11 years, 9 days

## rangeasview = true...

implicitplot(x^2 + y^2 = 1, x=-2..2,y=-2..2, rangeasview = true);

## fprintf not the problem...

Your problem is e^x which must be replaced with exp(x) and then all work

## K3...

replace K3 with k3

## output=procedurelist...

I think the problem is in dsolve when you use output=procedurelist. I don't know why.

I changed to

dsol2 := dsolve(dsys1, numeric, method = rkf45, output = Array([0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24]));

It work

The same for

dsol3 := dsolve(dsys1, numeric, method = dverk78, output = Array([60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70]));

it work

## solve...

b=0 is a parameter and not an equation with no variables

restart:b:=0:Eqs := [x*b = y*b]; Vars := [x, y];

solution := solve(Eqs, Vars);

solution := [[x = x, y = y]]

## Two curves in one graphe...

restart: with(plots): spacecurve({[cos(t),sin(t),t],[cos(t)^2,sin(t)^2,t]},t=0..4*Pi);﻿

## labels...

inequal({x[1] >= 0, x[2] >= 0, x[1]-3*x[2] >= -3, 2*x[1]+3*x[2] <= 6}, x[1] = 0 .. 3.5, x[2] = 0 .. 2.5, labels = ["x[1] values", "x[2] values"]);﻿

## Dirichlet condition...

You must introduce your Dirichlet condition into the Matrix A[341,341] and the vector B[341] befor solving the system with LinearSolve.

There are some methods to introduce those type of conditions into A and B.

One methode is to force nodes in diagonal rigidity Matrix A and B with high numbers.

## Plotting solution of differntial system ...

restart: with(plots):sys := diff(x(t),t)=2*x(t)-x(t)*y(t),diff(y(t),t)=-y(t) + 0.4*x(t)*y(t):
> fcns := {x(t), y(t)}:
> p:= dsolve({sys,y(0)=1,x(0)=5},fcns,type=numeric,method=classical):
> odeplot(p, [[t,x(t)],[t,y(t)]],-4..4);

## printlevel:=0:...

Try this:

restart: printlevel:=0:max1:=60:
q:=1/2:

for w1 from 1/27 by 7/27 to 1/2 do

ww:= [seq( [n, add(1/(n+1), k=ceil(  max(0,(q-w1)/(1-w1)*n)  )..floor( min(n, q/(1-w1)*n)  ))], n=1..max1)]:
plot(ww, style=point):
plot(w1/(1-w1),0..max1,color=blue):
print(plots:-display(%,%%)); print(evalf(w1), evalf(1/w1));
end do;

correct.mw

## Sturm Liouville pde...

Those type of pde are solved symbolically (not numerically because parameter lambda must be found) as :

restart: with(LinearAlgebra):dsolve(diff(y(x),x,x)+lambda^2*y(x)=0,y(x)): y:=unapply(rhs(%),x);
y := x -> _C1 sin(lambda x) + _C2 cos(lambda x)

> bc:=[y(0)=0, y(1)=0];

bc := [_C2 = 0, _C1 sin(lambda) + _C2 cos(lambda) = 0]

> with(linalg):Cco:=genmatrix(bc,[_C1,_C2]);

[     0                  1           ]
Cco := [                                      ]
[sin(lambda)    cos(lambda)]

> Chareqn:=det(Cco)=0;

Chareqn := -sin(lambda) = 0

> lambda:=solve(Chareqn,lambda,AllSolutions);

lambda := Pi _Z1~

> lambda:=subs(_Z1=n,lambda);
﻿y:=x->sin(n*Pi*x);

Here, the boundary are of type dirichlet.

You can do the same study with newman and mixed boudaries

## Save as or Export as...

You can save or export your maple document as an RTF (word document) or Latex or text

## dsolve symbolically and numerically...

You can solve it symbolically as:

restart: with(plots):

dsys1 := diff(y(x), `\$`(x, 2))+(lambda*lambda)*y(x) = 0;

bc1 := y(0) = 0, D(y)(0) = 1; # just two boundary conditions are required

dsolve({bc1,dsys1});

Or numerically with assigning a value to lambda as:

lambda:=1: p:=dsolve({bc1,dsys1},numeric); # example with lambda:=1

odeplot(p);

## (fog)(x)=f(g(x))...

See

?composition

and

restart: with(Student[Precalculus]): f:=x->x^2; g:=x->cos(x); CompositionTutor(f(x), g(x));

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