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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 represent 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).

User Charles R
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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

User Tulkkas
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