calculus – Solve this differential equation when $y(0)=1$

Let $y”-xy=0$, $y(0)=alpha$, $y'(0)=beta$. Also, $y=sum_{n=0}^{infty}a_n x^n$ is a solution to y.

Is this differential equation first order linear? And if $alpha=1$, $beta=0$, then does $a_0=1, a_1=0$? (recall that $a_n$ is the series)

For the first question, my answer is yes. Also if $alpha=0, beta=0$, then $y(0)=1, y'(0)=0$. I’m not really sure how $alpha, beta$ are related to the equation, but my intuition tells me that this is true. Intuitions are not really that reliable so I wanted to verify if this is true or not.

equation – Evaluating a time dependent variable out of NDsolve/in NDsolve without affecting the result

this is my first question here.
I am having trouble, I might be able to express what I want in Matlab but I suffer to do that in Mathematica.
So I have a direction cosine matrix that is solved in NDsolve. I have a vector, say vec(0)={1,0,0}. I want to have these initial conditions and do a time marching to this vector (i.e. multiply it with my direction cosine matrix which is also time-dependent, and as mentioned solved in NDsolve). In the end, I want to create an animation of that vector.
I tried to put x’ (first component of vector), y’, and z’ expression in the NDsolve where I get the components of direction cosine (please do not confuse it with built-in direction cosine it has nothing to do with my assignment); but it turned out that putting x’, y’, z’ somehow managed to mess up my solution.
I am open to brand new approach to obtain this vector.
I add last part of my code below.

tmax = 10
m = 4
lx = 0.3
ly = 0.1
lz = 0.1
eqnplotorig = {Derivative(1)((Theta)x)(t) == 
   Sec((Theta)y(t)) (Cos((Theta)z(t)) (Omega)x(t) - 
      Sin((Theta)z(t)) (Omega)y(t)), 
  Derivative(1)((Theta)y)(t) == 
   Sin((Theta)z(t)) (Omega)x(t) + Cos((Theta)z(t)) (Omega)y(t), 
  Derivative(1)((Theta)z)(
    t) == -Cos((Theta)z(t)) Tan((Theta)y(t)) (Omega)x(t) + 
    Sin((Theta)z(t)) Tan((Theta)y(t)) (Omega)y(t) + (Omega)z(t), 
  Derivative(1)((Omega)x)(
    t) == (36 lx^2 ly lz (ly^2 - lz^2) (Omega)x(t)^2 - 
      3 lx (Omega)x(
        t) ((9 ly^4 lz + 22 ly^2 lz^3 - 3 lz^5 - 
            lx^2 (11 ly^2 lz + 3 lz^3)) (Omega)y(t) + 
         ly (3 ly^4 - 22 ly^2 lz^2 - 9 lz^4 + 
            lx^2 (3 ly^2 + 11 lz^2)) (Omega)z(t)) - 
      4 (-3 ly (lx^2 - 2 ly^2) lz (lx^2 + lz^2) (Omega)y(t)^2 + 
         2 (lx^2 + ly^2) (ly^2 - lz^2) (lx^2 + lz^2) (Omega)y(
           t) (Omega)z(t) + 
         3 ly (lx^2 + ly^2) lz (lx^2 - 2 lz^2) (Omega)z(
           t)^2))/(5 lx^4 (ly^2 + lz^2) + 32 ly^2 lz^2 (ly^2 + lz^2) +
       lx^2 (5 ly^4 - 62 ly^2 lz^2 + 5 lz^4)), 
  Derivative(1)((Omega)y)(
    t) == (3 lx (5 lx^2 - 16 ly^2) lz (ly^2 + lz^2) (Omega)x(
        t)^2 + (Omega)x(
        t) (3 ly lz (lx^2 (-23 ly^2 + lz^2) + 
            12 (ly^2 + lz^2)^2) (Omega)y(
           t) + (lx^2 + ly^2) (5 lx^2 + 12 ly^2 - 20 lz^2) (ly^2 + 
            lz^2) (Omega)z(t)) + 
      3 lx (3 ly^2 lz (-4 lx^2 + 3 ly^2 + 7 lz^2) (Omega)y(t)^2 + 
         ly (3 ly^4 + 20 ly^2 lz^2 + 9 lz^4 + 
            lx^2 (3 ly^2 - 13 lz^2)) (Omega)y(t) (Omega)z(
           t) + (lx^2 + ly^2) lz (7 ly^2 - 5 lz^2) (Omega)z(
           t)^2))/(5 lx^4 (ly^2 + lz^2) + 32 ly^2 lz^2 (ly^2 + lz^2) +
       lx^2 (5 ly^4 - 62 ly^2 lz^2 + 5 lz^4)), 
  Derivative(1)((Omega)z)(
    t) == (-3 lx ly (5 lx^2 - 16 lz^2) (ly^2 + lz^2) (Omega)x(
        t)^2 - (Omega)x(
        t) ((lx^2 + lz^2) (ly^2 + lz^2) (5 lx^2 - 20 ly^2 + 
            12 lz^2) (Omega)y(t) + 
         3 ly lz (lx^2 (ly^2 - 23 lz^2) + 
            12 (ly^2 + lz^2)^2) (Omega)z(t)) + 
      3 lx ((lx^2 + lz^2) (5 ly^3 - 7 ly lz^2) (Omega)y(t)^2 - 
         lz (9 ly^4 + 20 ly^2 lz^2 + 3 lz^4 + 
            lx^2 (-13 ly^2 + 3 lz^2)) (Omega)y(t) (Omega)z(t) - 
         3 ly lz^2 (-4 lx^2 + 7 ly^2 + 3 lz^2) (Omega)z(
           t)^2))/(5 lx^4 (ly^2 + lz^2) + 32 ly^2 lz^2 (ly^2 + lz^2) +
       lx^2 (5 ly^4 - 62 ly^2 lz^2 + 5 lz^4)), (Theta)x(0) == 
   0, (Theta)y(0) == 0, (Theta)z(0) == 0, (Omega)x(0) == 
   2 (Pi)/180, (Omega)y(0) == 2 (Pi)/180, (Omega)z(0) == 
   2 (Pi)/180}
solplot3Dorig = 
 Flatten@NDSolve(
   eqnplotorig, {(Theta)x, (Theta)y, (Theta)z, (Omega)x, 
(Omega)y, (Omega)z}, {t, 0, tmax})
Plot(Evaluate({(Theta)x(t), (Theta)y(t), (Theta)z(t)} /. 
   solplot3Dorig), {t, 0, tmax}, 
 PlotLegends -> 
  Placed({(Theta)x(t), (Theta)y(t), (Theta)z(t)}, {0.85, 0.55}))
Plot(Evaluate({(Omega)x(t), (Omega)y(t), (Omega)z(t)} /. 
   solplot3Dorig), {t, 0, tmax}, 
 PlotLegends -> 
  Placed({(Omega)x(t), (Omega)y(t), (Omega)z(t)}, {0.85, 0.55}))

