For second order des, the roots of the characteristic equation may be real or complex. if the roots are real, the complementary solution is the weighted sum of real exponentials. use c1 and c2 for the weights, where c1 is associated with the root with smaller magnitude. if the roots are complex, the complementary solution is the weighted sum of complex conjugate exponentials, which can be written as a constant times a decaying exponential times a cosine with phase. use c1 for the constant and phi for the phase. (note: some equations in the text give the constant multiplying the decaying exponential as 2c1. this was done for the derivation. the constant for this problem should be c1 alone.)

all numerical angles(phases) should be given in radian angles (not degrees). given the differential equation y′′+10y′+61y=6u(t)

write the functional form of the complementary solution, yc(t)
yc(t)=(c1e^(-5t)cos(6t+phi))u(t) < - solved correct

find the particular solution, yp(t).
yp(t)=(6/61)u(t) < -solved, correct

find the total solution, y(t)y(t) for the initial conditions y(0)=9 and y′(0)=4
y(t)=

all numerical angles(phases) should be given in radian angles (not degrees).

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.

slope =   (y2 -y1) ÷ (x2 -x1)

slope = (0 -0) / (-5, -7)

slope = 0 / -12 = 0

the slope of the line is zero.   it is perfectly horizontal.

step-by-step explanation: