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).

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

step-by-step explanation:

; fx

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752