Why are plots different on Wolfram|Alpha and Mathematica? [closed]

Question

I got to do a simple numerical solving of this non-linear and second order differential equation.

$\begin{align*} x''(t)&=-3x'(t)\sqrt{\frac{(x´(t))^2}{6}+\frac{(1-e^{-x(t)\cdot\sqrt{2/3}})^2}{4}} \end{align*}-\sqrt{\frac{3}{2}}e^{-x(t)\cdot\sqrt{2/3}}(1-e^{-x(t)\cdot\sqrt{2/3}})$

with initial conditions $x(0)=5.352$ and $x'(0)=-1.033\cdot10^{-2}$. Therefore, I first tried using WolframAlpha, getting the next Plot Solutions.

However, I want to know more about the numerical values on the axis, and since I couldn't add them on WolframAlpha, I decided to do the problem on Mathematica, using $\text{NDSolve}$ with $\{t,0,150\}$. What I got is the following plot.

The image looks similar to the one plotted on WolframAlpha as the Plot Solution. However, wanting to extend the graph, I found Mathematica with a negative exponential solution, which is not the general behaviour I am looking for. Therefore, is there any problem with the $\text{NDSolve}$ or why are these plots of the exact same problem different?

Another doubt is how to get the second plot from WolframAlpha on Mathematica? Since Mathematica does not give an explicit solution on this case, how do I make an $x'(t)\ \text{vs}\ x(t)$ plot on Mathematica using the solutions from the $\text{NDSolve}$?

The code was

NDSolve[{x''[t] + 
    3*x'[t]*Sqrt[(x'[t])^2/6 + 1/4 - Exp[-x[t]*Sqrt[2/3]]/2 + 
      Exp[-2*x[t]*Sqrt[2/3]]/4] + 
    Sqrt[3/2]*Exp[-x'[t]*Sqrt[2/3]]*(1 - Exp[-x'[t]*Sqrt[2/3]]) == 0, 
  x'[0] == -0.010327936857938202, x[0] == 5.352343521285111}, x, {t, 
  0, 150}]

Plot[InterpolationFunction[x], {x, 0., 150.}]

Answer 1

Your code does not match the equation at the top of the post.

sol = NDSolve[{x''[t] + 
     3 x'[t] Sqrt[(x'[t])^2/6 + (1 - Exp[-x[t] Sqrt[2/3]])^2/4] + 
     Sqrt[3/2] Exp[-x[t] Sqrt[2/3]] (1 - Exp[-x[t] Sqrt[2/3]]) == 0, 
   x'[0] == -0.010327936857938202, x[0] == 5.352343521285111}, 
  x, {t, 0, 150}]

Plot[Evaluate[x[t] /. sol], {t, 0., 150.}, PlotRange -> All]

ParametricPlot[{x[t], x'[t]} /. sol, {t, 1, 150}, PlotRange -> All, 
 AspectRatio -> .6]