Take the initial value problem
$$y'=\cases{x\bigl(1+2\log|x|\bigr)\quad &$(x\ne0)$ \cr 0&$(x=0)$\cr}\ ,\qquad y(0)=0\ .$$
This example obviously fulfills the assumptions of the existence and uniqueness theorem, so there is exactly one solution. As is easily checked this solution is given by
$$y(x)=\cases{x^2\>\log|x|\quad&$(x\ne0)$\cr 0&$(x=0)$\cr}\ .$$
This function is not analytic in any neighborhood of $x=0$.
If you can assume that $\gamma$ is a small parameter, then write
$$x(t) = x_0(t) + \gamma x_1(t) + \gamma^2 x_2(t) + \cdots$$
$$y(t) = y_0(t) + \gamma y_1(t) + \gamma^2 y_2(t) + \cdots$$
i.e., assume such convergent series exist. A zeroth order solution is $y_0(t)=y(0)=y_0$, a constant, and the equation for $x_0(t)$ becomes
$$\frac{dx_0}{dt} = 2 (W+y_0) x_0-4 x_0^3$$
This equation is integrable:
$$\int \frac{dx_0}{2 (W+y_0) x_0-4 x_0^3} = t \implies \frac{1}{4 (W + y_0)} \log{\left (\frac{x_0(t)}{2 x_0(t)^2-W-y_0}\right)} = t+C$$
Solve for $x_0(t)$, then plug into the first order equation for $y_1(t)$:
$$\gamma \frac{dy_1}{dt} = \gamma(x_0(t)^2-y_0) \implies \frac{dy_1}{dt} = x_0(t)^2-y_0$$
Integrate with respect to $t$ to get $y_1(t)$, then plug into $x$ equation:
$$\frac{d}{dt} (x_0+\gamma x_1) = 2 (W + y_0+\gamma y_1) (x_0+\gamma x_1) - 4 (x_0+\gamma x_1)^3$$
Note that $(x_0+\gamma x_1)^3 = x_0^3 + 3 \gamma x_0^2 x_1 + O(\gamma^2)$. Coefficient of $\gamma^0$ is zero because of above equation. Equating coefficients of $\gamma^1$, we get
$$\frac{d x_1}{dt} = 2 (W+y_0) x_1 + x_0 y_1 - 12 x_0^2 x_1$$
This is an inhomogeneous 1st order equation for $x_1$, which may be solved using known techiques (e.g., integration factor).
At this point, you may repeat this process to get higher powers of $\gamma$. I do not have a proof that the resulting series converges.
Best Answer
The solution of the homogeneous equation
$$y'(t)=-by(t)$$ is easily found to be $y(t)=Ce^{-bt}$.
Then by variation of the constant, $y(t)=c(t)e^{-bt}$ yields
$$c'(t)e^{-bt}-bc(t)e^{-bt}=av(t)-bc(t)e^{-bt}$$
or
$$c'(t)=av(t)e^{bt},$$
$$c(t)=\int av(t)e^{bt}dt+C,$$
$$y(t)=\left(\int av(t)e^{bt}dt+C\right)e^{-bt}.$$
Obviously, the equation has an analytical solution when the integral has one.
If the unknown is the function $v$, then
$$v(t)=\frac{y'(t)+by(t)}a.$$