5.5-6
==========================================

The characteristic equation 2r^2+4r+7=0 has roots r=-1+wi, r=-1-wi,
w=sqrt(5). The atoms of the homogeneous equation are e^{-x}cos wx,
e^{-x}sin wx.

Because f(x)=x^2, the initial atom list for f(x) is x^2. Add all
lower-power related atoms to obtain the list 1,x,x^2.

Because none of the atoms 1,x,x^2 are solutions of the homogeneous 
equation, then no change is required. The corrected trial solution is

     y=d1 + d2 x + d3 x^2.

Substitute the trial solution into the DE, solve for d1,d2,d3. The
book's answer is correct.

5.5-12
==========================================

The characteristic equation is r^3 + r = 0. The roots are r=0,i,-i. The
atom list for the homogeneous problem y''' + y' = 0 is 1,cos(x),sin(x).

Because the right side of the differential equation is f(x)=2-sin(x), then 
the atoms in the trial solution are initially 1 and sin(x). Differentiation 
of this list of atoms leads to the modified list 1, cos(x), sin(x).

Because 1, cos(x), sin(x) appear in the homogeneous solution, then the
corrected trial solution uses the atom list x, x cos(x), x sin(x).
 
    Book: multiply atom 1 by x^s where s=1=multiplicity of the root r=0 in the
          characteristic equation r^3 + r = 0. Multiply atoms cos(x) and sin(x) 
          by x^s where s=1=multiplicity of the root r=i in the characteristic 
          equation r^3 + r = 0.

    Lectures: Identify three groups of related atoms: group 1: 1, group 2: cos(x), 
              group 3: sin(x). For each group, cross out the homogeneous solution(s) 
              and append the next atom in that group.

The trial solution with fewest atoms is y = d1 (x) + d2 (x cos(x)) + d3 (x sin(x)).

Substitute the trial solution into the DE, solve for the d-symbols. The
book's answer is correct.


5.5-14
==========================================

The characteristic equation is r^4-2r^2+1=0. The roots are r=1,-1,1,-1.
The homogeneous DE y^(4) - 2y'' + y = 0 has atom list e^x, xe^x, e^{-x}, xe^{-x}.

Because f(x)=xe^x, then the initial atom list for f(x) is xe^x. Differentiation expands
the list to e^x, xe^x. This list forms a single group of related atoms.

Because e^x and xe^x appear in the homogeneous solution, then both must be crossed-out
from group 1, then atoms x^2e^x and x^3e^x are appended. The corrected trial solution is

      y=d1 (x^2e^x) + d2 (x^3e^x). 
   
   Book: multiply the initial atom list by by x^s where s=2=multiplicity of the root r=1
         in the characteristic equation r^4-2r^2+1=0.

   Lestures: Cross-out from group 1 both homogeneous solutions and add on the next atoms 
             in the group, namely, x^2e^x and x^3e^x.

Substitute the trial solution y into the DE, solve for the d-symbols. The
book's answer is correct.

5.5-22
==========================================

The characteristic equation is r^5 - r^3 = 0. The roots are r=0,0,0,1,-1. The homogeneous 
equation y^(5) - y''' = 0 then has atom list 1, x, x^2, e^x, e^{-x}.

Because f(x)=e^x + 2x^2 + 2, then the initial list of atoms is e^x, x^2, 1.
Expand this list to 1, x, x^2, e^x [add all lower-power related atoms].

Because 1, x, x^2, e^x are atoms of the homogeneous solution, then the corrected 
atom list is 

   x^3, x^4, x^5, xe^x

  Book: The trial solution is

      y = x^{s1}(d1 + d2 x + d3 x^2) + x^{s2}(d4 e^x)

        where s1=3 is the root multiplicity for root r=0 of r^5 - r^3 = 0,
        and s2=1 is the root multiplicity for root r=1 of r^5 - r^3 = 0.

  Lectures: Doing this construction from basic rules in class, we would alter the
            initial atom list as follows

     group 1:  1,x,x^2   Three homogeneous sols, cross out each, add  x^3,x^4,x^5
     group 2:  e^x       Homogeneous sol, cross it out, add  xe^x

            Then the trial solution is

     y =  d1 (x^3)+ d2 (x^4) + d3 (x^5) + d4 (x e^x)

Substitute the [corrected] trial solution into the DE, solve for the
d-symbols. The book's answer is correct.


5.5-27
==========================================

The characteristic equation is r^4+5r^2+4=0. The roots are complex,
r=2i,-2i,i,-i. The homogeneous equation y^(4) + 5y'' + 4y = 0 then has atom
list cos 2x, sin 2x, cos x, sin x.

Because f(x)=sin x + cos 2x, then the initial list of atoms is sin(x), cos(2x).
Expand this list to cos(x), sin(x), cos(2x), sin(2x).

Because cos x and cos 2x are atoms of the homogeneous solution, then the corrected 
atom list is 

   x cos(x),  x sin(x),  x cos(2x),  x sin(2x).

  Book: The trial solution is

      y = x^{s1}(d1 cos x + d2 sin x) + x^{s2}(d3 cos 2x + d4 sin 2x)

        where s1=1 and s2=1 are root multiplicities for roots of r^4+5r^2+4=0.

  Lectures: Doing this construction from basic rules in class, we would alter the
            initial atom list as follows

     group 1:  cos x    Homogeneous sol, cross it out, add  (x cos x) 
     group 2:  sin x    Homogeneous sol, cross it out, add  (x sin x) 
     group 3:  cos 2x   Homogeneous sol, cross it out, add  (x cos 2x)
     group 4:  sin 2x   Homogeneous sol, cross it out, add  (x sin 2x)

            Then the trial solution is

     y =  d1 (x cos x)+ d2 (x sin x) + d3 (x cos 2x) + d4 (x sin 2x)

Substitute the [corrected] trial solution into the DE, solve for the
d-symbols. The book's answer is correct.


