is(diff(abs(x), x) = x/abs(x))  assuming x < 0;

is(diff(abs(x), x) = x/abs(x))  assuming x > 0;

                                  true

                                  true

Slightly more programmatically:

op(-1, eval(splcurve) assuming v < 4.2);

If you want to get a normal manual solution, you can use  Student[Calculus1][ShowSolution]  command.

Example:

Student[Calculus1][ShowSolution](Int(sin(x)*(x + sin(x)),x));

roll:=rand(1..6):

L:=[seq([seq(roll(), j=1..10)], i=1..30)];

NumbersOfThrees:=map(t->ListTools[Occurrences](3, t), L);

The number  pi  should be coded as  Pi .

All curves are in the same plot:

restart;

Eq:=ln(2*s*t+2*s*x*sqrt(t)/sqrt(Pi)+x^2)-2*s*(arctan((s+x*sqrt(Pi/t))/sqrt(2*Pi*s-s^2))-(1/2)*Pi)/sqrt(2*Pi*s-s^2) = 0;

S:=[0.01,0.005,0.003,0.002];

plots[implicitplot]([seq(eval(Eq,s=S[i]), i=1..nops(S))], t=0..5, x=-2..2, color=[red,blue,green,yellow], gridrefine=5);

It is easy to verify identity

is(sin(2*x)^2 - cos(8*x)^2 = - cos(10*x)*cos(6*x));

                                         true

Therefore, the equation splits into two equations, which are easy to solve

solve({x >= 0, `or`(cos(10*x) = 0, cos(6*x)+1/2 = 0), x <= (1/2)*Pi}, AllSolutions, explicit);

Consider the function which is defined by the equation and its symmetries:

restart;

f:=x->sin(2*x)^2-cos(8*x)^2-(1/2)*cos(10*x):

is(f(Pi-x)=f(Pi+x)) and is(f(Pi/2-x)=f(Pi/2+x));

                                        true

Therefore it is sufficient to solve this equation in the range  0 .. Pi/2

RootFinding[Analytic](f(x), x, re = 0 .. (1/2)*Pi, im = -1 .. 1);

identify([%]);

You can enter a variable (the counter), first assign it the value  0 , and then at each step of the loop is increased by 1 .

Example: find all Pythagorean triangles with hypotenuse not exceeding 100 . Solution in the for loop:

restart;

k:=0: # The counter for the steps of the loop

n:=0: # The counter for solutions

for c from 2 to 100 do

for a from 1 to c-1 do

k:=k+1:

b:=sqrt(c^2-a^2):

if type(b,integer) and b>a then n:=n+1; L[n]:=[a,b,c] fi;

od:

od:

L:=convert(L, list):

k;  n;  L;

Addition:  The example of a purely illustrative nature. There are far more effective methods of generating Pythagorean triples.

restart;

 L:=[seq(seq(x^i*y^j, j=0..3-i), i=0..3)];

 add(a[k]*L[k+1], k=0..nops(L)-1);

The procedure  Euler  solves your problem:

restart;

Euler:=proc(Eq,Ivp,h,n)  # n is the number of steps

local Y, k;

Y[0]:=Ivp[2];

for k to n do

Y[k]:=Y[k-1]+h*eval(rhs(Eq),{x=Ivp[1]+h*(k-1), y(x)=Y[k-1]});

od;

[seq([Ivp[1]+i*h,Y[i]], i=0..n)];  # List of points of the approximate solution

end proc:

Example of use:

Eq:=diff(y(x),x)=(y(x)-x)^2; Ivp:=[0,0];

Euler(Eq, Ivp, 0.1, 10);

plot([%, x-tanh(x)], x=0..1, color=[red, blue]); # Graphs of approximate (red) and exact (blue) solutions

The code of the procedure was edited.

