Trascendental equation in Mathematica

I cannot make my Mathematica solve with Solve the following trascendental equation with coefficient A,B,C,D,F uniform but symbolic:

-Aexp[-C]-Bexp[+D] = (1 – Ax^[-C] – Bx^[+D])/log[x] – exp[2]*(F/x)^2

Would someone please help to find a solution, if an analytical exists? I do not have hypothetical solution x where start from. I could try to guess the numerical value of the constants A,B,C,D,F, but not trivial

Using output from Solve or Reduce as a value in subsequent equation

I’m trying to run a simulation. I will number sentences to make response easier. (1) Here is my first equation:

𝑗[LBar_, y_, x1_, σ_, X2M_, w_, X1M_]:=(LBar(𝜎−1)⁄𝜎 + 𝑦(𝜎−1)⁄𝜎−x1(𝜎−1)⁄𝜎)𝜎⁄(𝜎−1)−(𝑦−X2M+𝑤(LBar+X1M)−𝑤x1) 

(2) I insert some parameter values and Reduce:

Reduce[𝜕x1(𝑗[100,𝑦,x1,13⁄,42,.23,80])==0 && x1>0, x1, Reals] 

(3) This produces output:

x1 == 119.575 

(4) When I try to call the output value (119.575) I run into problems. (5) For instance

2 % 

results in:

2 (x1 == 119.575) 

(6) and

j[100, 80, %[[2]], 1/3, 42, .23, 80] 

produces this output:

-79.4 + 1/Sqrt[41/160000 - 1/239.149[[2]]^2] + 0.23 239.149[[2]] 

How do i use Streamplot to plot a non homogenous differential equation

I have the following equations:

x’ = x-y y’ = x+y-2xy

I used the following code to do the streamplot: I apologize for only having an image, as I am using Mathematica through my school servers I cannot copy and paste the data.

enter image description here

But I do not get any results, so then I linearize the equations and plot them separately, but I cannot figure out how to combine the plots to show in one plot.:

enter image description here

Is there any way to use Streamplot with nonhomogenous DE’s or a better way to combine the equations?

Why both DSolve and NDSolve are unable to solve a second-order differential equation?

I am trying to solve a recurrence relation using generating functions method. After some long calculations, I have arrived to this second-order differential equation: $ 0.5 x^5 y”(x)+(2x^4+x^3)y'(x)+\left(x^3+x^2+x-1\right)y(x)+1=0$

and these conditions: $ y(0)=1, y'(0)=1$ . $ y(x)$ is the function that needs to be expanded as Taylor Series at $ x=0$ to obtain the sequence from the coefficients. However, when I try to solve it both using DSolve and NDSolve, I have no luck. With DSolve it just returns the request itself:

$ $ \text{DSolve}\left[\left\{0.5 x^5 y”(x)+(2. x+1) x^3 y'(x)+\left(1. x^3+x^2+x-1\right)y(x)+1=0,y(0)=1,y'(0)=1\right\},y,x\right]$ $

And with NDSolve I just receive errors and no equation:

Power::infy: Infinite expression 1/0.^5 encountered. Infinity::indet: Indeterminate expression 0. ComplexInfinity encountered. NDSolve::ndnum: Encountered non-numerical value for a derivative at x == 0.`. 

$ \text{NDSolve}\left[\left\{0.5 x^5 y”(x)+(2. x+1) x^3 y'(x)+\left(1. x^3+x^2+x-1\right)y(x)+1=0,y(0)=1,y'(0)=1\right\},y,\{x,0,1\}\right]$

How could I resolve this problem?

Solving the heat equation using Laplace Transforms

I am trying to solve the 1-D heat equation using Laplace Transform theory. The equation is as follows. I don’t have the capability to write the symbols so I will write it out.

                     partial u/partial t = 2(partial squared u/ partial x squared) -x    boundary conditions are partial u/partial x(0,t)=1, partial u/partial x(2,t)=beta. 

The problem asks the following: (a). For what value of beta does there exist a steady-state solution? (b). if the initial temperature is uniform such that u(x,0)=5 and beta takes the value suggested by the answer to part (a), derive the equilibrium temperature distribution.

I was able to get an equation that looks like U(x,s)=c e^(s/2)^1/2 -(1/s)((x/s)-u(x,0)). But I am not sure how to go from here to solve for beta using the boundary conditions. I need some assistance from someone.

Steady state solution (1D) of nonlinear dispersal equation

Now I’m interested in the equation $ $ \frac{\partial }{\partial x}\Bigl(\text{sgn}(x) u \Big) +\frac{\partial}{\partial x} \Bigl[ u^2 \frac{\partial u}{\partial x} \Bigr] =0$ $ with boundary conditions $ u(-5)=u(5)=0$

Since $ \text{sgn}(x)$ is not differentiable at $ x=0$ , I expectd ND solve to have some problems. I tried

sol = NDSolveValue[{   0 == D[Sign[x]*u[x],x] + D[u[x]^2 D[u[x], x], x],    u[-6] == 0, u[6] == 0}   , u, {x, -7, 7}] 

but I can’t even plot it and I think that I’m writing it in the wrong way. Could someone confirm I wrote the right snippet and show the plot I should obtain?

  • I asked a related question three days ago, where the equation was the PDE $ \partial_t u = \partial_x (\text{sign}(x) u) + \partial_x (u^2\partial_x u)$ . The one I have above it’s the steady state solution, and I want to compute it directly, instead of integrating in time.

Finding a valid equation for fixed point problem

I currently am working on learning more about fixed point method. Finding equations that satisfy the constraints of a g function can sometimes require a bit of engineering. I have come across one that many would consider simple. Yet, I have been stuck on it for some time now.

Here it is $ f(x) = x^2 – x – 2 = 0 $ on $ [1.5,3]$ .

I have tried many things; however, I have yet to successfully discover one that maps domain to range for both $ g$ and $ g^\prime$ .

Would anyone be able to give me a guiding hand?

What can I do to get Mathematica to solve this equation?

I am trying to solve an equation on Mathematica 12.1 to get angles in radians as part of some analysis:

forceToStrain[in_] := in/(51.76*10^9); strainFromTheta[in_] := ((in/(2*Sin[in/2])) - 1) // N (*thickness = 0.28 * 16 ply = 4.48 mm = 4.48*10^(-2 m)*)  Solve[((x/(2*Sin[x/2])) - 1) == forceToStrain[10], x] 

But I am getting this error:

Reduce::inex: Reduce was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Reduce require exact input, providing Reduce with an exact version of the system may help. 

It doesn’t seem complicated as an equation as I can solve it by hand:

strainFromTheta[.00007] forceToStrain[10] Output: 2.04167*10^-10 and 1.93199*10^-10 

Can anyone tell me whether I could use a different function or rearrange the equation differently to get the angles?