Updated 2013

1.4-2:   y'+2xy=0
======

 There are two answers in the answer report. Show the steps for the
 equil sol y=0 and the non-equil sol y=???. An answer check is
 required. The easiest answer check is for the explicit sol.

 MAPLE ANSWER CHECK
 # y'+2xy=0
 de:=diff(y(x),x)+2*x*y(x)=0;
 dsolve(de);
 # Answer y(x) = _C1*exp(-x^2)
 # Symbol _C1 is an arbitrary constant c, so we write the maple 
 # answer as y=c exp(-x^2). It so happens that c=0 produces the 
 # required equilibrium solution y=0, so the book's answer is 
 # correct and maple is correct.
 
1.4-6:  y'=3 sqrt(xy)
======

 The book's answer is wrong.
 
 The problem splits into two problems
  (1) y' = 3sqrt(x)sqrt(y)  for x>0, y>0, and
  (2) y' = 3sqrt(-x)sqrt(-y) for x <0, y<0
 There is one equil sol y=0.
 Show the sol steps for the equil sol, and for each of (1),(2).
 That makes 3 panels of solution, which makes for 3 answers in the
 answer report. Check the answers to (1), (2). The equil sol y=0
 has reversible steps (document your ans check accordingly).

 Ans equilibrium sol:  y=0
 Ans quadrant I: B.o.B. ans y = (x^(3/2)+c1)^2
 Ans quadrant III: y = -((-x)^(3/2)+c2)^2

 MAPLE ANSWER CHECK
 # y'=3 sqrt(xy)
 de:=diff(y(x),x)=3*sqrt(x*y(x));
 dsolve(de,y(x)); # What a mess!
 solve(%,y(x));   # The answer was implicit, got 2 answers
                  # but the equilibrium solution y=0 was skipped
 # Answers? Only 2 from maple, each nearly unintelligible.

1.4-8:  y'=2x sec(y)
======

 The separable form is 
  cos(y)y'=2x
 Equil solutions arise from 
  2x sec(y)=0
 which has no solution y=K=constant.
 The answer report has one answer, which is implicit.
 An answer check for the implicit solution is expected.

 WRONG ANSWERS.
 An explicit solution y=Arcsin(x^2+c) has been found by many people,
 but it is not equivalent to the implicit solution: the answer check
 only works with restrictions on x,c. 
 
 To understand the difficulty, consider the equation sin(u+4Pi)=0.67
 to be solved for u>0. Do you get the answer u+4Pi=Arcsin(0.67)?
 Does not this equation imply that u+4Pi is between -Pi/2 and Pi/2?
 The symbol arcsin (lower case A) has a different meaning. The
 mathematical symbols for inverse trig functions are problematic:
 the "arcsin" on a calculator is actually Arcsin.

 MAPLE ANSWER CHECK
 # y'=2x sec(y)
 de:=diff(y(x),x)=2*x*sec(y(x));
 dsolve(de,y(x));    
 # Answer y(x) = arcsin(x^2+2*_C1)
 # Maple's answer uses arcsin, not Arcsin. The argument of arcsin
 # is not restrained to -Pi/2 to Pi/2. 
 # The DE has no equilibrium solution, so maple is correct.
 
1.4-10: (1+x^2)y' = (1+y)^2
=======

 The book's answer is wrong, it is missing a term of -1.

 The correct answer is two equations,

                       y = -1
 and
                       y = -1 + (x+1)/(1+(x+1)c).

 An answer check is required for the second solution above. The
 equilibrium solution y=-1 answer check consists of the phrase
 "reversible steps."

 MAPLE ANSWER CHECK
 # (1+x^2)y' = (1+y)^2
 de:=(1+x^2)*diff(y(x),x)=(1+y(x))^2;
 dsolve(de,y(x));    
 # Answer y(x) = -(arctan(x)+1+_C1)/(arctan(x)+_C1)
 # Dividing, y(x) = -1 + 1/(arctan(x)+_C1)
 # Maple skipped the answer y = -1, which cannot be obtained from 
 # the reported maple answer. 
 
1.4-18:  x^2y' = 1 - x^2 + y^2 - x^2y^2
=======

 Let y'=f(x,y), f(x,y)=(1-x^2+y^2-x^2y^2)/x^2.

 Then f(2,0)=-3/4 and F(x)=f(x,0)/f(2,0), G(y)=f(2,y) converts
 y'=f(x,y) into y'=F(x)G(y), the separable form. Divide by G(y) to
 get the separated form y'/G(y)=F(x), then apply quadrature.

 Avoid all this algebraic work by looking for the factorization
 y'=F(x)G(y) directly in the equation. If you find it, then F and G
 are known and you proceed with separation of variables, applying
 quadrature to the separated equation y'/G=F.

 Equil solutions y=K arise from G(K)=0. Solve this equation for K
 and report all equil sols. A full answer check is required, if the
 steps are not obviously reversible.

 The implicit answer involves arctan(y). Answer check the implicit
 non-equil solution, if you wish (see below).

 Is there an explicit solution? Yes, and you can find it by applying
 the tangent to each side of the implicit equation. It works,
 because the tan(arctan(u))=u and sec^2(u)=1+tan^2(u). Please report
 the explicit solution and do an answer check.

 MAPLE ANSWER CHECK
  # x^2y' = 1 - x^2 + y^2 - x^2y^2
  de:=(x^2)*diff(y(x),x)=1-x^2+(y(x))^2-x^2*(y(x))^2;
  dsolve(de,y(x));    
  # Answer y(x) = -tan((x^2+1+x*_C1)/x)
  # There is no equilibrium solution, because
  # 1 - x^2 + y^2 - x^2y^2=0 is (1-x^2)(1+y^2)=0 which has no 
  # constant solution y (a real number y that is a root).
  # Maple's answer happens to be correct.
 
1.4-22:  y'=4x^3y-y, y(1)=-3
=======

 There are two equations (DE+IC) and hence two answer checks. The
 equil sol is y=0 and since it cannot satisfy y(1)=-3, then only the
 non-equil solution needs to be derived and checked. The answer that
 results has a constant c in it, which is determined by y(1)=-3.

 Show all steps in obtaining the non-equil sol and do an answer
 check. Then find c and check that the proposed solution satisfies
 y(1)=-3.

 Is there an explicit solution? Yes, because ln|y|=expression in x.
 Use properties of log and exp to find the explicit solution. That
 formula should be reported with c evaluated, followed by an answer
 check of the DE and the IC.

 MAPLE ANSWER CHECK
  # y'=4x^3y-y, y(1)=-3
  de:=diff(y(x),x)=4*x^3*y(x)-y(x);
  ic:=y(1)=-3;
  dsolve([de,ic],y(x));    
  # Answer  y(x) = -3*exp(x*(x-1)*(x^2+x+1))
  # By Picard's theorem, the solution is unique. Maple found it.