According to StefanSolution Starting from dTT4 Tm4 k dt Le

Solution

Starting from dT/(T^4 - (T_m)^4) = k dt: Let a = T_m for simplicity. By partial fractions, 1/(T^4 - a^4) = A/(T - a) + B/(T + a) + (CT + D)/(T^2 + a^2). Clearing denominators: 1 = A(T + a)(T^2 + a^2) + B(T - a)(T^2 + a^2) + (CT + D)(T^2 - a^2) Letting T = a yields 1 = (2a^3)A ==> A = 1/(2a^3) Letting T = -a yields 1 = (-2a^3)B ==> B = -1/(2a^3) Letting T = ai yields 1 = (Cai + D)(-2a^2) ==> C = 0 and D = -1/(2a^2). Hence, (1/(2a^3)) [1/(T - a) - 1/(T + a) - a/(T^2 + a^2)] = k dt ==> 1/(T - a) - 1/(T + a) - a/(T^2 + a^2) = 2ka^3 dt. Integrate both sides: ln(T - a) - ln(T + a) - arctan(T/a) = 2ka^3 t + C. ==> ln [(T - a)/(T + a)] - arctan(T/a) = 2ka^3 t + C. Assuming that T(0) = T0 (initial temperature), we find that ln(T0 - a)/(T0 + a)] - arctan(T0/a) = C. ---------------- For your second part: Assume that T - T_m is negligible compared to T_m itself. Working with the original DE, dT/dt = k(T^4 - (T_m)^4) .........= k[((T - T_m) + T_m)^4 - (T_m)^4], expanding in powers of T - T_m .........= k[(T - T_m)^4 + 4T_m (T - T_m)^3 + 6(T_m)^2 (T - T_m)^2 + 4(T_m)^3 (T - T_m)] Any power of T - T_M is especially negligible, so ignoring these terms yields dT/dt = k[0 + 4(T_m)^3 (T - T_m)] ........= (4k(T_m)^3) * (T - T_m), which is essentially Newton\'s Law of cooling with constant 4k(T_m)^3.