First get more awkward answer, but after a few conversions manages to simplify it.

restart;

dsolve({diff(y(x),x)=(y(x)-x)^2, y(0)=0}, y(x));

assign(%):

convert(expand(simplify(convert(y(x), trig))), tanh);

restart;

A:=(n,a,b,c,p)->((1+c*p)*a/(1-p)+(2+c*p)*b)*mul((1+c*(1-p)^k)*p, k=1..n-1)+add((a+b+k*b*(1-p))*mul((1+c*(1-p)^j)*p,j=k..n-1), k=2..n);

for n do

a[n]:=A(n,0.2,0.09,0.57,0.3);

if a[n]>=1  then Sol:=[[n-1,a[n-1]], [n,a[n]]]; break; fi;

od:

Sol;

You choose what  answer to you is more appropriate  n=6  or  n=7

The code is edited.

The simple procedure A  constructs the characteristic matrix with your properties of any size. For example displayed  A(10). Then we find the characteristic polynomial for  n = 50, and is set to 0 the coefficient of the imaginary part. We find that it is equal to  0  only for  x = 0. Thus we have a symmetrical real matrix that have only real roots:

A:=n->Matrix(n, {(1,1)=-I*x-lambda, (n,n)=-I*x-lambda, seq((i,i)=-lambda, i=2..n-1), seq((i,i-1)=-1, i=2..n), seq((i-1,i)=-1, i=2..n)}):

A(10);

B:=sort(LinearAlgebra[Determinant](A(50)), lambda);

x=solve(coeff(B,I)=0, x);

Verification:

C:=subs(x=0, B):

fsolve(C=0, lambda, complex);

We got only real roots.

restart;

eliminate({(x-1)^2+(y-sqrt(3))^2+z^2 = 4, (x-1)^2+(y-(1/3)*sqrt(3))^2+(z-2*sqrt(6)*(1/3))^2 = 4}, z):

y1 := solve(op(%[2]), y)[2]:

eliminate({(x-2)^2+y^2+z^2 = 4, (x-1)^2+(y-(1/3)*sqrt(3))^2+(z-2*sqrt(6)*(1/3))^2 = 4}, z):

y2 := solve(op(%[2]), y)[1]:

eliminate({x^2+y^2+z^2 = 4, (x-1)^2+(y-(1/3)*sqrt(3))^2+(z-2*sqrt(6)*(1/3))^2 = 4}, z):

y3 := solve(op(%[2]), y)[1]:

Seq := 2*sqrt(6)*(1/3)-sqrt(4-(x-1)^2-(y-(1/3)*sqrt(3))^2), x = 0 .. 2, y = y1 .. piecewise(`and`(x >= 0, x <= 1), y2, `and`(x >= 1, x <= 2), y3), style = surface, numpoints = 10000, scaling = constrained:

A := plot3d(Seq, color = red):

B := plot3d(Seq, color = blue):

C := plot3d(Seq, color = green):

Bottom := plot3d(Seq, color = yellow):

Side1 := plottools[rotate](A, 2*Pi*(1/3), [[0, 0, 0], [1, (1/3)*sqrt(3), (1/6)*sqrt(6)]]):

Side2 := plottools[rotate](B, 4*Pi*(1/3), [[0, 0, 0], [1, (1/3)*sqrt(3), (1/6)*sqrt(6)]]):

Side3 := plottools[rotate](C, (-2*Pi)*(1/3), [[2, 0, 0], [1, (1/3)*sqrt(3), (1/6)*sqrt(6)]]):

plots[display](Side1, Side2, Side3, Bottom, axes = none, orientation = [80, 80, -160]);

Maple can solve your system only numerically:

Sys:=-diff(p(z),z)=(0.855*(1+2*x(z)))/p(z), diff(x(z),z)=((6.39*10^(-3)*p(z)*(1-x(z)))/(1+2*x(z)));

Inc:=p(0)=5, x(0)=0;

Sol:=dsolve({Sys, Inc}, numeric);

plots[odeplot](Sol, [[z,p(z)], [z,x(z)]], z=0..10, color=[red,blue], thickness=2);

Addition: using the procedure  Sol , you can not only build graphics of solutions as above, but also to find solutions values at certain points, such as

Sol(5);

                           [z = 5., p(z) = 3.91451233089194739, x(z) = .120154610510706167]