Plotting – Plot the solution from DSolve

I try to solve a differential equation as in the following code:

FullSimplify(DSolve(x'(t) == a + b E^(g t) + (c + d E^(-g t)) x(t), x(t), t))

what generates

Enter image description here

Now I want to draw it with certain assigned parameter values, for example: a = 1; b = 2; c = 3; d = 4; g = 0.1; A = 1 where I replaced the integration constant c_1 With A,

Here is my code for plotting x(t):

a = 1; b = 2; c = 3; d = 4; g = 0.1; A = 1
x(t_) := E^(-((d E^(-g t))/g) + c t) (A + Integrate(E^((d E^(-g K(1)))/g - c K(1)) (a + b E^(g K(1))), {K(1), 1, t}))
Plot(x(t), {t, 1, 10})

It works forever. To check if Mathematica performs calculations, I tried

x(1)

and it gave way

3.83926*10^-15

which is nice. But when I tried

x(2)

I have

Enter image description here

It seems that Mathematica can only compute the integral if the integration region is $ int_1 ^ 1 $, Is that because the integrand is too complicated? Is there a way to let Mathematica calculate this? Many Thanks!

Differential Equations – Using NDSolve on Wave PDE on String if the solution is given at 2 different times instead of the initial velocity?

This is a PDE from a Maple document. Mathematica DSolve can not solve it right now.

I wanted to review the Maple solution with NDSolve. This is a string of length 1, which is fixed on the left and can move freely to the right. Give a starting position and let go.

Here are the specifications of the PDE

Solve for $ 0 <x<1, t>0 $ the wave PDE
$$
-u_ {tt} + u (x, t) = u_ {xx} + 2e ^ {- t} left (x – frac {1} {2} x ^ 2 + frac {1} {2} t – 1 right)
$$

With boundary condition

begin {align *}
u (0, t) & = 0 \
frac { partial u (1, t)} { partial x} & = 0
end {align *}

And initial conditions

begin {align *}
u (x, 0) & = x ^ 2-2 x \
u (x, 1) & = u left (x, frac {1} {2} right) + e ^ {- 1} left ( frac {1} {2} x ^ 2-x right ) – left ( frac {3} {4} x ^ 2- frac {3} {2} x right) e ^ { frac {-1} {2}}
end {align *}

The tricky thing is that no initial speed is given. But only starting position $ t = 0 $, and then instead a relation to the solution is given at 2 different times.

NDSolve Complain with this dreaded mistake

The boundary condition is not indicated on a single edge of the border
the computational domain.

And I do not know how to get rid of it. Here is the code

Delete everything[u, x, t];
pde = -D[u[x, t], {t, 2}]+[x, t] ==
D[u[x, t], {x, 2}]+ 2 * exp[-t]* (x - (1/2) * x ^ 2 + (1/2) * t - 1);

bc = {u[0, t] == 0, derivative[1, 0][u][1, t]    == 0};

ic = {u[x, 0] == x ^ 2 - 2 * x,
u[x, 1] == u[x, 1/2] + ((1/2) * x ^ 2 - x) * Exp[-1] - ((3 * x ^ 2) / 4 - (3/2) * x) * Exp[-2^(-1)]};

sol = NDSolve[{pde, ic, bc}, u, {x, 0, 1}, {t, 0, 1}]

Here's the Maple code and the analytic solution it contains

pde: = -diff (u (x, t), t, t) + u (x, t) =
diff (u (x, t), x, x) + 2 * exp (-t) * (x- (1/2) * x 2 + (1/2) * t-1);
ic: = u (x, 0) = x ^ 2-2 * x,
u (x, 1) = u (x, 1/2) + (1/2) * x ^ 2-x) * exp (-1) - (3/4 * (x ^ 2) -3/2 * x) * exp (-1/2);
bc: = u (0, t) = 0, eval (diff (u (x, t), x), {x = 1}) = 0;
pdsolve ([pde, ic, bc], u (x, t))

$$
u (x, t) = – frac {e ^ {- t}} {2} (x ^ 2-2 x) (t-2)
$$

Here is an animation of the Maple solution that I wanted to review

mapleSol[x_, t_] : = - (Exp[-t]/ 2) (x ^ 2 - 2 x) (t - 2)
Manipulate[
plot[mapleSol[mapleSol[mapleSol[mapleSol[x, t], {x, 0, 1}, PlotRange -> {{0, 1}, {-1, .1}}].
{{t, 0, "time"}, 0, 10, .1}
]

Enter image description here

Any suggestion how to fix the bug of NDSolve?

