Inserting the constants, the equation becomes
x'' + 2*c*x' + x = F*cos(W*t)
The general solution form is
x(t)=A*cos(W*t)+B*sin(W*t)+exp(-c*t)*(C*cos(w*t)+D*sin(w*t))
w^2=1-c^2
for the particular solution one gets
-W^2*(A*cos(W*t)+B*sin(W*t))
+2*c*W*(B*cos(W*t)-A*sin(W*t))
+ (A*cos(W*t)+B*sin(W*t))
=F*cos(W*t)
(1-W^2)*A + 2*c*W*B = F
-2*c*W*A + (1-W^2)*B = 0
For the initial conditions it is needed that
A+C = 1
W*B-c*C+w*D=0
In python code thus
F=0.2; c=0.1; W=1.4
w=(1-c*c)**0.5
A,B = np.linalg.solve([[1-W*W, 2*c*W],[-2*c*W,1-W*W]], [F,0])
C = 1-A; D = (c*C-W*B)/w
print(f"w={w}, A={A}, B={B}, C={C}, D={D}")
with the output
w=0.99498743710662, A=-0.192, B=0.056, C=1.192, D=0.04100554286257586
continuing
t = np.linspace(0, 100, 1000)
u=odeint(lambda u,t:[u[1], F*np.cos(W*t)-2*c*u[1]-u[0]], [1,0],t, atol=1e-13, rtol=1e-13)
plt.plot(t,u[:,0],label="odeint", lw=3);
plt.plot(t,A*np.cos(W*t)+B*np.sin(W*t)+np.exp(-c*t)*(C*np.cos(w*t)+D*np.sin(w*t)), label="exact");
plt.legend(); plt.grid(); plt.show();
results in an exact fit