Here I solve theta values that will be used in my direction cosine matrix. Which is defined as :

Dirmat = {{1, 0, 0}, {0, Cos((Theta)x(t)), -Sin((Theta)x(t))}, {0, 
    Sin((Theta)x(t)), Cos((Theta)x(t))}} . {{Cos((Theta)y(t)), 0, 
    Sin((Theta)y(t))}, {0, 1, 0}, {-Sin((Theta)y(t)), 0, 
    Cos((Theta)y(t))}} . {{Cos((Theta)z(t)), -Sin((Theta)z(t)), 
    0}, {Sin((Theta)z(t)), Cos((Theta)z(t)), 0}, {0, 0, 1}}

So what I need is a “vec” vector variable time march that has some initial values like {1,0,0} (or whatever) and time march it so that I can animate it using “frames” or something like that.

Sorry, it was a long question.
Any help will be appreciated.

equation solving – Why does Reduce give False on SameQ with its copy-pasted result?

If I evaluate

Reduce(#, x) & /@ {{4 x > 1, 2 x < 5}, {5 x > 1, x < 5}, {4 x > 1, x < 5}}

the result is

{1/4 < x < 5/2, 1/5 < x < 5, 1/4 < x < 5}

If I select the result and copy-paste it on a new line (that is, I am not typing by hand the inequalities) to compare it with SameQ

(Reduce(#, x) & /@ {{4 x > 1, 2 x < 5}, {5 x > 1, x < 5}, {4 x > 1, 
     x < 5}}) === {1/4 < x < 5/2, 1/5 < x < 5, 1/4 < x < 5}

I get False

Why is that so?

