Math 54(R) Homework #19: selected answers

 =============================  B&dP section 3.3  =============================

19.
    SOLUTION ONE
      Derive the result directly from Abel's formula.
      To use Abel's formula you must put the equation in the form
          y'' + p(t)y' + q(t)y = 0
      So expand the given ODE with the product rule:
          0 = [p(t)y']' + q(t)y
            = p(t)y'' + p'(t)y' + q(t)y
        ==>   y'' + (p'/p)y' + (q/p)y = 0
      Since p(t)>0, (p'/p) & (q/p) are continuous. So Abel's formula applies--
      W(t) = c Exp[-Int p'/p dt]. p'/p can be integrated by substitution, so
      W(t) = c Exp[-ln(p(t))] = c p(t)^(-1).

    SOLUTION TWO
      Modify the proof of Abel's formula (3.3.2).
      If y1 and y2 are solutions, then
          0 = [p(t)y1']' + q(t)y1
          0 = [p(t)y2']' + q(t)y2
      ==> 0 = -y2[(p(t)y1')' + q(t)y1] + y1[(p(t)y2')' + q(t)y2]
            = y1[p(t)y2']' - y2[p(t)y1']'
            = y1[p(t)y2'' + p'(t)y2']  -  y2[p(t)y1'' + p'(t)y1']
            = p(t)[y1y2'' - y1''y2]  +  p'(t)[y1y2' - y1'y2]
            = p(t)W'(t) + p'(t)W(t)
      Noticing that this looks like a product rule, or else integrating by
      parts, you then get
          c = p(t)W(t)
      and so W(t)=c/p(t).


 =============================  B&dP section 4.1  =============================

18.
       L[c1 y1 + c2 y2]
     = a0 (c1 y1 + c2 y2)^(n) + ... + an (c1 y1 + c2 y2)
     = c1 (a0 y1^(n) + ... + an y1) + c2 (a0 y2^(n) + ... + an y2)
     = c1 L[y1] + c2 L[y2].
     Thus L is a linear operator.
     If L[y1] = ... = L[yk] = 0, then
     L[y1 + ... + yk] = L[y1] + ... + L[yk] = 0.
     So a sum (in fact, a linear combination) of solutions is a solution.