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Kitonum

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17 years, 107 days
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These are answers submitted by Kitonum

Solution of the problem 1...

Kitonum 21540
November 25 2019
2 2

To calculate the areas integrals are used. If this is not permitted, then use the formula for the area of a circular segment.

> 

with(plots): with(plottools):
Square:=curve([[0,0],[0,4],[4,4],[4,0],[0,0]], color=black):
Circle_green:=plot([4*cos(t),4*sin(t),t=0..Pi/2], color=green, thickness=2):
Circle_blue:=plot([2*cos(t),2*sin(t)+2,t=-Pi/2..Pi/2], color=blue, thickness=2):
Circle_red:=plot([2*cos(t)+2,2*sin(t)+4,t=-Pi..0], color=red, thickness=2):
Region_yellow:=inequal({y<sqrt(4^2-x^2),y>sqrt(2^2-x^2)+2}, x=0..2, y=2..4, color=yellow, nolines):
Region_pink:=inequal({y<sqrt(4^2-x^2),y>-sqrt(2^2-(x-2)^2)+4}, x=2..4, y=2..4, color=pink, nolines):
display(Square,Circle_green,Circle_blue,Circle_red, Region_yellow, Region_pink, gridlines);
Area_yellow:=int(sqrt(4^2-x^2)-(sqrt(2^2-x^2)+2), x=0..2);
evalf(%);
Area_pink:=int(sqrt(4^2-x^2)-(-sqrt(2^2-(x-2)^2)+4), x=2..eval(x,solve({sqrt(4^2-x^2)=-sqrt(2^2-(x-2)^2)+4,x>2})));
evalf(%);
Area_yellow+Area_pink;
evalf(%);

 

2*3^(1/2)+(1/3)*Pi-4

  

.511299167

  

-2*3^(1/2)-(4/3)*Pi+2*arcsin(3/5)+8*arcsin(4/5)

  

1.052472142

  

-Pi-4+2*arcsin(3/5)+8*arcsin(4/5)

  

1.563771308

 (1)
> 

 


 

Download Plottings_and_calculations.mw

Adjustment...

Kitonum 21540
November 23 2019
0 2

See the corrected

permanouuuuuuuuuuuuuuuuuun_new.mw

Or (manually)...

Kitonum 21540
November 23 2019
0 0 
Eq := a*x+b*y+c = 0; 
expand((Eq-a*x-c)/b);

 

intersect...

Kitonum 21540
November 23 2019
0 2 
L1:=[1,2,3,4,5]:
L2:=[0,5,2,3,7]:
L3:=[7,5,2,3,2]:
convert~([L1,L2,L3], set);
`intersect`(%[]);

                     [{1, 2, 3, 4, 5}, {0, 2, 3, 5, 7}, {2, 3, 5, 7}]
                                             {2, 3, 5}

If necessary, we can take any subsets of this set.

Selection of differentiable part...

Kitonum 21540
November 22 2019
0 1

You did not present your example in an editable form, so I am showing the solution with a simple example (terms with derivatives are left on the left side of the equation, the rest are moved to the right):

restart;
Eq:=2*x(t)+1/3*diff(x(t),t)-5=0;
Terms:=[op(lhs(Eq))];
(select,remove)(has,Terms,diff);
add(%[1])=-add(%[2]);


Of course, for use, this code should be written as a procedure, and then you can apply it to each of your differential equations.

Student:-Precalculus:-CompleteSquare...

Kitonum 21540
November 21 2019
1 0 
restart;
Student:-Precalculus:-CompleteSquare(x^2 + y^2 - 10*x - 75 = 0, [x,y]);
% + 100;

                                    (x-5)^2+y^2-100 = 0
                                     (x-5)^2+y^2 = 100

is...

Kitonum 21540
November 21 2019
3 0

Unfortunately, the  simplify  command does not always cope with the task. If you want to check the equivalence with one command, then you can use the  is  command:

 

> 

restart;
> 

expr1:=(-exp(n*Pi*(2*b - y)/a) + exp(n*Pi*y/a))/((exp(2*n*Pi*b/a) - 1)):
expr2:= sinh(n*Pi/a*y)/tanh(n*Pi/a*b)-cosh(n*Pi/a*y):
is(expr1-expr2=0);

true

 (1)
> 

 


 

Download q_new.mw

Solution...

Kitonum 21540
November 20 2019
2 0

I wrote down all the code in 1d math (I hate 2d math input) and added the plotting of  x(t)  and  y(t) :

 

> 

restart;
> 

Phi:=<x(t), y(t)>;
Sys:=Equate(diff(Phi,t), <6,1; 4,3>.Phi+<6*t, -10*t+4>);
Sol:=dsolve(Sys);
plot(eval([x(t),y(t)], eval(Sol,[_C1=1,_C2=1])), t=0..0.5, color=[red,blue]);

Vector(2, {(1) = x(t), (2) = y(t)})

  

[diff(x(t), t) = 6*x(t)+y(t)+6*t, diff(y(t), t) = 4*x(t)+3*y(t)-10*t+4]

  

{x(t) = exp(7*t)*_C2+exp(2*t)*_C1-2*t-4/7, y(t) = exp(7*t)*_C2-4*exp(2*t)*_C1+10/7+6*t}

  

 


 

Download Solving_ODE_new.mw

plot3d...