5.5-48
==========================================

There are two terms in the hand-generated solution, which uses formula (33) of 5.5. 
One looks like y_h(x). Removing it leaves the shortest expression for y_p(x): -(x/2)exp(-2x).

The answer can be tested from the maple function odetest() as follows:

        ODE := diff(y(x),x,x)-2*diff(y(x),x)-8*y(x)=3*exp(-2*x);
        sol := y(x)=(-1/2)*x*exp(-2*x);
        odetest(sol,ODE,y(x));


5.5-50
==========================================

The atoms for characteristic equation r^2-4=0 are e^{2x}, e^{-2x}. Use
these for y1, y2 in the method of variation of parameters, formula (33)
in the textbook section 5.5. Then

        W=Wronskian(y1,y2)= y1 y2' - y1' y2 = -2 - 2 = -4,
        f(x)=RHS of the DE = sinh 2x = (1/2)(e^{2x}-e^{-2x})
        yp=C1 y1 + C2 y2 [formula (33) in section 5.5]
        C1 = int(-f y2/W)=int((1/8)-(1/8)e^{-4x})=x/8 + e^{-4x}/32
        C2 = int( f y1/W)=int((-1/8)e^{4x}+(1/8))=x/8 - e^{4x}/32
        yp=(1/8)xe^{2x}+(1/8)xe^{-2x}+(1/32)e^{-2x}-(1/32)e^{2x}

Drop the yh terms to get the shortest solution
        yp=(1/8)xe^{2x}+(1/8)xe^{-2x}+(yh terms to be dropped)
        yp=(1/8)xe^{2x}+(1/8)xe^{-2x}

5.5-52
==========================================

The answer obtained by integration by hand contains three terms, two of
which collect to (1/18)sin(3x)(sin^2(3x)+cos^2(3x)) which reduces to
(1/18)sin(3x), a special case of y_h(x). Since y_h(x) should not appear
in the shortest possible y_p(x), we remove it to obtain a single term
for the answer: -(x/6)cos(3x).

5.5-54
==========================================

The atoms for characteristic equation r^2+1=0 are cos(x), xin(x). Use
these for y1, y2 in the method of variation of parameters, formula (33)
in the textbook section 5.5. Then

        W=Wronskian(y1,y2)= y1 y2' - y1' y2 = cos^2 x + sin^2 x = 1,
        f(x)=RHS of the DE = csc^2(x),
        yp=C1 y1 + C2 y2 [formula (33) in section 5.5]
        C1 = int(-f y2/W)=int(-csc^2(x)sin(x),x)=-ln|csc(x)-cot(x)|,
        C2 = int( f y1/W)=int( csc^2(x)cos(x),x)=int(csc(x)cot(x),x)=-csc(x),
        yp=-cos(x)ln|csc(x)-cot(x)| - csc(x)sin(x) Drop the yh terms to get the shortest solution
        yp=-cos(x)ln|csc(x)-cot(x)| - 1

5.5-58
==========================================

The std form of the DE, required to directly use formula (33) in section 5.5, is

                 y'' +(-4/x)y'+ (6/x^2)y=x

The functions y1, y2 in (33) are a basis for the solution space of the
homogeneous DE, which is given to have general solution 

                        yh = c1 (x^2) + c2 (x^3).
                                    
Take y1, y2 to be the partials on symbols c1, c2:

                          y1 = x^2,  y2 = x^3

Then
        W = y1 y2' - y1' y2 = x^2(3x^2)-x(2x)(x^3) = x^4
        f(x)=x^3/x^2=x [RHS of DE in standard form]
        yp = C1 y1 + C2 y2  [variation of parameters (33)]
        C1 = int(-f y2/W)=int(-x(x^3)/x^4,x)=-x,
        C2 = int(f y1/W)=int(x(x^2)/x^4,x)=ln|x|
        yp = C1 y1 + C2 y2 = -x^3 + x^3 ln|x|

5.5-60
==========================================
The problem statement has a typo but the book's answer in BOB is correct.
The formula for y_c given in the problem statement should be

   y_c = c1 (x^(1/2)) + c2 (x^(3/2))

Use y1:=sqrt(x); and y2:=x*sqrt(x); when trying to do integrals of the
variation of parameters formula

                yp := y1*int(y2*(-f(x))/w,x) + y2*int(y1*f(x)/w,x);

where 

                 w:=y1*diff(y2,x)-y2*diff(y1,x);

Hint from Jon Engle, 10:45 class. If you got a coefficient of 72/7, then likely
you used the formula for y_c given in the problem, which means you should
change y_c to the one above.


5.5-62
==========================================

The std form of the DE, required to directly use formula (33) in section 5.5, is

                 y'' +((-2x/(x^2-1))y'+ (2/(x^2-1))y=1

The functions y1, y2 in (33) are a basis for the solution space of the
homogeneous DE, which is given to have general solution 

                        yh = c1 x + c2 (1+x^2).
                                    
Take y1, y2 to be the partials on symbols c1, c2:

                          y1 = x,  y2 = 1+x^2

Then
        W = y1 y2' - y1' y2 = 2x^2-1-x^2 = x^2-1
        f(x)=(x^2-1)/(x^2-1)=1 [RHS of DE in standard form]
        yp = C1 y1 + C2 y2  [variation of parameters (33)]
        C1 = int(-f y2/W)=-x-ln|x-1|+ln|x+1|
        C2 = int(f y1/W)=(1/2)ln|x-1|+(1/2)ln|x+1|
        
======= end of problem notes 5.5 =======