## 13999 Reputation

11 years, 189 days

## is...

Unfortunately, the  simplify  command does not always cope with the task. If you want to check the equivalence with one command, then you can use the  is  command:

 > restart;
 > expr1:=(-exp(n*Pi*(2*b - y)/a) + exp(n*Pi*y/a))/((exp(2*n*Pi*b/a) - 1)): expr2:= sinh(n*Pi/a*y)/tanh(n*Pi/a*b)-cosh(n*Pi/a*y): is(expr1-expr2=0);
 (1)
 >

## Solution...

I wrote down all the code in 1d math (I hate 2d math input) and added the plotting of  x(t)  and  y(t) :

 > restart;
 > Phi:=; Sys:=Equate(diff(Phi,t), <6,1; 4,3>.Phi+<6*t, -10*t+4>); Sol:=dsolve(Sys); plot(eval([x(t),y(t)], eval(Sol,[_C1=1,_C2=1])), t=0..0.5, color=[red,blue]);

## plot3d...

Side surface and bases of a cylinder can be specified parametrically using 2 parameters (as any 2D surface).

```Side:=plot3d([cos(t),sin(t),z], t=0..2*Pi, z=0..2, style=surface, color=green):
Base1:=plot3d([R*cos(t),R*sin(t),0], t=0..2*Pi, R=0..1, style=surface, color=green):
Base2:=plot3d([R*cos(t),R*sin(t),2], t=0..2*Pi, R=0..1, style=surface, color=green):
plots:-display(Side,Base1,Base2);
```

## solve...

```tax := 0.3*profit:
profit := 0.1*totalsalesx:
solve(totalsalesx = 60.1 + tax + profit);
```

69.08045977

If you need an absolutely accurate (symbolic solution) then use fractions:

```tax := 3/10*profit:
profit := 1/10*totalsalesx:
solve(totalsalesx = 60+1/10 + tax + profit);
floor(%) %+ frac(%); # Isolation of the whole part```

6010/87
69 + 7/87

## Solution...

 > restart; lambda := 2*(1/10);                               mu := -1; beta := 10;                               alpha := -25;                               C := 1;                                k := (1/12)*sqrt(6)/sqrt(beta*lambda*mu);                       w := alpha/((10*sqrt(-lambda*mu))*beta);                            A[0] := (1/2)*alpha/((10*sqrt(-lambda*mu))*((1/12)*beta*sqrt(6)/sqrt(beta*lambda*mu))); A[1] := -(1/10)*alpha/((1/12)*beta*mu*sqrt(6)/sqrt(beta*lambda*mu));       A[2] := -(12*((1/12)*sqrt(6)/sqrt(beta*lambda*mu)))*lambda^2*alpha/(10*sqrt(-lambda*mu));                     H := ln(sqrt(lambda/(-mu))*tanh(sqrt(-lambda*mu)*(xi+C)));                xi := k*x-t*w;                     u[0] := A[0]+A[1]*exp(-H)+A[2]*exp(-H)*exp(-H); plot3d(Im(u[0]), x = -10 .. 10, t = -10 .. 10, view=-50..50);
 > A:=plots:-contourplot3d(Im(u[0]), x = -10 .. 10, t = -10 .. 10, color=red, thickness=3, contours=[seq(C,C=-40..40,10)], coloring=[white,blue], view=-50..50, filledregions=true, grid=[100,100]):
 > B:=plots:-contourplot(Im(u[0]), x = -10 .. 10, t = -10 .. 10, color=red, contours=[seq(C,C=-40..40,10)], grid=[100,100]):
 > f:=plottools:-transform((x,y)->[x,y,-50]): plots:-display(A,f(B));
 >

 > restart; ContoursWithLabels := proc (Expr, Range1::(range(realcons)), Range2::(range(realcons)), Number::posint := 8, S::(set(realcons)) := {}, GraphicOptions::list := [color = black, axes = box], Coloring::`=` := NULL) local r1, r2, L, f, L1, h, S1, P, P1, r, M, C, T, p, p1, m, n, A, B, E; uses plots, plottools; f := unapply(Expr, x, y); if S = {} then r1 := rand(convert(Range1, float)); r2 := rand(convert(Range2, float)); L := [seq([r1(), r2()], i = 1 .. 205)]; L1 := convert(sort(select(a->type(a, realcons), [seq(f(op(t)), t = L)]), (a, b) ->is(abs(a) < abs(b))), set); h := (L1[-6]-L1[1])/Number; S1 := [seq(L1[1]+(1/2)*h+h*(n-1), n = 1 .. Number)] else S1 := convert(S, list)  fi; print(Contours = evalf[2](S1)); r := k->rand(20 .. k-20); M := []; T := []; for C in S1 do P := implicitplot(Expr = C, x = Range1, y = Range2, op(GraphicOptions), gridrefine = 3); P1 := [getdata(P)]; for p in P1 do p1 := convert(p[3], listlist); n := nops(p1); if n < 500 then m := `if`(40 < n, (r(n))(), round((1/2)*n)); M := `if`(40 < n, [op(M), p1[1 .. m-11], p1[m+11 .. n]], [op(M), p1]); T := [op(T), [op(p1[m]), evalf[2](C)]] else if 500 <= n then h := floor((1/2)*n); m := (r(h))(); M := [op(M), p1[1 .. m-11], p1[m+11 .. m+h-11], p1[m+h+11 .. n]]; T := [op(T), [op(p1[m]), evalf[2](C)], [op(p1[m+h]), evalf[2](C)]] fi; fi; od; od; A := plot(M, op(GraphicOptions)); B := plots:-textplot(T); if Coloring = NULL then E := NULL else E := ([plots:-densityplot])(Expr, x = Range1, y = Range2, op(rhs(Coloring)))  fi; display(E, A, B); end proc: z := -y + sech(x - 3*t); w := 10*sech(x - 3*t); with(plots): P1 := plot(eval(w, t = 0), x = -10 .. 10): P2 := contourplot(eval(z, t = 0), x = -10 .. 10, y = -eval(w, t = 0) .. eval(w, t = 0), contours = 5, grid = [101, 101]): display(P1, P2); Q2:=ContoursWithLabels( eval(z, t = 0), -10 .. 10, -10..10, {-7,-3,1,5,9}, [color=blue,axes=box]): display(P1,Q2);
 >

## Solution...

I do not understand why you got solutions in terms of Bessel functions. The direct solution below is expressed through the function  erfi :

