4u
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