solve for two variables for each n related to Collatz conjecture

For this code, for each x I would like to solve for all value ranges for c1 and c2 in a bounded range ie c1 and c2 in the range of real numbers +-100 for c1 and c2 for each x, which combined give “Length[stepsForEachN] == nRangeToCheck – 1”. Here is the code so far, I am not sure how to solve for the two variables c1 and c2 for each x:

(*stepsForEachN output is A006577={1,7,2,5,8,16,3,19} if c1=c2=1*) c1 = 1;  c2 = 1; nRangeToCheck = 10; stepsForEachNwithIndex = {}; stepsForEachN = {}; stepsForEachNIndex = {}; maxStepsToCheck = 10000;  c1ValuesForEachN = {};  For[x = 2, x <= nRangeToCheck, x++,   n = x;   For[i = 1, i <= maxStepsToCheck, i++,   If[EvenQ[n], n = Floor[(n/2)*c1],    If[OddQ[n], n = Floor[(3*n + 1)*c2]]    ];    If[n < 1.9,    AppendTo[stepsForEachN, i];    AppendTo[stepsForEachNIndex, x];    AppendTo[stepsForEachNwithIndex, {x, i}];    i = maxStepsToCheck + 1    ]   ]  ] Length[stepsForEachN] == nRangeToCheck - 1 

How to solve the ambiguity of selected state in a segmented control with just two segments?

I’m working on a web application that has two similar data grids in a page. User should be able to switch between these two data grids within the same page.

I have used segmented buttons to navigate between these two grids. But I feel this component visually becomes a toggle, and the current state of the toggle becomes ambiguous.

As of now I’m solely relying on the color / hue to show the selected state but I feel it’s kinda unclear which state is selected.

Is there a better approach to solve this problem?

Segmented buttons circled in red enter image description here

How I can solve below pde (Reynolds equation) with mathematica?

I want to solve below equation in Mathematica. I know a little bit about Mathematica. Can somebody help me?

\begin{array}{l} \frac{1}{{{R^2}}}\frac{\partial }{{\partial \theta }}\left( {{P_0}\frac{{\partial \left( {{P_X}} \right)}}{{\partial \theta }} – \Lambda {R^2}{P_X}} \right) + \frac{1}{R}\frac{\partial }{{\partial R}}\left( {R{P_0}\frac{{\partial \left( {{P_X}} \right)}}{{\partial R}} + R{P_X}\frac{{\partial \left( {{P_0}} \right)}}{{\partial R}}} \right) – i2\Lambda \Gamma \left( {{P_X}} \right)\ + \underbrace {\frac{1}{R}\frac{\partial }{{\partial R}}\left( { – 3R{P_0}\frac{{\partial {P_0}}}{{\partial R}}} \right) + i2\Lambda\left( {{P_0}} \right)}_f = 0\ {P_0}\left( R \right) = {\rm{available}}\ \Lambda = 1 \ {P_X}\left( {{R_1},\theta } \right) = {P_X}\left( {{R_1},\theta } \right) = 0\ {P_X}\left( {R,0} \right) = {P_X}\left( {R,2\pi } \right) \end{array}

How to Solve $y'(x)=x\ln(y(x))$ with $y(1)=1$ using DSolve

The solution to $ y’=x \ln(y)$ with initial conditions $ y(1)=1$ is $ y=1$ .

How to persuade DSolve to obtain this solution?

ClearAll[y, x]; ode = y'[x] == x Log[y[x]]; ic = y[1] == 1; sol = DSolve[{ode, ic}, y[x], x] 

Mathematica graphics

One can see that $ y=1$ is solution that also satisfies the ic by looking at direction field.

ClearAll[x, y]; fTerm = x Log[y]; StreamPlot[ {1, fTerm}, {x, -1, 3}, {y, 0, 2},  Axes -> True,  Frame -> False,  PlotTheme -> "Classic",  AspectRatio -> 1 / GoldenRatio,  StreamPoints -> {{{{1, 1}, Red}, Automatic}},  Epilog -> {{Red, PointSize[.025], Point[{1, 1}]}},  PlotLabel -> Style[Text[Row     [{"Solution curve with initial conditions at {", 1, ",", 1,"}"}]], 14]  ] 

enter image description here

V 12.0 on windows 10

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