The following code simulates the manual solution:

Eq:=sin(x*y(x)) = x + y(x):

diff(Eq, x);

subs(y(x)=y, solve(%, diff(y(x), x)));

subs(W[0]=Re*V[0], V[0]/W[0] * diff(u(t),t));

Re  is an unfortunate designation because in Maple  Re  means the real part of a complex number.

You can use the construction  ``(...)  to prevent the automatic disclosure of parentheses:

A:=simplify( (a1/2+a2/2)*ln(a1/2+a2/2+sqrt((a1/2+a2/2)^2+(b1/2-b2/2)^2)+d^2)+(a1/2-a2/2)*ln(a1/2-a2/2+sqrt((a1/2-a2/2)^2+(b1/2-b2/2)^2+d^2)), {a1-a2=deltaa,a1+a2=sigmaa}):

a:=-2*b1*b2+b2^2+b1^2;  ``(factor(a));

algsubs(a=%, A);

Different way - to convert the exponent into the symbol:

A:=simplify( (a1/2+a2/2)*ln(a1/2+a2/2+sqrt((a1/2+a2/2)^2+(b1/2-b2/2)^2)+d^2)+(a1/2-a2/2)*ln(a1/2-a2/2+sqrt((a1/2-a2/2)^2+(b1/2-b2/2)^2+d^2)), {a1-a2=deltaa,a1+a2=sigmaa}):

a:=-2*b1*b2+b2^2+b1^2; subs(2=`2`, factor(a));

algsubs(a=%, A);

simplify( (a1/2+a2/2)*ln(a1/2+a2/2+sqrt((a1/2+a2/2)^2+(b1/2-b2/2)^2+d^2))+(a1/2-

a2/2)*ln(a1/2-a2/2+sqrt((a1/2-a2/2)^2+(b1/2-b2/2)^2+d^2)), {a1-a2=deltaa, a1+a2=sigmaa});

The parts of the cube below and above the surface  z=x^2*y:

plot3d(x^2*y , x=0..1, y = 0 .. 1, filled, color=grey, scaling=constrained, axes=normal, orientation=[25,60], lightmodel=light4);

plottools[translate](plot3d(x^2*y-1 , x=0..1, y = 0 .. 1, filled, color=grey, scaling=constrained, axes=normal, orientation=[25,60], lightmodel=light4), 0, 0, 1);

Сonstruction  ``(...)  was used to prevent premature simplification.  Compound inequality  a<b<c  was replaced by the equivalent  a<b and b<c

1) First step (uncovering brackets):

A:=4477.25 <= ``(expand(4477.25+.25*(t-32450)))  and  ``(expand(4477.25+.25*(t-32450)))<= 16042.25;

                     A := 4477.25 <= ``(-3635.25+.25*t) and ``(-3635.25+.25*t) <= 16042.25

2) Second step (adding  of 3635.25 to both sides of the inequality):

B:=lhs(op(1,A))+3635.25<=``(expand(rhs(op(1,A))+3635.25)) and ``(expand(lhs(op(2,A)))+3635.25)<=rhs(op(2,A))+3635.25;

                                  B := 8112.50 <= ``(.25*t) and ``(.25*t) <= 19677.50

3) Third step (dividing by 0.25 both sides of the inequality):

map(simplify@(x->x/0.25)@expand, B);

                                                         32450. <= t and t <= 78710.

Your original equation:

lambda:=0.1:  k:=0.2:

dsolve({diff(f(eta),eta$3)+f(eta)*diff(f(eta),eta$2)-(diff(f(eta),eta))^2+lambda*(2*f(eta)*(diff(f(eta),eta))*(diff(f(eta),eta,eta))-f(eta)^2*(diff(f(eta),eta,eta,eta)))-K* (diff(f(eta),eta)-1)=0, f(0) = 0, D(f)(0) = 1, D(f)(10) = 0}, f(eta), numeric);

      Error, (in dsolve/numeric/bvp/convertsys) too few boundary conditions: expected 4, got 3

After adding the extra condition  no problems:

lambda:=0.1:  k:=0.2:

dsolve({diff(f(eta),eta$3)+f(eta)*diff(f(eta),eta$2)-(diff(f(eta),eta))^2+lambda*(2*f(eta)*(diff(f(eta),eta))*(diff(f(eta),eta,eta))-f(eta)^2*(diff(f(eta),eta,eta,eta)))-K* (diff(f(eta),eta)-1)=0, f(0) = 0, D(f)(0) = 1, f(10)=0, D(f)(10) = 0}, f(eta), numeric);

plots[odeplot](%, [eta, f(eta)], 0..10, thickness=2); 

restart; with(plots):

A := animate(spacecurve, [[(t+1)*cos(phi)-1, t^2, (t+1)*sin(phi)], t = 0 .. 1, color = red, thickness = 3], phi = 0 .. 2*Pi, frames = 100):

