1. **Stating the problem:** We are given the heat balance differential equation for temperature $T_1$: $$\rho_1 c_{p,1} \frac{e_1}{2n_1} \frac{dT_1}{dt} = \alpha_{sol} I_{tot} + h_{ext}(T_{ext} - T_1) + h_{rad,ext}(T_{sky} - T_1) - \frac{\lambda_1}{\Delta x_1}(T_1 - T_2)$$ 2. **Understanding the terms:** - $\rho_1$ is density. - $c_{p,1}$ is specific heat capacity. - $e_1$, $n_1$ are geometric or material parameters. - $\frac{dT_1}{dt}$ is the time derivative of temperature $T_1$. - $\alpha_{sol} I_{tot}$ is absorbed solar radiation. - $h_{ext}(T_{ext} - T_1)$ is convective heat exchange with external temperature. - $h_{rad,ext}(T_{sky} - T_1)$ is radiative heat exchange with sky temperature. - $\frac{\lambda_1}{\Delta x_1}(T_1 - T_2)$ is conductive heat transfer to adjacent layer at temperature $T_2$. 3. **Rearranging the equation:** Group terms involving $T_1$ on the right: $$\rho_1 c_{p,1} \frac{e_1}{2n_1} \frac{dT_1}{dt} = \alpha_{sol} I_{tot} + h_{ext} T_{ext} + h_{rad,ext} T_{sky} - T_1 (h_{ext} + h_{rad,ext} + \frac{\lambda_1}{\Delta x_1}) + \frac{\lambda_1}{\Delta x_1} T_2$$ 4. **Expressing as a first-order ODE:** Divide both sides by $\rho_1 c_{p,1} \frac{e_1}{2n_1}$: $$\frac{dT_1}{dt} = \frac{1}{\rho_1 c_{p,1} \frac{e_1}{2n_1}} \left( \alpha_{sol} I_{tot} + h_{ext} T_{ext} + h_{rad,ext} T_{sky} + \frac{\lambda_1}{\Delta x_1} T_2 - T_1 \left(h_{ext} + h_{rad,ext} + \frac{\lambda_1}{\Delta x_1} \right) \right)$$ 5. **Interpretation:** This is a linear first-order ODE in $T_1$ with forcing terms from solar radiation, external and sky temperatures, and adjacent layer temperature $T_2$. 6. **Summary:** The equation models the rate of change of temperature $T_1$ considering heat gains and losses by convection, radiation, conduction, and solar absorption. Final form: $$\boxed{\frac{dT_1}{dt} = \frac{2 n_1}{\rho_1 c_{p,1} e_1} \left( \alpha_{sol} I_{tot} + h_{ext} T_{ext} + h_{rad,ext} T_{sky} + \frac{\lambda_1}{\Delta x_1} T_2 - T_1 \left(h_{ext} + h_{rad,ext} + \frac{\lambda_1}{\Delta x_1} \right) \right)}$$