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10

If I understand the question correctly, you wish to obtain a parameterized solution {U2[U1], W2[W1]} from the equation in the question, so that you can vary that parameter to obtain a "nice" solution. One approach is as follows. Define exp = 32 + 8 a^2 (2 + b) + 4 a (2 + b) (6 + b) - b (-4 (8 + U1 - W1 + U2[U1]) + b (-8 + (-2 + U1) U1 + W1^2 - 2 U2[...

6

Introduce a new variable $z=x^py^q$. eqs1 = {(y + 1) (a - z) == 0, (y + 1) z - c == 0} res1 = Solve[eqs1, {y, z}] eq2 = (z == x^p y^q /. First[res1]) Solve[eq2, x] The answer is $y=\frac{c-a}{a}$ and $x=\left(a \left(\frac{c-a}{a}\right)^{-q}\right)^{1/p}$.

6

You can add a new variable in Reduce do get your multiplication, just like this In[13]:= Reduce[ ans == x y && x*y > 50 && x*y < 100 && x < 100 && x > 1 && y > 1 && y < 50, ans, {x, y}, Integers] Out[13]= ans == 51 || ans == 52 || ans == 54 || ans == 55 || ans == 56 || ans == 57 || ans ...

3

Here's my attempt to construct a non-analytic solution. The idea is to use one of the analytic solutions in the accepted answer when x belongs to a set of points reachable through successive application of either analytic solution or their inverses, and the other solution otherwise. This set must not be equal the set of reals, otherwise the solution just ...

2

Clear[x] solv = NDSolve[{ D[u[t, x], {t, 2}] - D[u[t, x], {x, 2}] == 0, u[t, 30] == 0, u[0, x] == E^(-x^2), Derivative[1, 0][u][0, x] == 0}, u[t, x], {t, 0, 2000}, {x, -30, 30}, Method -> {"MethodOfLines", "SpatialDiscretization" -> {"TensorProductGrid", "MinPoints" -> 50, "MaxPoints" -> 50, "DifferenceOrder" ->...

2

This is a solution to the problem with exactly the boundary conditions as stated: Dc = 1460 Kc = 9.41/(10^2) LaplacEquation = Dc*D[u[x, y], {x, 2}] + Dc*D[u[x, y], {y, 2}] - Kc*u[x, y] == 0 v[x] /. First[DSolve[{LaplacEquation /. u -> (v[#] &), v[0] == 1}, v[x], x]] (* ==> E^(-0.0080282 x) (1. - 1. C[1] + E^(0.0160564 x) C[1]) *) All I did ...

2

EDIT Using to rules will be better: x y/.{ToRules@Reduce[x*y > 50 && x*y < 100 && x < 100 && x > 1 && y > 1 && y < 50, {x, y}, Integers]} @Michael thanks! in your situation, the result will be in the form x==2&&y==26||x==2&&y==27...... so we can use replacement rules to make it ...

1

You can use Floor and Solve for part a. Solve[Floor[5 x] == 2 x + 1, x, Reals] x -> 1/2 FindInstance for part b (or a). FindInstance[Floor[1/5 x] > 1 + 2 x, {x}, Reals] x -> -(109/5)

1

Inspired by @obsolesced, it is simple to find solutions based on any set that is mapped to itself by f1 and f2. For example, let S = {q1 + q2 Sqrt[2]} for {q1, q2} rational then f[x] = f1[x] x in S f[x] = f2[x] x not in S satisfies the required functional relation. It is continuous nowhere (except x = -7/2) and S is countably infinite and its ...

1

Or use my favorite command: FindInstance[ {(x - 1)^2 + (y - 1)^2 + (z - 1)^2 == 49, x != 1, y != 1,z != 1}, {x, y, z},Integers, 100]

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