A1 := animate(plot3d, [[(t+1)*cos(phi)-1, t^2, (t+1)*sin(phi)], t = 0 .. 1, phi = 0 .. s, style = wireframe], s = 0 .. 2*Pi, frames = 100):

B := animate(spacecurve, [[(t+1)*cos(phi)-1, sqrt(t), (t+1)*sin(phi)], t = 0 .. 1, color = red, thickness = 3], phi = 0 .. 2*Pi, frames = 100):

B1 := animate(plot3d, [[(t+1)*cos(phi)-1, sqrt(t), (t+1)*sin(phi)], t = 0 .. 1, phi = 0 .. s, style = wireframe], s = 0 .. 2*Pi, frames = 100):

C:=spacecurve([-1, t, 0], t=-0.4..1.4, linestyle=3, color=green, thickness=2):

display(A, A1, B, B1, C, axes = normal, scaling=constrained, orientation = [20, 70], labels = [x, y, z]);

Rotation around the axis  y= -1  can be done similarly.

x:=proc(n)

option remember;

Digits:=20;

sqrt(5+x(n-1))/6/x(n-2);

end proc:

x(0):=2.25:   x(1):=2.5:

evalf[10](x(366));   evalf[10](x(733));

                         0.3848385779

                          3.553804582

is((1/2)*Pi < 3.535344555);

is(3.535344555 <= (3/2)*Pi);

                             true

                             true

Let  a  and   b  are real numbers, b> 0. Then there exist  unique numbers  n  and  r, such that  a=n*b+r  and  n  is an integer and  0<=r<b . By analogy with integers  r  is the remainder of the division of  a  by  b . 

The following simple procedure  R  finds this remainder:

R:=proc(a,b)

expand((a/b-floor(a/b))*b);

end proc:

Examples:

R(5, 7/2),  R(10, 2*Pi),  R(20, 2*Pi);

plot(R(x,2*Pi), x=-4*Pi..4*Pi, discont=true, scaling=constrained);

P := proc (k)

dsolve({diff(x(t), t, t)+k*x(t) = 0, x(0) = 1, (D(x))(0) = 0}, numeric);

end proc:

plot3d((k, t)->rhs((P(k))(t)[2]), 0 .. 10, 0 .. 5, axes = normal, view = [-1 .. 10.5, -1 .. 5.5, -1.5 .. 1.4]);

Yes, you are right. Maple in general weakly solves the examples with parameters. Maple got the solution only for general case, when  m <> -1 / 2 .

Here is the complete solution found with Mathematica (in output used notations for logical  and  and  or ):

Reduce [{-x3 * x5 /m - x3 * x5 - 2 * x1 * x5 + 2 * x1 == 0,
x3 * x5 /m - 2 * x2 * x5 + 2 * x2 ==
0,-х3 + 2 * x5 - x3 * x5 + 1 - 3 * x1 - 3 * x2 + x4 ==
0, -2 * x4 - 2 * x4 * x5 == 0,-x3 * x5 ^ 2 / (x1 * m) - 4 * х5 ^ 2 + 4 * x5 ==
0}, {x1, x2, x3, x4, x5}] / / TraditionalForm

It is strange that in yourcritical lines parameter  m  is missing. Also, when writing functions you missed the multiplication sign. I got the following result:

f := (x1, x2, x3, x4, x5)->-x3*x5/m-x3*x5-2*x2*x5+2*x1:

g := (x1, x2, x3, x4, x5)->x3*x5/m-2*x2*x5+2*x2:

h := (x1, x2, x3, x4, x5)->-x3+2*x5-x3*x5+1-3*x1-3*x2+x4:

k := (x1, x2, x3, x4, x5)->-2*x4-2*x4*x5:

l := (x1, x2, x3, x4, x5)->-x3*x5^2/(x1*m)-4*x5^2+4*x5:

solve({f(x1, x2, x3, x4, x5) = 0, g(x1, x2, x3, x4, x5) = 0, h(x1, x2, x3, x4, x5) = 0, k(x1, x2, x3, x4, x5) = 0, l(x1, x2, x3, x4, x5) = 0}, {x1, x2, x3, x4, x5});

L := [[1,2,3,2,3],[4,5,6,2,3],[7,8,9,2,3]]:

map(`[]`@f@op, L);

                         [[f(1,2,3,2,3)], [f(4,5,6,2,3)], [f(7,8,9,2,3)]]

More specific example:

L := [[1,2,3,2,3],[4,5,6,2,3],[7,8,9,2,3]]:

f:=(v,w,x,y,z)->`+`(v,w,x,y,z):

map(`[]`@f@op, L);

                                            [[11], [20], [29]]