Question

$$\begin{gathered} \frac{d^{2}x}{d{t}^2}\&+2x-x^2=0 \\ x\left(0\right)\&=1,x^{\text{'}}\left(0\right)=1;x\left(0\right)=\frac{3}{2},x^{\text{'}}\left(0\right)=-1 \end{gathered}$$

Short answer

For the boundary conditions$x\left(0\right)=x^{\text{'}}\left(0\right)=1$The solution is a periodic one and the period of oscillations is. For the boundary conditions

$x\left(0\right)=\frac{3}{2}and{x}^{\text{'}}\left(0\right)=-1$

The solution is not a periodic one. And there is no such a period of oscillations for that kind of behaviour.

Step-by-step solution

To Find the equation

$\mathrm{We}\mathrm{have}\mathrm{n}\&\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}+2\mathrm{x}-{\mathrm{x}}^2=0\mathrm{where}\mathrm{x}\left(0\right)=\mathrm{l},{\mathrm{x}}^{\text{'}}\left(0\right)=1$

$and$

$\mathrm{x}\left(0\right)=\frac{3}{2},{\mathrm{x}}^{\text{'}}\left(0\right)=-1$

Solving numerically, using Mathematic, as follows

$In\left[58\right]:=NDSolve[\left\{Derivative\left[2\right]\left[x\right]\left[t\right]+2x\left[t\right]-x{\left[t\right]}^2=0,x\left[0\right]=1,x^{\text{'}}\left[0\right]=1\right\},x\left[t\right],\{t,0,14\}]$

Out[58] $=\{\{x\left[t\right]\to$Interpolating Function $\left.\left[\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,14.\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}\right]\left[\mathrm{t}\right]\}\}\right\}$

$ln\left[59\right]:=$$NDSolve\left[\left\{Derivative\left[2\right]\left[x\right]\left[t\right]+2x\left[t\right]-x{\left[t\right]}^2=0,x\left[0\right]=\frac{3}{2},x^{\text{'}}\left[0\right]=-1\right\}\right],$

$\mathrm{x}\left[\mathrm{t}\right],\{\mathrm{t},0,14\}]$

$\mathrm{t}==7.{268794944467533}^{\text{'}}$, step size is effectively zero; singularity or stiff system suspected.

Out[F9] $=\{\{x\left[t\right]\to$Interpolating Function$\begin{array}{l}\mathrm{Domain}:\left\{\right\{0.,7.27\left\}\right\}\\ \mathrm{Output}:\mathrm{scalar}\end{array}$$\left\{\right[t\left]\right\}\}$

The two sets of the intial condition

And we plot the results, for the two sets of the intial conditions, yields

$\mathrm{In}\left[63\right]:\mathrm{Plot}\left[\right\{\mathrm{Evaluate}\left[\mathrm{x}\right[\mathrm{t}]/.\mathrm{\%}58],\mathrm{Evaluate}\left[\mathrm{x}\right[\mathrm{t}]/.\mathrm{\%}59]\},\{\mathrm{t},0,7\},$

Plot Legends$\to \left\{"x\left(0\right)=x^{\text{'}}\left(0\right)=1^{"},"x\left(0\right)=3/2,x^{\text{'}}\left(0\right)=-1^{"}\right\}]$

Final Answer

For The first boundary conditions

The solution is a periodic one. And from the graph, we can deduce that the half period of the oscillations is

$\frac{1}{2}\mathrm{T}=4.1-1.2=2.9\mathrm{s}\Rightarrow \mathrm{T}=5.8\mathrm{s}$

For The Second Boundary Conditions

The solution is not a periodic one. And there is no such a period of oscillations for that kind of behaviour.

See graphs. For the boundary conditions $x\left(0\right)=x^{\text{'}}\left(0\right)=1$.The solution is a periodic one and the period of oscillations is $5.8s$. For the boundary conditions

$x\left(0\right)=\frac{3}{2}and{x}^{\text{'}}\left(0\right)=-1$

The solution is not a periodic one. And there is no such a period of oscillations for that kind of behaviour.