restart;
with(DEtools);
ode := diff(y(x), x) = epsilon - y(x)^2;
                       d                       2
               ode := --- y(x) = epsilon - y(x) 
                       dx                       

sol := dsolve(ode);
                  /           (1/2)            (1/2)\        (1/2)
sol := y(x) = tanh\_C1 epsilon      + x epsilon     / epsilon     

P := particularsol(ode);
                          (1/2)                 (1/2)  
       P := y(x) = epsilon     , y(x) = -epsilon     , 

                /    y(x)    \            (1/2)          
         arctanh|------------| - x epsilon      + _C1 = 0
                |       (1/2)|                           
                \epsilon     /                           

i am looking for finding all solution of this equation like this picture below

Dear all,

I would like to find how I can calculate covariant derivative of Einstein tensor for an arbitrary metric ds_2=-A(r)*dt^2+B(r)*dr^2+dtheta^2+sin(theta)^2*dphi^2

with best

Dear all,

I have this polynomial function

G(x, y) := (-0.14*y^3 + 1.20000000000000*y^2 - 1.26000000000000*y + 0.200000000000000)*x^3 + (1.20*y^3 - 10.0800000000000*y^2 + 10.0800000000000*y - 1.20000000000000)*x^2 + (-8.82*y + 10.08*y^2 - 1.26*y^3)*x + 1. - 1.2*y^2 + 0.2*y^3

I don't understand why the command

minimize(G(x,y),x=0..1,y=0..1);

produces the error

Error, (in RootOf/RootOf:-algnum_in_range) invalid input: RootOf/RootOf:-rootof_in_range expects its 1st argument, rt, to be of type ('RootOf')(polynom(rational,_Z),identical(index) = posint), but received RootOf(7*_Z^3-93*_Z^2+327*_Z-187)

Thanks for your advices, Nicola

Using ScatterPlot (or ErrorPlot), one can add error bars to a 2d point plot of data. However, the bars are single lines. I wish to create a plot with H-type error bars in both the horizontal and vertical directions.  Below is an example showing how the bars should appear. (This image is taken from a previous question about adding error bars.) 

I do not need to reproduce this figure exactly. The location of the data points and the size of the error bars are irrelevant. The closest I have seen is using BoxPlot.

Has this question been asked and answered? If so, I cannot find it. 

there is any way for define conformable fractional derivative in partial differential equation

restart;
with(PDEtools);
pde := a*diff(psi(x, t), x $ 2) + (b*abs(psi(x, t))^(-2*n) + c*abs(psi(x, t))^(-n) + d*abs(psi(x, t))^n + f*abs(psi(x, t))^(2*n))*psi(x, t) = 0;
pde + i*diff(u(x, t), [t $ beta]) = 0;

how define a  fractional derivative in sense of conformable derivative

I don't understand where the the csgn(L) comes from in the solution below:

all the variables are defined as real:

coupled_network_vbat.mw

Thanks in advance for any help.

Jorge

I notice that the command gcd(a,b), if a and b are large degree polynomials, takes too much time and often crashes Windows (not only Maple).

As the euclidean algorithm is very efficient even for large numbers,why not for polynomials?

And how could I calculate the gcd between polynomials with a large degree?

Thanks Michele

Hi everyone:

I watn to create expressions like below expressions in "printf" command:

%15s

%15s%15s

%15s%15s%15s

%15s%15s%15s%15s

...........................................

Can I write these form expressions with "for' command for i=1..n or "seq" command? infact I need those in print command so that I do not type manually. 

i did a solution of this ODE equation but the solution of paper is different from mine also in other some equation i have same problem i can't get exactly and pretty solution

how  define a function for computing multi-variable adomian polynomial  what is wrong with this? what i did mistake

Dear all

I have a function like 

F[1] := (x, y) -> x*y/(1 + 10.35841093*(1 - x)*((-1)*0.9*x + 1)*(1 - y)*((-1)*0.9*y + 1))

This function is continuous on D = [0,1]x[0,1]. I'm interested in the (approximate) value of the double integral over D.

Unfortunately the entry

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

produces Float(infinity). 

