LTspice: Simulating a Van Der Pol Oscillator
Regarding the implementation in LTspice  

Don't use the options "start up" or "uic" when simulating such circuits. Only use the initial conditions.
Using "uic" makes V(v1) still start at 0V.



Next, your circuit shows the initial condition .ic V(V1)=2 but the condition for i(L1) seems not implemented correctly:



I tested/confirmed by simulation that you made the inductance of L1 equal to 1 IC=-6A.
So, change the inductance value back to 1. And add the initial condition for L1 to the one for V1 like .ic V(V1)=2 i(L1)=-6A 

However, this will still not give the same result as Mathematica(2V).

Regarding the implementation of Van der Pol oscillator
I wasn't sure if OP's circuit was a good implementation of a Van der Pol oscillator, but a quick KCL shows it does:

$$i_C + i_R + i_L = 0$$
$$ C v1' + \frac{1}{R}v1 + \frac{1}{L} \int v1 = 0$$
$$ C v1'' + \frac{1}{R}v1' + \frac{1}{L} v1 = 0$$
Substituting 
$$ C=1 \text{, } L=1 \text{ and } R = -\frac{1}{\mu(1-v1^2)} = \frac{1}{\mu(v1^2-1)}  $$
yields the Van der Pol equation:

EDIT 
Regarding the implementation of Van der Pol oscillator
Noting the other answer I found the deduction above is incorrect.
R is dependent to v1, so substition afterwards is incorrect.

$$i_C + i_R + i_L = 0$$
$$ C v1' + \frac{1}{R}v1 + \frac{1}{L} \int v1 = 0$$

At this point the substition should be done because, next step, the derivative to v1 is taken, while R depends on v1. For that reason R should be defined as follows:

$$ R = -\frac{1}{\mu(1 - \frac{1}{3} v1^2)} $$

$$ C v1' - \mu(1 - \frac{1}{3} v1^2) v1 + \frac{1}{L} \int v1 = 0$$
$$ C v1' - \mu v1 + \mu \frac{1}{3} v1^3 + \frac{1}{L} \int v1 = 0$$
Taking the derivative to v1:
$$ C v1'' - \mu \text{ } v1' + \mu \text{ } v1^2 \text{ } v1' + \frac{1}{L} v1 = 0$$
$$ C v1'' - \mu (1 - v1^2)v1' + \frac{1}{L} v1 = 0$$

Substituting 
$$ C=1 \text{, } L=1 \text{ and } \mu=1$$
yields the Van der Pol equation:
$$ v1'' - \mu(1-v1^2)v1' + v1  = 0$$

Doing the simulation with .ic V(V1)=2 I(L1)={-(2*2/3-1)} will give the correct response.