## 749 Reputation

11 years, 87 days

## Xstar...

actually Z and R are different directions and thus indendepnt parameters,

Xstar is a paramatere dependet to Boundary Y
from the last BC we conclude :
Y(T,1)=(Xstar)^n -> Xstar=Y(T,1)^(1/n) .

can not solve this problem using maple ?! what is th idea for solving this problem with and witouth maple !?

## example...

@Carl Love for example ( not surely correct example ) i say if -0.12a<a<0.08  and -0.44<c<-0.02 and b in sth like them, how many inervals can be found for my three variables that my answers are real ? the varibale are allowed to vary between -1 and 1 , but what are the intervals for each varibale that in those intervals,the answers are real ?
how many intervals exist in which the result is real ? can we find group of intervals for each varibale that in those intervals the answers are real ? for example
group one ={ <a< , <b< , <c< } -> the answers are real

group two = { .... } sth like this, -> the answers are real

and ....
what are the groups of intervals in which the answers are real .

## find the intervals in which results are ...

@acer @Carl Love can i find the exact intervals of each variable for which it has real answer ?
the varibale can change between -1 and 1 , but what are intervals for each varibale for which the answer is real ? how many intervals are there for the vrabiles for which the answer in them are surely real ?

## please correct and specify the variables...

please change e to exp() and specify the varibales, are your variables theta[1](t) or theta[1] ?
please make correction about the varibales . and then speciify the equation which are going to be solved simultaneously. after all . u can use the command like this :
dsolve( {eq[1],eq[2],eq[3]});

## thnx...

@Kitonum it is exactly what i needed

## what is your problem exactly ?...

what is your problem exactly ? post your mw code so that we can help

## nooo...

@Axel Vogt as u can see,the integration has not been solved , ! i want the integration with respect to x, to be solved. in your answers,there is no x, but in mine there are . whay i can not do that , what is your maple version ?

## are u sure ?...

@Axel Vogt this does not answere in my question ! :(

 > restart:
 > l := 16; m := .75; EIy := 0.2e5; EIz := 0.400e7; GJ := 0.1e5; mj := .1; Pz := 5000;
 (1)
 > eq1 := m*(diff(diff(w(x, t), t), t))+diff(diff(EIy*(diff(diff(w(x, t), x), x))+(EIz-EIy)*(diff(diff(w(x, t), x), x))*(theta(x)+phi(x, t))^2, x), x)-Pz = 0;
 (2)
 > w := unapply(q[1](t)*cosh(1.8751*x/l),x, t);
 (3)
 > theta := proc (x) options operator, arrow; 0 end proc;
 (4)
 > #theta(x);eq1;
 > phi := unapply(q[4](t)*sin(1.5708*x/l),x, t);
 (5)
 > eq2:=unapply(lhs(eq1),x,t);
 (6)
 > f[1] := value(expand(int(expand(eq2(x,t)*cosh(1.8751*x/l)), x = 0 .. l)));
 (7)
 > expand(eq2(x,t)*cosh(1.8751*x/l)):
 > Int(%, x = 0 .. l);
 > expand(%);
 > value(%);
 >
 (8)
 >
 >

## what do u mean by plot in term of 1/x?...

f(x)=x^2= 1/(1/x)^2 , thus what does this plot will have that the other one won't ?

## yess,...

@Preben Alsholm but now it is optional, and u should subscribe for receiving email notifications.

## i have tried sth like this...

i have tried sth like this, but it has some problems, for example for an unkown number of rows and columns,i should update Matrix dimensions ! but i do not know them. i want the code be general .

 > restart:
 > M:=Matrix(2,9):N:=Matrix(3,4):
 > file1:="ELIST.txt":fclose(file1);
 > file2:="NLIST.txt":fclose(file2);
 > for i from 1 to 2 do
 > M[i]:=(fscanf(file1,"%{9}fr")[1]);
 > od:
 > for i from 1 to 3 do
 > N[i]:=(fscanf(file2,"%{4}fr")[1]);
 > od:
 > N;M;
 (1)

## actually this does not worked !...

@Mac Dude it only imports one row ! not more !

 > restart:
 (1)
 (2)
 > M:=ImportMatrix("NLIST.txt", source=delimited, datatype=float, delimiter=" ");
 > M:=ImportMatrix("NLIST.txt", source=delimited, datatype=float);
 (3)
 >

@schloemilch please insert your original question here, and maybe i can help u and upload a working version !
actually i do not know u can not use these. until u insert your worksheet here. tnx

## some explanations...

 > restart:
 > df1:=a-y(x)=diff(y(x),x)*bc;
 (1)
 > dsolve(df1);
 (2)
 > df2:=diff(y(x),x)*bc-a+y(x);
 (3)
 > dsolve(df2);
 (4)
 > # do u think there is a difference between a=c and a-c=0 ? and according to one theorem in differential equation,if y1 and y2 are two answer to an ODE,c1*y1+c2*y2 is also an answer ! so if y(x) is an answer,-y(x) is also an answer,please note that _C1 is a constant number and u can assume _C2= - (_C1) .
 >