Unfortunately Maple does not solve your equation in the general case. But it is easy to write a procedure that solves the equation   x^2+y^2 = z^2+t^2 = n   for any specified  n .  Here is the such procedure:

Sol := proc(n)

local S, L, k, i, j;

S := [isolve(x^2+y^2 = n)];

k := 0;

for i in S do

for j in S do

k := k+1; L[k] := [op(i), z = rhs(j[1]), t = rhs(j[2])]

end do end do;

L := convert(L, list);

if op(L)::symbol then return [] else L end if;

end proc:

Example of use:

S := Sol(1000):  nops(S);  seq(S[25*i], i = 1 .. 10);  # Total 256 solutions. Displayed  10  ones

restart;

combinat[permute]([0$4,1$4], 4):

L:=%[2..-1]:

P:=combinat[permute]([0$3,1$3], 3):

k:=0:

for l in L do

for p in P do

k:=k+1: M[k]:=[l,l*p[1],l*p[2],l*p[3]]:

od: od:

M:=convert(M, list):

S:={seq(seq([op(M[i,2..k-1]),M[i,1],op(M[i,k..4])], k=2..5), i=1..nops(M))}:  # Each matrix is defined as the list of lists

nops(S);

seq(Matrix(S[20*i]), i=1..10);  # As example 10 matrices from 225 ones

Edited. The code is optimized (removed unnecessary permutations)

Maybe you just want to bring your equation of order 2 to the canonical form? This is an imaginary ellipse:

eq:=25*(y1-3)^2+200+100*(x1-1)^2=0:

 (eq-200)/200;

solve(simplify(25*(y1-3)^2+200+100*(x1-1)^2=0, {(y1-3)^2+(x1-1)^2=a}), a);

 or

isolate(simplify(25*(y1-3)^2+200+100*(x1-1)^2=0, {(y1-3)^2+(x1-1)^2=a}), a);

eq:=diff(y(x), x)+4*y(x)^3-3*y(x)=0:

DEtools[DEplot](eq, y(x), x=-1 .. 0.1, y=-1 .. 0.1);

1) For the numerical solution you must specify numerical values for all parameters  (C_d, rho, r, m, g )

2) I do not understand the notation  d2v_x . If this is the second derivative, then it should be written as  diff(v_x, t,t)  

Using this procedure, you can find the values of the unknown function at certain points and build its plot:

dy4:=diff(y(x),x):

eqn4:=dy4=sin(x*y(x)):

ic1:=y(0)=1:

ans3:=dsolve({eqn4,ic1}, y(x), numeric);

ans3(1);

plots[odeplot](ans3, [x,y(x)], x=0..10, thickness=2, numpoints=1000);

The example - all on the same plot:

A := plot3d(x^2-y^2, x = -1 .. 1, y = -1 .. 1, shading = zhue):

B := plot3d([1, y, z], y = -2 .. -3/2, z = -1 .. 1, shading = zhue):

plots[display](A, B, axes = frame, orientation = [40, 75], lightmodel=light4);

map(simplify@sum, op(sum(a*u[k]+b*v[k], k=1..n)));

combine(%);

restart;

f:=(x,y)->x*(x+y)*exp(y-x);

extrema(f(x,y), {}, {x,y}, `s`);

Points:=`s`;

Student[MultivariateCalculus][SecondDerivativeTest](f(x,y), [x,y]=[0,0]);

Student[MultivariateCalculus][SecondDerivativeTest](f(x,y), [x,y]=[1/2,-3/2]);

You have forgotten  expand  f  in the Taylor series in the neighborhood  x=4. Should be

f := x^(6*ln(x));

Digits:=15;

taylor(f, x=4, 7);

T2 := convert(%, polynom);

f_value := evalf(subs(x = 5, T2));

Let   F(x,y,t,s) = K(x,y,t,s)*h_m(x)*h_n(y)*h_p(t)*h_q(s) . 

int(F(x,y,t,s), [x=0..1, y=0..1, t=0..1, s=0..1]);

Example:

Max:=-infinity:

S:=[2, -5, 4, 0, 4, 1]:

for s in S do

if  s>Max  then  Max:=s  fi;

od:

Max;

                                                        4

To find the positive roots of your equations, I first put them to the reduced form of quadratic equation, and then find the positive roots of these equations using well-known formula, taking the plus sign in front of the square root:

eq1 := 1.6*10^(-7)*R*sin(t)-4.4*10^(-14)*R^2*cos(t)*sin(t)-1.6*10^(-14)*R^2*cos(t)^2+4.2*10^(-14)*R^2-1.3+2.1*10^(-9)*R*cos(t) = 0:

eq2 := 8.3*10^(-8)*R*sin(t)-1.2*10^(-13)*R^2*cos(t)*sin(t)-2.9*10^(-44)*R^2*cos(t)^2+7.1*10^(-14)*R^2-1.3+8.3*10^(-8)*R*cos(t) = 0:

eq3 := 8.3*10^(-8)*R*sin(t)-1.2*10^(-13)*R^2*cos(t)*sin(t)-2.2*10^(-44)*R^2*cos(t)^2+7.1*10^(-14)*R^2-1.3+8.3*10^(-8)*R*cos(t) = 0:

eq4 := 2.1*10^(-9)*R*sin(t)-4.4*10^(-14)*R^2*cos(t)*sin(t)+1.6*10^(-14)*R^2*cos(t)^2+2.6*10^(-14)*R^2-1.3+1.6*10^(-7)*R*cos(t) = 0:

L := [eq1, eq2, eq3, eq4]:  # List of original equations

L1 := map(collect, map(t->t/coeff(lhs(t), R^2), L), R):   # List of reduced equations

Sol := map(t->-(1/2)*coeff(lhs(t), R)+sqrt((1/4)*coeff(lhs(t), R)^2-coeff(lhs(t), R, 0)), L1):   # List of positive roots of this equations  

A := plot(Sol, t = 0 .. 2*Pi, color = [blue, red, red, green], thickness = 3, coords = polar):  #  Required curves in polar coordinates

B := plot(min(op(Sol)), t = 0 .. 2*Pi, color = yellow, coords = polar, filled = true): # The intersection of the regions

plots[display](A, B);  # All together

 Addition:   min(f(t), g(t))  -  the minimum function of  f(t)  and  g(t)

Obviously the equality  holds

x=<0.3603, 1.3279, 1.6882> = 0.3603*<1, 0, 1> + 1.3279*<0, 1, 1> 

From this equality it follows that  x  is in the rowspace  of the vectors  <1,0,1>  and  <0,1,1>