## I am not able to solve these equation, can anyone ...

restart;
Pr:=0.71: n:=-1:

eta0:=0.0699;

EQ1:=diff(H(x), x ) - x*diff(F(x), x ) ;

EQ2:=(1+x^2)*diff(F(x), x\$2) + (3*x + x*F(x)-H(x))*diff(F(x), x) + F(x)^2 + G(x)^2 +2*P(x) + x*diff(P(x), x) ;

EQ3:=(1+x^2)*diff(G(x), x\$2) + (3*x + x*F(x)-H(x))*diff(G(x), x) ;

EQ4:=(1+x^2)*diff(H(x), x\$2) + (3*x + x*F(x)-H(x))*diff(H(x), x) + (1+F(x))*H(x)- diff(P(x), x);

EQ5:=(1+x^2)*diff(theta(x), x\$2) + x*(1-2*n)*diff(theta(x), x) + n^2*theta(x) - Pr*( n*F(x)*theta(x) + ( H(x)-x*F(x) )*diff(theta(x), x)  ) ;

EQ:={EQ1=0, EQ2=0,EQ3=0,EQ4=0 ,EQ5=0}:

IC:={ F(0)=0, G(0)=12, H(0)=0, theta(0)= 1, F(eta0)=0, G(eta0)=12, H(eta0)=0, theta(eta0)= 0, P(0)=0};

sol:= dsolve(EQ union IC,numeric,output=Array([0,0.0699]));

ques.mw

## Problem in IntegrationTools:-Change ?...

Just wanted to ask, what the issue here is:

restart;
Int(1/(1 - x*ln(x)), x);
IntegrationTools:-Change(%,u=1-x*ln(x),u);

doesn't give the proper transformation. It gives

Int(1/u,u)

Solving for x and writing the transformation in terms of LambertW gives something else, if I'm not mistaken.

## 3D graphics on Ubuntu 20.04...

I cannot view 3d graphics with my version of Ubuntu 20.04. I've updated all my computer's graphics card drivers and the problem persists. If I run without hardware acceleration, nothing changes; no visualization and no production possible.
Do you have any ideas for solving this problem? Maple uses OpenGL libraries for 3D production and visualization, and these libraries are installed on my computer. Would installing mesa solve the problem, for example?

## Undefined integral of zero....

The following 2D integrals of 0 are seemingly trivial and one would expect them to evaluate to zero, but Maple evaluates them to undefined

int(0, x=0..infinity, y=0..1) # undefined
int(0, x=0..1, y=0..infinity) # undefined

When the 2D integral is split into two 1D integrals, it does evaluate to zero, as the following examples show

int(0,x=0..infinity) # 0
int(int(0,x=0..infinity), y=0..1) # 0
int(int(0,x=0..1),y=0..infinity) # 0

If infinity is replaced by a variable (say 'c'), the first two integrals are also evaluated to zero.

It may be connected by the following

int(a, x=0..infinity, y=0..1) # a*infinity
int(a, x=0..1, y=0..infinity) # a*infinity
int(a,x=0..infinity) # signum(a)*infinity

So for the 1D integrals the signum is applied to 'a' when the interval is infinite, but not for the 2D integrals. I'm not sure about this difference.