Page 323, 5.3: 8, 10, 16, 32, 40

5.3-8
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 y'' - 6y' + 13y = 0
 r^2 - 6r + 13 = 0, char equation
 (r-3)^3 + 4 = 0, complete the square
 (r-3 + 2i)(r-3 - 2i) = 0, factor sum of squares with complex numbers
 roots 3 + 2i, 3 - 2i
 exp(3x)cos(2x), exp(3x)sin(2x), atoms by Euler's theorems
 y=linear combination of the atoms
    
5.3-10
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 5 D^4 y + 3 D^3 y = 0
 5r^4 + 3r^3 = 0, char equation
 r^3(5r+3)=0, factored
 r=0,0,0 and r=-3/5, roots according to multiplicity
 exp(0x), x exp(0x), x^2 exp(0x) three atoms for the 3 roots r=0,0,0
 exp(-3x/5), atom for the root r=-3/5, by Euler's theorems
 y = linear combination of the atoms
 
5.3-16
======
 D^4 y + 18 D^2 y + 81y = 0
 r^4 + 18r^2 + 81 = 0, char equation
 u^2 + 18u + 81 = 0, where u=r^2 
 (u+9)(u+9)=0, factored by inverse FOIL
 (r^2+9)^2 = 0, substitute u=r^2
 r = 3i, 3i, -3i, -3i, roots according to multiplicity
 cos(3x), x cos(3x),
 sin(3x), x sin(3x), the four atoms, by euler's theorems
 y=linear combination of the atoms
 
5.3-32
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 D^4 y + D^3 y - 3 D^2 y -5 D y - 2y = 0
 r^4 + r^3 - 3r^2 - 5r - 2 = 0, char equation
 You are supposed to find the roots using Theorey of Equations from
 college algebra. This involves the following theorems:
  Root-factor theorem
  Rational root theorem
  Division algorithm for polynomials
 PROCEDURE. Start by applying the rational root theorem to distill the
 possible rational roots into the finite list of all factors of the
 constant term (-2) divided by all possible factors of the leading
 coefficient (1). This gives the root list
      2/1, -2/1, 1/1, -1/1
 Try each of the four numbers in the characteristic equation, to
 determine if it is a root. For example, r=1 is not a root, because
           r^4 + r^3 - 3r^2 - 5r - 2 = 1+1-3-5-2=-8, not zero
 On the other hand, r=-1 is a root, because   
           r^4 + r^3 - 3r^2 - 5r - 2 = 1-1-3+5-2=0.
 Then (r-(-1)) is a factor of the characteristic polynomial, meaning r+1
 perfectly divided it. Use long division of polynomials to find the
 quotient after division, which is a cubic. Then
      r^4 + r^3 - 3r^2 - 5r - 2 = (r+1)(cubic)
 The project is then reduced to factoring the cubic, which can be
 attacked by the same set of theory of equations tools cited above.
 Finally, the roots are known, which happen to be r = -1, -1, -1, 2.
 Then the atoms are exp(-x), x exp(-x), x^2 exp(-x) and exp(2x) by
 Euler's theorems. The solution y is a linear combination of the atoms.

5.3-40
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 y =A exp(2x) + B cos(2x) + C sin(2x), find the DE
 Like problems in section 5.1, the idea is to differentiate on the
 symbols A, B, C to identify the basis functions, which are atoms. Then
    exp(2x), cos(2x), sin(2x)
 are the three atoms. Euler's theorem produces these atoms from the
 three roots r=2, r = 2i, r = -2i. Then r-2, r-2i, r+2i are factors of
 the characteristic equation, by the root-factor theorem of college
 algebra. Finally, the product of these three factors must give the
 characteristic equation. From it, we construct the DE.
                (r-2)(r-2i)(r+2i)=0,
                (r-2)(r^2+4)=0,
                r^3 - 2r^2 + 4r - 8 = 0,
                y''' - 2y'' + 4y' - 8y = 0.

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