1. **Problem statement:** Solve the heat equation $$\frac{\partial u}{\partial t} = a^2 \frac{\partial^2 u}{\partial x^2}$$ for an infinite insulated rod with initial condition $$u(x,0) = \begin{cases} T_0, & x > 0 \\ -T_0, & x < 0 \end{cases}$$ using separation of variables. 2. **Method:** The heat equation is linear and homogeneous. We use the method of separation of variables and the fundamental solution (heat kernel) for the infinite domain. 3. **Key formula:** The solution for initial condition $u(x,0) = f(x)$ on the infinite line is given by the convolution with the heat kernel: $$u(x,t) = \frac{1}{2a\sqrt{\pi t}} \int_{-\infty}^\infty f(\xi) e^{-\frac{(x-\xi)^2}{4a^2 t}} d\xi$$ 4. **Apply initial condition:** Here, $$f(\xi) = \begin{cases} T_0, & \xi > 0 \\ -T_0, & \xi < 0 \end{cases}$$ 5. **Split integral:** $$u(x,t) = \frac{T_0}{2a\sqrt{\pi t}} \left( \int_0^\infty e^{-\frac{(x-\xi)^2}{4a^2 t}} d\xi - \int_{-\infty}^0 e^{-\frac{(x-\xi)^2}{4a^2 t}} d\xi \right)$$ 6. **Change variables:** Let $$\eta = \frac{x-\xi}{2a\sqrt{t}}$$, then $$d\xi = -2a\sqrt{t} d\eta$$. 7. **Rewrite integrals:** $$u(x,t) = \frac{T_0}{\sqrt{\pi}} \left( \int_{\frac{x}{2a\sqrt{t}}}^{-\infty} e^{-\eta^2} (-d\eta) - \int_{\infty}^{\frac{x}{2a\sqrt{t}}} e^{-\eta^2} (-d\eta) \right)$$ 8. **Simplify:** $$u(x,t) = \frac{T_0}{\sqrt{\pi}} \left( \int_{-\infty}^{\frac{x}{2a\sqrt{t}}} e^{-\eta^2} d\eta - \int_{\frac{x}{2a\sqrt{t}}}^{\infty} e^{-\eta^2} d\eta \right)$$ 9. **Use error function definition:** $$\operatorname{erf}(z) = \frac{2}{\sqrt{\pi}} \int_0^z e^{-s^2} ds$$ 10. **Express solution:** $$u(x,t) = T_0 \left( \operatorname{erf}\left( \frac{x}{2a\sqrt{t}} \right) \right)$$ **Final answer:** $$\boxed{u(x,t) = T_0 \operatorname{erf}\left( \frac{x}{2a\sqrt{t}} \right)}$$ This solution satisfies the initial condition and the heat equation for all $x$ and $t>0$.