Thanks Nicola

Dear Maple users Help me to  get the desire graph for this codes. 

restart:
with(plots):
with(IntegrationTools):
h:=z->piecewise( z<=d+1,   1,
                 z<=d+4,   1-(delta/(2))(1 + cos(2(Pi)*(z - 1 - 1/2))),                                                           z<=d+6,   1 ):
w0:=(-c*h(z)^2/4)+(3/64)(b*c-4*a)*h(z)^4+(19/2304)*b(b-4*a)*h(z)^6:
w1:=(c/4)+(1/16)*(4*a-b*c)*h(z)^2:
w2:=(1/256)(4(b*c-4*a)-b*h(z)^2):
w3:=(1/2304)b(b-4*a):
a:=(x4*S*Gr)*sin(alpha)/(4*x1*x5):
b:=(1/Da)+(x3*M/(x1*(1+m^2))):
c:=(1/x1)*Dp:
Dp:=96*x1/((6-b*h(z)^2)h(z)^4)(F+(a*h(z)/24)-((11/6144)b(b-4*a)*h(z)^8)):
x1:=1/((1-phi1)^2.5*(1-phi2)^2.5):
x2:=(1-phi2)((1-phi1)+phi1*Rs1/Rf)+phi2(Rs2/Rf):
x3:=(shnf)/(sf):
x4:=(1-phi2)((1-phi1)+phi1(RBs1)/(RBf))+phi2*((RBs2)/(RBf)):
x5:=khnf/kf:
shnf:=sbf*((ss2+2*sbf-2*phi2*(sbf-ss2))/(ss2+2*sbf+phi2*(sbf-ss2))):
sbf:=sf*((ss1+2*sf-2*phi1*(sf-ss1))/(ss1+2*sf+phi1*(sf-ss1))):
khnf:=kbf*((ks2+2*kbf-2*phi2*(kbf-ks2))/(ks2+2*kbf+phi2*(kbf-ks2))):
kbf:=kf*((ks1+2*kf-2*phi1*(kf-ks1))/(ks1+2*kf+phi1*(kf-ks1))):
RBs1:=(8933*16.7*10^6):
RBf:=(1063*1.8*10^6):
RBs2:=6320*18*10^6:
kf:=0.492:
sbf:=6.67*10^(-1): ss2:=2.7*10^(-8):
sf:=6.67*10^(-1):ss1:=59.6*10^(6):
ks2:=76.5:kf:=0.492: ks1:=401:
phi1:=0.01: phi2:=0.02:alpha:=Pi/4:m:=0.5:Da:=0.1:Gr:=5:delta:=1:S:=0.5:  d:=1:
                                      
W1:=w0+w1*r^2+w2*r^4+w3*r^6:

by varing M =2,5,7 and r varies from 0 to 1 i want this type of graphs.  please see the sample graphs

Why Maple return 0 when I try to find coefficients of different power of lambda's.coeff.mw

Hello,

I need to check if Maple can solve a specific PDE. Since I don't know much about the PDEtools package, I wonder if a user familiar with it and experienced in solving PDEs could help me.

with(PDEtools);
declare(u(x,y,z,w));
PDE1:=alpha*(y+b*(w))*diff(u(x,y,z,w),x)+(x+z-b*(w))*diff(u(x,y,z,w),y)-c*y*diff(u(x,y,z,w),z)+d*(y-x)*diff(u(x,y,z,w),w)=0;
Sol1:=pdsolve(PDE1);

Maple returns NULL as the solution. Any ideas on how to obtain a solution, if possible? In other similar PDEs, u(x,y,w,z) has a quadratic form.

Many thanks,

When you look at negative horiztonal values on a basic plot, the axis labels can be obscured by gridlines or the plot itself. I have been told many times to keep axis labels and gradation labels outside of the plotting area to avoid adding more information than is neccessary in the actual plotting area. Is there a way to move the label to the other side and rotate it by 180 degree? How do people normally deal with this issue?

See my example of how the axis labels can be obscured by gridlines.

axis_label_on_wrong_side_of_axis.mw