Using V 12 on Windows 10. ps. I also solved this by hand, but I can not find a Maple solution, and my solution looks wrong. I still have to find out why.

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Equation Solution – Definition of BCs for complex shape areas

is it possible that we can do such BCs for complex form with NDsolve,

The initial value for a disk is 1. Then we have to define a circle with such Dirichlet BCs: i.

code

(* Initial parameters *) needs["NDSolve`FEM`"];
Kappa = 1 / 10.0;
xmin = -1.0;
xmax = 1;
ymin = -1;
ymax = 1.0;

[CapitalOmega] = Rectangle[{xmin, ymin}, {xmax, ymax}];
RegionPlot[[[[[CapitalOmega], AspectRatio -> Automatic]mesh = ToElementMesh[[[[[CapitalOmega], "MaxCellMeasure" -> 1/1000];
mesh["Wireframe"]
n = length[mesh["Coordinates"]]u0 = ElementMeshInterpolation[{mesh}, RandomReal[{1, 1}, n]];
Plot3D[u0[x, y], {x, y} [Element] mesh][CapitalOmega]1 = hard disk[{0, 0}, xmax/10];
RegionPlot[[[[[CapitalOmega]1, aspect ratio -> automatic]mesh1 = ToElementMesh[[[[[CapitalOmega]1, "MaxCellMeasure" -> 1/1000];
mesh1["Wireframe"]
(* n = length[mesh1["Coordinates"]]u01 = ElementMeshInterpolation[{mesh1},0.5-RandomReal[{0,1},n]];
Plot3D[u01[x,y], {x, y} [Element]mesh1]*)


u = NDSolveValue[{d[x, y] - 1. -
4 * Kappa * Kappa * Laplace[d[x, y], {x, y}]== 0,
DirichletZustand[d[x, y] == 0, [CapitalOmega]1]d[x, y] == 0},
d, {x, y} [Element] [CapitalOmega]];

Equation Solution – Should Reduce specify all cases where $ sqrt {x y} = sqrt x sqrt y $?

My understanding is that To reduce Specifies all conditions (with or) where the input is true.

Now, $ sqrt {xy} = sqrt x sqrt y $, from where $ x, y $ are realunder the following three conditions / cases

$$
begin {align *}
x geq 0, y geq0 \
x geq0, y leq0 \
x leq0, y geq 0 \
end {align *}
$$

but not when $ x <0, y <0 $

This is verified by this

Delete everything[x,y]
Provided[Element[{x,y},Reals]&& x> = 0 && y <= 0, simplify[ Sqrt[x*y] - sqrt[x]* Sqrt[y]]]Provided[Element[{x,y},Reals]&& x<= 0&&y>= 0, simplify[ Sqrt[x*y] - sqrt[x]* Sqrt[y]]]Provided[Element[{x,y},Reals]&& x<= 0&&y>= 0, simplify[ Sqrt[x*y] - sqrt[x]* Sqrt[y]]]Provided[Element[{x,y},Reals]&& x <= 0 && y <= 0, simplify[ Sqrt[x*y] - sqrt[x]* Sqrt[y]]]

Mathematica Graphics

Why then?

    To reduce[ Sqrt[x*y] - sqrt[x]* Sqrt[y]== 0, {x, y}, reales]

Do you just indicate one of the three cases mentioned above?

Mathematica Graphics

Is my understanding of To reduce wrong or should To reduce did the other two cases exist?

V 12 under Windows.

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How much solution to fill Paterson Universal Tank during film development?

Moving with Paterson tanks is done by rotating the roller, so only the film needs to be covered. For other systems where stirring may be by inversion or otherwise, the tank must be present nearly full (some air is needed).

The necessary chemical reactions are concentration- and time-dependent. More or less solution in the tank does not affect the results, as long as the film is sufficiently covered.