Computer Algebra Recipes

4.3. VARIATIONAL CALCULUS MODELS 197
> sol:=dsolve(ode,y(x)); #implicit solution
q
y(x) ¡ K1
sol := x +
2 y(x) 2
¡
K1
1
0p
K1 B
arctanB
@
2
2
μ
y(x) ¡ 1 1
2 K1 2
q
y(x) ¡ K1 2 y(x) 2
1
C
A
p
K1 K1 2
¡ C1 =0;
q
y(x) ¡ K1
x ¡
2 y(x) 2 0p
K1 B
arctan B
@
+ 1
K1 2
2
μ
y(x) ¡ 1 1
2 K1 2
q
y(x) ¡ K1 2 y(x) 2
1
C
A
p
K1 K1 2
¡ C1 =0
A formidable-looking implicit solution has been produced, with two forms
present. It turns out that we must select the form having the positive square
root (the ¯rst solution for this particular run). For later convenience, I will also
replace the awkward constant symbols K1 and C1 with1= p A and B.
> ans:=subs(fK[1]=1/sqrt(A),_C1=Bg,sol[1]); #select + root
ans := x + p r
A y(x) ¡ y(x)2
0r
1 A
p B (y(x) ¡
A arctan B A 2
@
1
¡
A 2
)
r
y(x) ¡ y(x)2
1
C
A
r
A
¡ B =0
1
A
To proceed any further with the implicit answer, its square-root structure suggests
a trigonometric substitution of the form y = A sin 2 μ,whereμisan angular
parameter not connected to any geometrical feature in our picture. Here A and
B are assumed to be positive, as are sinμ and cos μ as well.
> assume(A>0,B>0,sin(theta)>0,cos(theta)>0):
> Y:=y(x)=A*sin(theta)^2;
Y := y(x) =A sin(μ) 2
Then Y is substituted into the answer, and simpli¯ed with the trig option.
> sol1:=simplify(subs(Y,ans),trig);
sol1 := x + A sin(μ)cos(μ)+ 1
μ
1 ¡1+2cos(μ)
A arctan
2 2
2
¡ B =0
sin(μ)cos(μ)
Then we solve sol1 for x and combine with respect to trig terms.
> x:=combine(solve(sol1,x),trig);
x := ¡ 1
μ
1
cos(2 μ)
A sin(2 μ) ¡ A arctan
+ B
2 2 sin(2 μ)
So, now we have expressions for the coordinates x and y of our sought-after
curve in terms of A, B, andμ. If A and B can be found and the range of μ