mehdi jafari

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11 years, 229 days

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These are replies submitted by mehdi jafari


restart:

df1:=a-y(x)=diff(y(x),x)*bc;

a-y(x) = (diff(y(x), x))*bc

(1)

dsolve(df1);

y(x) = a+exp(-x/bc)*_C1

(2)

df2:=diff(y(x),x)*bc-a+y(x);

(diff(y(x), x))*bc-a+y(x)

(3)

dsolve(df2);

y(x) = a+exp(-x/bc)*_C1

(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) .

 


Download dsolve.mw

@wo0olf maybe u want to notice Mr.Kitonum's suggestion about your input correctness.

khahesh mikonam aga, deggat kon bebin oun 1/1 boode ya 1/() kolle ebarat bode? good luck.

what these expressions means ? 
Max(f1-M,0)
max(g1,0) ?
are they maximize ? or maximum elements ?!

@fereydoon_shekofte tnx Mr.shekofte for these links,very nice. thank u.

@ecterrab tnx for this comprehensive and acutally perfect answer. i learned many thing from these and actually as a Ph.d candidate in mechanical engineering field,i appreciate maple physics package to be very useful and applicable in mechenics. 'actually i face this problem when my friend and i do the same thing but me in maple and my friend mathematrica ! actullay maple do the whole things super faster than mathematica ! i got some differentials using physics package and some integrals ( i also do it by parts ) in just a few seconds ! but my friend do it in mathematica in more than some hours ! and he asked me why this takes too many hours ! and thus i make decision to ask why maple or why mathematica or why not !
anyway,i am really thankfull for your comprehensive answer ! really thank u .

for more information about breakpoint,i have a link :

http://www.mathworks.com/help/matlab/matlab_prog/debugging-process-and-features.html


i think maple will be more perfect and flexible by adding this feature in new versions . hope maple good staff see this post.

 

 

@Preben Alsholm break point in matlab,is a tool that help u break in any line of the code ( for example in your loop) and continue your code manually ( by pressing F10 i think ) . thus the code up to the line which has a break point will go automatically and after that,u can continue computing using f10 line by line ! and if you are into the loop,thus you can continue loop manually .

please upload your worksheet here,tnx



restart:

a := Matrix([1, 2, 3, 4, 5]);

 

for k to 5 do

 

x[k]:=rhs(op(1,op(3,DirectSearch:-SolveEquations([a(1, k)*x+2+y = 0, x+y = 0]))));
y[k]:=rhs(op(2,op(3,DirectSearch:-SolveEquations([a(1, k)*x+2+y = 0, x+y = 0]))));

 

end do;

 

a := Matrix(1, 5, {(1, 1) = 1, (1, 2) = 2, (1, 3) = 3, (1, 4) = 4, (1, 5) = 5})

 

HFloat(-0.9449969839679876)

 

HFloat(-0.05500301603201337)

 

HFloat(-2.0)

 

HFloat(2.0)

 

HFloat(-1.0)

 

HFloat(1.0)

 

HFloat(-0.6666666666666666)

 

HFloat(0.6666666666666666)

 

HFloat(-0.5)

 

HFloat(0.5)

(1)

 



Download similar.mw

@hacker9130368 

@Alejandro Jakubi i think it will take me months to know what is going on in your complicated,but perfect code. could you please explain more ?i am really appreciate it.tnx

@Axel Vogt could you please explain more about this great code ? i actually do not know what has happend in your code ?

@Carl Love it was exactly what i needed,thank u. and thank u for thinking and giving time for my first and second question.

please write down an example of your matrix manually,tnx

u can also find roots using solve and fsolve commands,

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