```E:=1/2:
Eq := -(1/2)*(diff(psi(x), x, x))+(1/2)*x^2*psi(x) = E*psi(x);
Sol := dsolve(Eq);
plot(eval(rhs(Sol),[_C1=2,_C2=1]), x=0..2);

```

## Convolution...

There are different versions of this concept. See the wiki for this. For example, for functions defined on the positive half-axis (for the negative half-axis they are considered equal to 0), the problem reduces to calculating the integral over a finite interval, i.e. in Maple we can use the  int   function.

Example:

```restart;
f:=t->sin(t):
g:=t->cos(t):
int(f(t)*g(x-t), t=0..x);
plot(%, x=0..10);
```

## Explanation...

You did not specify what values the function  f  takes outside the segment  [0,1] . (Maple ignores your original assumption  assume(0 <= x, x <= 1)). By default, in this case outside the segment  [0,1]  the function  f  is considered equal to 0. If you specify in the definition of the function that it is undefined outside the segment  [0,1] , then everything is OK (see the function  g  below):

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You also wrote: "Whats is the relationship between this example and Existence and uniqueness theorem for fist order ode". I do not see any relationship here.

## Solution...

If I understand the question correctly, then here is the solution (x(t)  is a vector-function  x(t)=<x1(t),x2(t)>):

```restart;
A:=<1,2; 3,4>;
ODE:=Equate(diff(<x1(t),x2(t)>, t),A.<x1(t),x2(t)>);
dsolve({ODE[],x1(0)=0,x2(0)=1});
```

## Visual verification of Joachimsthal's t...

Here is a visalization of Joachimsthal's theorem using the ellipse  x^2/5^2+y^2/3^2=1  as an example. I took the point  P0=[1,1]  inside the ellipse and found 4 normals to the ellipse passing through this point (P0A, P0B, P0C, P0E  in the pic.). The point  E1  is diametrically opposite the point  E . I found the equation of the circle passing through points  A, B, C. We see in the pic. that this circle also passes through the point E1.

 > restart; Ellipse:=plots:-implicitplot(x^2/5^2+y^2/3^2=1, x=-6..6,y=-6..6, color=blue): P0:=[1,1]: Sys:={x^2/5^2+y^2/3^2=1, (x-P0[1])*(y/3^2)-(x/5^2)*(y-P0[2])}; Sol:=[solve(Sys,explicit)]: L:=map(t->eval([x,y],t),simplify(fnormal~(evalf(Sol),9), zero)); Points:=plots:-pointplot([L[],-L[2],P0], symbol=solidcircle, color=[red\$5,green], symbolsize=15): a:=L[1]: b:=L[3]: c:=L[4]: e:=L[2]: e1:=-e: P0A:=plot([P0,a], color=green): P0B:=plot([P0,b], color=green): P0C:=plot([P0,c], color=green): P0E:=plot([P0,e], color=green): geometry:-circle(cc,[geometry:-point(A,a),geometry:-point(B,b),geometry:-point(C,c)]): Eq:=geometry:-Equation(cc, [x,y]); Circle:=plots:-implicitplot(Eq, x=-2..7, y=-4..4, color=red): T:=plots:-textplot([[P0[],"P0"],[a[],A],[b[],B],[c[],C],[e[],E],[e1[],E1]],font=[times,18], align={below,right}): plots:-display(Ellipse,Points, Circle, P0A, P0B, P0C, P0E, T, view=[-5.7..5.7,-3.7..3.7], scaling=constrained, size=[800,500]);

When solving numerically, all parameters must be specified.

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## intsolve...

See help on the  intsolve  command. Help quote: "When the solution contains integrals, they are represented with the inert Int." You can try to calculate these integrals by any methods, for example by  evalf(Int(...), method=...) .

## normal...

Use the normal command which is not in the "prohibited" list of commands or just solve by hand (without Maple):

normal(-1/(y-1)-y/(y-1)-y^2-2*y-1+y^3/(y-1)+y^2/(y-1));

0

## lambda as protected name...

1. Use the  Number Theory package instead of deprecated numtheory one.

2. lambda becomes protected only after calling  Number Theory package, because this is the name of the built-in function from this package. You can use  unprotect command to remove this protection (but then you cannot use this function):

```restart;
with(NumberTheory):
lambda:=5;
unprotect(lambda);
lambda:=5;
lambda;
```

If you want to work with this package and use the symbol  lambda  at the same time, it is better not to call the whole package, but only the command from the package that you need:

```restart;
NumberTheory:-lambda(100);
lambda:=5;
lambda;
```

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