1. **Problem statement:** Solve the first-order linear differential equation $$y' + 2ty = 2te^{-t^2}$$ with initial conditions not specified. 2. **Formula and method:** This is a linear ODE of the form $$y' + P(t)y = Q(t)$$ where $P(t) = 2t$ and $Q(t) = 2te^{-t^2}$. We use the integrating factor method. The integrating factor $\mu(t)$ is given by: $$\mu(t) = e^{\int P(t) dt} = e^{\int 2t dt} = e^{t^2}$$ 3. **Multiply both sides by the integrating factor:** $$e^{t^2} y' + 2t e^{t^2} y = 2t e^{t^2} e^{-t^2} = 2t$$ which simplifies to $$\frac{d}{dt} \left(e^{t^2} y \right) = 2t$$ 4. **Integrate both sides with respect to $t$:** $$e^{t^2} y = \int 2t dt = t^2 + C$$ where $C$ is the constant of integration. 5. **Solve for $y$:** $$y = e^{-t^2} (t^2 + C)$$ 6. **Final solution:** $$\boxed{y = e^{-t^2} (t^2 + C)}$$ This is the general solution. If initial conditions are given, substitute to find $C$.