Kitonum 21540
November 20 2019
1 0

Side surface and bases of a cylinder can be specified parametrically using 2 parameters (as any 2D surface).

Side:=plot3d([cos(t),sin(t),z], t=0..2*Pi, z=0..2, style=surface, color=green):
Base1:=plot3d([R*cos(t),R*sin(t),0], t=0..2*Pi, R=0..1, style=surface, color=green):
Base2:=plot3d([R*cos(t),R*sin(t),2], t=0..2*Pi, R=0..1, style=surface, color=green):
plots:-display(Side,Base1,Base2);

 

solve...

Kitonum 21540
November 19 2019
0 1 
tax := 0.3*profit:
profit := 0.1*totalsalesx:
solve(totalsalesx = 60.1 + tax + profit);

                          69.08045977


If you need an absolutely accurate (symbolic solution) then use fractions:

tax := 3/10*profit: 
profit := 1/10*totalsalesx: 
solve(totalsalesx = 60+1/10 + tax + profit); 
floor(%) %+ frac(%); # Isolation of the whole part

                                    6010/87
                                   69 + 7/87  

Solution...

Kitonum 21540
November 19 2019
0 0


 

> 

restart;
lambda := 2*(1/10);                              
mu := -1;
beta := 10;                              
alpha := -25;                              
C := 1;                               
k := (1/12)*sqrt(6)/sqrt(beta*lambda*mu);                      
w := alpha/((10*sqrt(-lambda*mu))*beta);                           
A[0] := (1/2)*alpha/((10*sqrt(-lambda*mu))*((1/12)*beta*sqrt(6)/sqrt(beta*lambda*mu)));
A[1] := -(1/10)*alpha/((1/12)*beta*mu*sqrt(6)/sqrt(beta*lambda*mu));      
A[2] := -(12*((1/12)*sqrt(6)/sqrt(beta*lambda*mu)))*lambda^2*alpha/(10*sqrt(-lambda*mu));                    
H := ln(sqrt(lambda/(-mu))*tanh(sqrt(-lambda*mu)*(xi+C)));               
xi := k*x-t*w;
                   
u[0] := A[0]+A[1]*exp(-H)+A[2]*exp(-H)*exp(-H);
plot3d(Im(u[0]), x = -10 .. 10, t = -10 .. 10, view=-50..50);

1/5

  

-1

  

10

  

-25

  

1

  

-((1/24)*I)*6^(1/2)*2^(1/2)

  

-(1/4)*5^(1/2)

  

-((1/4)*I)*5^(1/2)*6^(1/2)*2^(1/2)

  

-((1/2)*I)*6^(1/2)*2^(1/2)

  

-((1/20)*I)*6^(1/2)*2^(1/2)*5^(1/2)

  

ln((1/5)*5^(1/2)*tanh((1/5)*5^(1/2)*(xi+1)))

  

-((1/24)*I)*6^(1/2)*2^(1/2)*x+(1/4)*t*5^(1/2)

  

-((1/4)*I)*5^(1/2)*6^(1/2)*2^(1/2)-((1/2)*I)*6^(1/2)*2^(1/2)*5^(1/2)/tanh((1/5)*5^(1/2)*(-((1/24)*I)*6^(1/2)*2^(1/2)*x+(1/4)*t*5^(1/2)+1))-((1/4)*I)*6^(1/2)*2^(1/2)*5^(1/2)/tanh((1/5)*5^(1/2)*(-((1/24)*I)*6^(1/2)*2^(1/2)*x+(1/4)*t*5^(1/2)+1))^2

  

 
> 

A:=plots:-contourplot3d(Im(u[0]), x = -10 .. 10, t = -10 .. 10, color=red, thickness=3, contours=[seq(C,C=-40..40,10)], coloring=[white,blue], view=-50..50, filledregions=true, grid=[100,100]):
> 

B:=plots:-contourplot(Im(u[0]), x = -10 .. 10, t = -10 .. 10, color=red, contours=[seq(C,C=-40..40,10)], grid=[100,100]):
> 

f:=plottools:-transform((x,y)->[x,y,-50]):
plots:-display(A,f(B));

 
> 

 


 

Download contours2d_3d.mw

Adjustment...

