Plot the numerical solution of the differential equation for 0 ≤ t ≤ 50: x 00 + 0.15x 0 − x + x 3 = 0.3 cost , x(0) = −1 , x 0 (0) = 1

What am I doing wrong here ???

In(195):= DSolve({x''(t) + 0.15 x'(t) - x(t) + x(t)^3 == 0.3 Cos(t), x(0) == -1,x'(0) == 0}, x(t), t)
Out(195)= DSolve({-x(t) + x(t)^3 + 0.15 Derivative(1)(x)(t) + (x^(Prime)(Prime))(t) == 0.3 Cos(t),x(0) == -1, Derivative(1)(x)(0) == 0}, x(t), t)

How do you find the vector equation for the line of intersection of two planes when only given their distances from the origin?

If given two non parallel planes with unit normals n and m and their closest distance from origin are 3 and 7 respectively, how does one find their line of intersection?

I’m just looking for any hints toward finding the answer.

Thanks

simplifying expressions – Solving an equation, and verifying that it is correct

I want to solve the equation

 Solve(e == 1/2 (-2 ef + hvk - Sqrt(9 hvk^2 + 4 j^2)), hvk)

I then obtain the solutions

{{hvk ->
    1/4 (-e - ef - Sqrt(9 e^2 + 18 e ef + 9 ef^2 - 8 j^2))}, {hvk -> 
   1/4 (-e - ef + Sqrt(9 e^2 + 18 e ef + 9 ef^2 - 8 j^2))}}

To verify that they are solutions I expect that the command

FullSimplify(
 Refine(1/2 (-2 ef + hvk - Sqrt(9 hvk^2 + 4 j^2)) /. 
   hvk -> 1/4 (-e - ef + Sqrt(9 e^2 + 18 e ef + 9 ef^2 - 8 j^2))))

Should produce the variable $e$. However, I instead obtain

1/8 (-e - 9 ef + Sqrt(9 (e + ef)^2 - 8 j^2) - 
   4 Sqrt(4 j^2 + 9/16 (e + ef - Sqrt(9 (e + ef)^2 - 8 j^2))^2))

which I can not simplify to $e$. What on earth is going on?

I have used the following assumptions in Refine

$Assumptions = {j >=  0, e > 0,
  Element(ef, Reals)}

fluid dynamics – Match inner and outer solutions of viscous Burgers equation

For the Burgers equation, with $0<epsilonll 1$:
begin{align*}
frac {partial}{partial t}u+ufrac {partial}{partial x}u=epsilonfrac {partial^2}{partial x^2}u,text{ }-infty<x<infty,text{ }t>0,text{ subject to the initial condition}
end{align*}

and
begin{align*}
u(x,0)=begin{cases}
1,text{ }x< 0\
-1,text{ }x>0.
end{cases}
end{align*}

I wonder how to get the inner solution $u^{inner}$ and determine the location of the shock. The outer solution I got is for $x<-t$, $u^{outer}(x,t)=1$, and or $x>t$, $u^{outer}(x,t)=-1$. Thanks for any hints.

Nonlinear Functional Equation

Here is the problem:

Suppose the function $f$ maps the set of positive integers onto itself. Moreover, the functional $f(x)$ satisfies the following

$f(x^2+y^2)=f(x)f(y)$ and $f(x^2)=f(x)^2$ for all positive integers $x$ and $y$.

Find all functions $f$.

equation solving – How to prove that this function has a unique root?

I want to show that f has always a unique root,

f(d_)= (1 - 
     beta)*(-(1/lamda)*
      Log((delta - beta*(1 - 2*d))/(lamda*delta)))^(rho - 1) - (alpha*
     A^rho/d^2) (1 + (1/(lamda*beta*d*(1 - d)))*
       Log((delta - beta (1 - 2 d))/(lamda*delta)))^(alpha*rho - 1) - 
  lamda*(delta - 
     beta (1 - 2*d)) (beta*
       A^rho (1 + (1/(lamda*beta*d*(1 - d)))*
           Log((delta - beta (1 - 2*d))/(lamda*delta)))^(alpha*
          rho) + (1 - 
         beta)*(-(1/lamda)*
          Log((delta - beta (1 - 2*d))/(lamda*delta)))^
        rho)/(beta*(1 - 2*d))

The constraint on the variable and parameters are:

max{(beta - delta)/(2 beta);0} < d < (beta - delta + 
     lamda*delta)/(2 beta)

0<beta< 1
0<delta<beta/(1-lamda)
0<lamda<1 
0<alpha<1
rho < 1
A>0

I would like Mathematica to study the sign of.
But,
Both Reduce and Solve don’t seem to work for this kind of problem.
It says: “This system cannot be solved with the methods available to Reduce.”

What are the other tools that one can use for this kind of problem ??

equation solving – Computing periodic points of some function

Instead of recursively determine the period points, you could first iterate the function and subsequently determine the period points.

E.g. let us define the n-iterated function fn({x_,y_}) for {b,c}=={1/100,-1} (note, it is better to use rational numbers if computing times allows):

hn({x_, y_}) = Nest(Simplify(h(#, 1/100, -1)) &, {x, y}, n)

E.g. for n=2 :

h2{x_, y_}) = Nest(Simplify(h(#, 0.01, -1)) &, {x, y}, 2)

enter image description here

This gives the following period points:

Solve(h2({x, y}) == {x, y}, {x, y})

enter image description here

Up to n==9 hn is calculated rather fast, but h10 suddenly takes much longer. The reason is not clear to me.

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!

equation solving – Why is Reduce unable to solve this system of inequalities?

I need to solve the following inequalities but the execution just keeps going..

Reduce((1 - 
       beta)*(-(1/lamda)*
        Log((delta - beta*(1 - 2 d))/(lamda*delta)))^(rho - 
        1) - (alpha*
       A^rho/d^2) (1 - (1/(lamda*beta*d*(1 - d)))*
         Log((delta - beta (1 - 2 d))/(lamda*delta)))^{alpha*rho - 
        1} - lamda*(delta - 
       beta (1 - 2 d)) (beta*
         A^rho (1 - (1/(lamda*beta*d*(1 - d)))*
             Log((delta - beta (1 - 2 d))/(lamda*delta)))^{alpha*
            rho} + (1 - 
           beta)*(-(1/lamda)*
            Log((delta - beta (1 - 2 d))/(lamda*delta)))^
          rho)/(beta*(1 - 2 d)) < 0 && 
  delta - beta*(1 - 2 d) > 
   0 && (delta - beta*(1 - 2 d))/(lamda*delta) < 1 && rho < 1 && 
  A > 0 && 0 < alpha < 1 && 1 > beta > 0 && 1 > lamda > 0 && 
  delta > 0 && 0 < d < 1/2, beta, Reals)

Is it a problem with Reduce ?

Working out an equation, given tangent points?

This is the question I have been stuck on, is anyone able to walk me through this?

C is a circle with centre the origin.
A tangent to C passes through the points (-20, 0) and (0, 10) Work out an equation of C.
You must show all your working.