In particular, I found that out when using Length on the expressions

{Length /@ (Reduce(#, x) & /@ {{4 x > 1, 2 x < 5}, {5 x > 1, 
      x < 5}, {4 x > 1, x < 5}}), 
 Length /@ {1/4 < x < 5/2, 1/5 < x < 5, 1/4 < x < 5}}

which, surprisingly to me, leads to different results

{{5, 5, 5}, {3, 3, 3}}

Linear equation – matricies

Hello I am Dr. William from Cansas and here is my question. Every answers opinion are welcome.

Consider the linear equation System Ax=b where;

Question Image

equation solving – Is it possible to ask Mathematica to give all the roots of the given function?

sol = x /. Solve(36 Cos((3 x)/4) Cos((27 x)/20) (Cos((3 x)/5) + 2 Cos((21 x)/10)) == 0&& 0 < x < 4 Pi)

Plot(36 Cos((3 x)/4) Cos((27 x)/20) (Cos((3 x)/5) + 
    2 Cos((21 x)/10)), {x, 0, 4 Pi}, 
 Epilog -> {Red, 
   Point(Transpose({sol, ConstantArray(0, Length@sol)}))})

enter image description here

enter image description here

If you desire only roots that are rational multiples of Pi, you can use (among many other ways):

Select(sol, #/Pi (Element) Rationals &)

Can Mathematica solve this exponential equation

My version (12.1.0.0 Student Edition) does not solve this equation:

Solve(Exp (-n^2/(2 m)) <= 3/4, m)

WolframAlpha finds the solutions, but only for integer n? https://www.wolframalpha.com/input/?i=exp%28-n%5E2+%2F+%282m%29%29+%3C%3D+3%2F4%2C+solve+for+m

Find the numerical value of the equation

B) Given that $2x^2-3/x^2-4x+4 = 2 + A/x-2 + B/(x-2)^2$

(i) Find the numerical value of A and B.

(ii) Hence, or otherwise, determine the stationary points and the nature.

(iii) Determine the point of inflexion on the graph.

equation solving – How to solve this inequality with Reduce or Solve?

I need to solve this inequality:

Reduce(-(1/(lamda*beta*(1 - d)))*
    Log((delta - beta (1 - 2 d))/(lamda*delta)) < d && 0 < lamda < 1 &&
   0 < beta < 1 && 0 < delta < beta/(1 - lamda) && 
  Max((beta - delta)/(2 beta); 0) < 
   d < (beta - delta + lamda*delta)/(2*beta), d, Reals)

But Reduce doesn’t work for this..neither does solve
(“This system cannot be solved with the methods available to Solve.”)

One can analytically show that there exists a d_0 in ) Max((beta – delta)/(2 beta); 0) ; beta – delta + lamdadelta)/(2beta)( such that:

-(1/(lamda* beta*(1 – d)))*
Log((delta – beta (1 – 2 d))/(lamda *delta)) < d on )d_0; beta – delta + lamda delta)/(2beta)(.

and

-(1/(lamda* beta*(1 – d)))*
Log((delta – beta (1 – 2 d))/(lamda *delta)) > d on ) Max((beta – delta)/(2 beta); 0) ; d_0(.

But,
What i actually want to understand is why neither Reduce or Solve works for this kind of problem ?

equation solving – Decomposing polynomial as a sum of polynomials multiples of two polynomials

Suppose that I have a polynomial $f(s,t)$ with coefficients in $R=mathbb{R}(x_{1},dots, x_{n})$ and in the variables $s,t$ and I know that $f(s,t)$ can be expressed as $(s-t)f(s,t)=p(s,t)h(s)+q(s,t)h(t)$ with $h$ having coefficients also in R (so these coeffcients are actually polynomials) but I do not know who are $p,qin R(s,t)$. How can I make Mathematica solve the equation $(s-t)f(s,t)=p(s,t)h(s)+q(s,t)h(t)$ finding the polynomials $p,q$ in the variables $s,t$ with polynomials coefficients in $R$? Thanks!

Solving an ordinary differential equation with boundary condition at infinity

I want to solve the following differential equation

$$R^{primeprime}(r)+frac{1}{r}R^{prime}-R(r)+R(r)^{3}=0$$

subject to the boundary conditions

$$R^{prime}(0)=0qquadlim_{rtoinfty}R(r)=0$$
The solution for R subject to these boundary conditions is known as Townes soliton. I have tried to use NSolve as shown in the example solved here: https://mathematica.stackexchange.com/a/156362/73726 but I am struggling to understand how the boundary condition at infinity is implemented. In addition, how do I make some plots of $R(r)$ afterwards.