i already did  a lot of function transform but this one give me some error can anyone determine what is mistake of this equation?

pde_to_ode.mw

I don't know how make my graph be beter for real part and imaginary part and abs part which part how work with parameter can any one explain on this example?

G.mw

i did two case of this equation and odetest is worked good but in this case the odetest is not worked well anyone can determine what is mistake ?

F_P_Correct_case_three.mw

I get my on results but the results are not the same please help me if i did any mistake in my code

symmetry_PDESYS_3_time_fraction[1].mw

This code is working for function f1 but not for f2
f2 := (x,y)->9*x^2-24*x*y+16*y^2+10*x-70*y + 175;
Why this code is not working for f2 ?
unprotect(D);
f1:= (x, y) -> 3*x^2 - 3*y*x + 6*y^2 - 6*x + 7*y - 9;
coeffs(f(x, y));
A, B, C, D, E, F := %;
theta := 1/2*arctan(B/(A - C));
solve({-2*A*xc - B*yc = D, -B*xc - 2*C*yc = E});
assign(%);
x := xcan*cos(theta) - ycan*sin(theta) + xc;
y := xcan*sin(theta) + ycan*cos(theta) + yc;
Eq := simplify(expand(f1(x, y)));
xcan^2/simplify(sqrt(-tcoeff(Eq)/coeff(Eq, xcan^2)))^`2` + ycan^2/simplify(sqrt(-tcoeff(Eq)/coeff(Eq, ycan^2)))^`2` = 1;

Thank you

how we can make pretty of changing equation or case in maple to latex but be more readble and be simplify and the scientific referee don't reject the writting for example in this picture when i do convert i get something like that

if you watch the result 1 it is better denomenator which is 2 come out will be better and then times multiply by radical so how we do something like that?

latex2.mw

What is the idea  of getting equation (48),(49), and (50)

I find thus equation of course i use a little bit seperation by hand, i have 11 equation but i can't get results of parameters i don't know why my code is a little bit old i think and i don't know there is another technique for finding them or not here is the file of mp, thanks for any help

F_P_Correct.mw

for apear each part  remove (:)

This expression is easy i think but i can't get any results with same shape  how i can simplify to the same equation in the paper

i am looking for equation 9 and every thing is clear how i can get

mp1.mw

intersections := proc(P, Q, T)
local R, W, w, t, a, b, sol, buff, v;
sol := NULL;
if T = Y then W := X; else
W := Y; end if;
R := resultant(P, Q, T);
print(`Résultant :`); print(R);
w := fsolve(R, W); t := NULL;
for v in [w] do t := t, fsolve(subs(W = v, P), T); end do;
for a in {w} do for b in {t} do if T = Y then
buff := abs(subs(X = a, Y = b, P)) + abs(subs(X = a, Y = b, Q));
printf(`X=%a,   Y=%a   --->  %a\\n`, a, b, buff); if buff < 1/100000000 then sol := sol, [a, b]; end if;
else buff := abs(subs(X = b, Y = a, P)) + abs(subs(X = b, Y = a, Q));
printf(`X=%a,   Y=%a   --->  %a\\n `, a, b, buff); if buff < 1/100000000 then sol := sol, [b, a]; end if; end if; end do; end do; printf(`Nombre de solutions :  %a\\n`, nops({sol})); print({sol}); end proc:
intersections(X^2 + Y^2 - 1, X - Y, X);
X=-.7071067812,   Y=-.7071067812   --->  0.\n 
I do not wish to find \n in the answer

 In the calculation process, I selected (num=5). How can I detect the maximum value of num to ensure the series solution would be converged?

seried.mw

restart;
with(geometry);
with(plots);
Bl := color = black;
y0 := x -> -ln(1 - exp(-x));
y0 := proc (x) options operator, arrow; -ln(1-exp(-x)) end proc

y1 := x -> -ln(-1 + exp(-x));
y1 := proc (x) options operator, arrow; -ln(-1+exp(-x)) end proc

y2 := x -> -ln(1 + exp(-x));
y2 := proc (x) options operator, arrow; -ln(1+exp(-x)) end proc

p := plot(y0(x), x = 0.02 .. 4, scaling = constrained, color = blue);
p1 := plot(y2(x), x = -4 .. 4, scaling = constrained, color = green);
p2 := plot(y1(x), x = -4 .. 0, scaling = constrained, color = red);
display({p, p1, p2}, view = [-4 .. 4, -4 .. 5]);
Calculate its area; Thank you.

On donne un cercle fixe de diamètre AB, un point M variable sur ce cercle et on construit un carré de sens direct AMNP; Trouver les lieux des points N et P.
restart;
with(plots);
r := 1;
A := [-r, 0];
B := [r, 0];
M := [r*cos(theta), r*sin(theta)];
N := [r*cos(theta) - r*sin(theta), r*sin(theta) + r*cos(theta)];
P := [-r*cos(theta) - r*sin(theta), r*cos(theta) - r*sin(theta)];
c1 := `~`[plottools]*circle([0, 0], r, color = blue);
plot1 := plot(c1, color = blue);
plot2 := animate([N[1], N[2], theta = 0 .. 2*Pi], color = red, thickness = 2);
plot3 := animate([P[1], P[2], theta = 0 .. 2*Pi], color = green, thickness = 2);
display(plot1, plot2, plot3);
Would you like to improve this code so it works. Thank you.
 

i have solution of ODE but again i want take derivative from solution function F then i want take reciprocal of derivative
if F'=G then i want 1/F'=1/G like that i want all solution by list and if possible don't give the parameter a sequence  it will be better

thanks for any help

K := diff(G(xi), xi $ 2) = -lambda*diff(G(xi), xi) - mu;
                 2                                    
                d                    / d        \     
          K := ----- G(xi) = -lambda |---- G(xi)| - mu
                   2                 \ dxi      /     
                dxi                                   

V:= [seq](-1..1, 1/2);
                          [    -1     1   ]
                     V := [-1, --, 0, -, 1]
                          [    2      2   ]

interface(rtablesize= nops(V)^3):
DataFrame(
    <seq(seq(<a | b | rhs(dsolve(eval(K, [lambda,mu]=~ [a,b])))>, a= V), b= V)>,
    columns= [lambda, mu, F]
);

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Hi
my odetest must give me zero everything is true but still not simplify the function with power include 1/n  not do cancelation even

test_sol_for_PDE.mw