DJJerome1976

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12 years, 89 days

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These are answers submitted by DJJerome1976

Try using the numpoints option in the plot command as follows:

 

plot(sin(x),x=0..2*Pi,numpoints=1000)

 

You may try a larger value than 1000 if necessary.

The inequality is true for all values of x.

Try this:
> with(DETools);
> DEplot((D(u))(t) = -u(t)^5+u(t)^3-u(t)^2, u(t), t = -5 .. 5, u = -5 .. 5, arrows = medium);

Maple does not recognize e^(y/x) as the exponential function. You have to use exp(y/x).

Since you mentioned having difficulty using, in particular, the map command, here is one way your problem can be solved:

 

F:=x->Matrix([[x,x^2],[exp(x),sin(x)]]):

FF:=unapply(map(diff,A(x),x),x):

 

This allows you to do, for example, F(0) and F'(0), and get the expected matrices.

Try this:

 

with(plots):

implicitplot(x^2+(y-3)^2=25,x=-10..10,y=-10..10,numpoints=1000)


simplify((3*h^2+12*h)/h)

restart

with(plots):

K:=(V,E)->E*log(V):

P:=(V,E)->E+V*log(V):

plot3d({K(V,E),P(V,E)},V=0..20,E=0..190)

The DEplot command is made available by loading the DETools package:

with(DETools):

A simply fix would be to change all instances of \[ ... \] in your LaTeX code to \( ... \) (or $ ... $) if you prefer. \[ ... \] creates displaymath which is centered, whereas \( ... \) creates inline math which is not centered.

The following command generates the phase portrait along with the trajectory of the solution satisfying the condition x(0)=-1, y(0)=2.

 

with(DETools):

DEplot({(D(x))(t) = x(t)-y(t)^2, (D(y))(t) = x(t)*(D(y))(t)+2*y(t)}, {x(t), y(t)}, t = 0 .. 5,x = -5 .. 5, y = -5 .. 5, arrows = medium, [[0, -1, 2]]);


Hope this gets you in the right direction.

with(plots):

A:=Array(1..3,1..1):

A[1,1]:=plot(x,x=0..0.25):

A[2,1]:=plot(x^2,x=0..0.25):

A[3,1]:=plot(x^3,x=0..0.25):

display(A)

plota:=plot(x,x=-1/2..0):

plotb:=plot(x^2,x=-1/2..0):

plotc:=plot(x^3,x=0..1/2):

plotd:=plot(x^4,x=0..1/2):

plots:-display({plota,plotb,plotc,plotd})


> dsolve({(D(x))(t) = 2*x(t)-2*z(t), (D(y))(t) = 2*y(t)+2*z(t), (D(z))(t) = -2*x(t)+2*y(t)+4*z(t)}, {x(t), y(t), z(t)});
/ 1
{ x(t) = _C2 - - _C3 exp(6 t) + exp(2 t) _C1,
\ 2

1
y(t) = - _C3 exp(6 t) - _C2 + exp(2 t) _C1,
2

\
z(t) = _C2 + _C3 exp(6 t) }
/

The following would work:

sol:=dsolve({f'''(x)-f(x)*f''(x),f(0)=0,f'(6)=1,f'(-6)=1},numeric):

This creates a numerical procedure defining the solution, which can be evaluated as follows:

sol(1)

Hope this, at least, gets you started.

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