Kitonum 21540
November 16 2019
1 4


 

> 

restart;

ContoursWithLabels := proc (Expr, Range1::(range(realcons)), Range2::(range(realcons)), Number::posint := 8, S::(set(realcons)) := {}, GraphicOptions::list := [color = black, axes = box], Coloring::`=` := NULL)

local r1, r2, L, f, L1, h, S1, P, P1, r, M, C, T, p, p1, m, n, A, B, E;

uses plots, plottools;

f := unapply(Expr, x, y);

if S = {} then r1 := rand(convert(Range1, float)); r2 := rand(convert(Range2, float));

L := [seq([r1(), r2()], i = 1 .. 205)];

L1 := convert(sort(select(a->type(a, realcons), [seq(f(op(t)), t = L)]), (a, b) ->is(abs(a) < abs(b))), set);

h := (L1[-6]-L1[1])/Number;

S1 := [seq(L1[1]+(1/2)*h+h*(n-1), n = 1 .. Number)] else

S1 := convert(S, list)  fi;

print(Contours = evalf[2](S1));

r := k->rand(20 .. k-20); M := []; T := [];

for C in S1 do

P := implicitplot(Expr = C, x = Range1, y = Range2, op(GraphicOptions), gridrefine = 3);

P1 := [getdata(P)];

for p in P1 do

p1 := convert(p[3], listlist); n := nops(p1);

if n < 500 then m := `if`(40 < n, (r(n))(), round((1/2)*n)); M := `if`(40 < n, [op(M), p1[1 .. m-11], p1[m+11 .. n]], [op(M), p1]); T := [op(T), [op(p1[m]), evalf[2](C)]] else

if 500 <= n then h := floor((1/2)*n); m := (r(h))(); M := [op(M), p1[1 .. m-11], p1[m+11 .. m+h-11], p1[m+h+11 .. n]]; T := [op(T), [op(p1[m]), evalf[2](C)], [op(p1[m+h]), evalf[2](C)]]

fi; fi; od; od;

A := plot(M, op(GraphicOptions));

B := plots:-textplot(T);

if Coloring = NULL then E := NULL else E := ([plots:-densityplot])(Expr, x = Range1, y = Range2, op(rhs(Coloring)))  fi;

display(E, A, B);

end proc:

z := -y + sech(x - 3*t);

w := 10*sech(x - 3*t);

with(plots):

P1 := plot(eval(w, t = 0), x = -10 .. 10):
P2 := contourplot(eval(z, t = 0), x = -10 .. 10, y = -eval(w, t = 0) .. eval(w, t = 0), contours = 5, grid = [101, 101]):
display(P1, P2);

Q2:=ContoursWithLabels( eval(z, t = 0), -10 .. 10, -10..10, {-7,-3,1,5,9}, [color=blue,axes=box]):
display(P1,Q2);

-y+sech(-x+3*t)

  

10*sech(-x+3*t)

  

 

Contours = [-7., -3., 1., 5., 9.]

  

 
> 

 


 

Download Contours.mw

Solution...

Kitonum 21540
November 16 2019
0 3

I do not understand why you got solutions in terms of Bessel functions. The direct solution below is expressed through the function  erfi :

E:=1/2:
Eq := -(1/2)*(diff(psi(x), x, x))+(1/2)*x^2*psi(x) = E*psi(x); 
Sol := dsolve(Eq);
plot(eval(rhs(Sol),[_C1=2,_C2=1]), x=0..2);

 

Convolution...

Kitonum 21540
November 16 2019
1 1

There are different versions of this concept. See the wiki for this. For example, for functions defined on the positive half-axis (for the negative half-axis they are considered equal to 0), the problem reduces to calculating the integral over a finite interval, i.e. in Maple we can use the  int   function.

Example:

restart;
f:=t->sin(t):
g:=t->cos(t):
int(f(t)*g(x-t), t=0..x);
plot(%, x=0..10);


 

 

Explanation...

Kitonum 21540
November 14 2019
0 0

You did not specify what values the function  f  takes outside the segment  [0,1] . (Maple ignores your original assumption  assume(0 <= x, x <= 1)). By default, in this case outside the segment  [0,1]  the function  f  is considered equal to 0. If you specify in the definition of the function that it is undefined outside the segment  [0,1] , then everything is OK (see the function  g  below):
 

NULL

> 

restart
> 

assume(0 <= x, x <= 1); ode := diff(y(x), x) = 0

diff(y(x), x) = 0

 (1)
> 

dsolve(ode)

y(x) = _C1

 (2)
> 

f := proc (x) options operator, arrow; piecewise(0 <= x and x < 1/2, 1, x <= 1 and 1/2 <= x, 0) end proc

proc (x) options operator, arrow; piecewise(0 <= x and x < 1/2, 1, x <= 1 and 1/2 <= x, 0) end proc

 (3)
> 

f(-1); f(2); plot(f, -1 .. 2, color = red, axes = box, discont)

0

  

0

  

 
> 

diff(f(x), x)

piecewise(x = 0, undefined, x = 1/2, undefined, 0)

 (4)
> 

g := proc (x) options operator, arrow; piecewise(0 <= x and x < 1/2, 1, 1/2 <= x and x <= 1, 0, undefined) end proc; diff(g(x), x); g(-1); g(2); plot(g, -1 .. 2, color = red, axes = box, discont)

g := proc (x) options operator, arrow; piecewise(0 <= x and x < 1/2, 1, 1/2 <= x and x <= 1, 0, undefined) end proc

  

piecewise(x = 1/2, undefined, 0)

  

undefined

  

undefined

  

 

NULL



You also wrote: "Whats is the relationship between this example and Existence and uniqueness theorem for fist order ode". I do not see any relationship here.

Download diff_piecewise_new.mw

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