# 2015年考研数二第20题解析：物理应用、微分、一阶线性微分方程

## 解析

$$\frac{{\rm d}T(t)}{{\rm d}t} = – k [T(t) – 20], k > 0.$$

[1]. 在下文中，有时会把 $T(t)$ 简写为 $T$.

$$\frac{{\rm d}T(t)}{{\rm d}t} = – k [T(t) – 20] \Rightarrow$$

$$\frac{{\rm d}T}{{\rm d}t} = – k [T – 20] \Rightarrow$$

$$\frac{{\rm d}T}{{\rm d}t} = – kT + 20k \Rightarrow$$

$$\frac{{\rm d}T}{{\rm d}t} + kT = 20k \Rightarrow$$

$$T^{‘} + k T = 20k, k > 0.$$

$$T(t) = [\int 20k \cdot e^{\int k {\rm d}t} {\rm d}t + C] \cdot e^{-\int k {\rm d}t} \Rightarrow$$

$$T(t) = [20k \int e^{\int k {\rm d}t} {\rm d}t + C] \cdot e^{-\int k {\rm d}t} \Rightarrow$$

$$T(t) = [20k \int e^{kt} {\rm d}t + C] \cdot e^{-kt} \Rightarrow$$

$$T(t) = [20k \cdot \frac{1}{k} e^{kt} + C] \cdot e^{-kt} \Rightarrow$$

$$T(t) = 20k \cdot \frac{1}{k} e^{kt} \cdot e^{-kt} {\rm d}t + C \cdot e^{-kt} \Rightarrow$$

$$T(t) = 20 + C \cdot e^{-kt}. ①$$

$$T(0) = 20 + C = 120 \Rightarrow$$

$$C = 100.$$

$$T(30) = 20 + 100 \cdot e^{-30 k} = 30 \Rightarrow$$

$$100 \cdot e^{-30 k} = 10 \Rightarrow$$

$$10 \cdot e^{-30 k} = 1 \Rightarrow$$

$$e^{-30 k} = \frac{1}{10} \Rightarrow$$

$$(-1) \cdot 30k = \log_{e}^{10^{-1}} \Rightarrow$$

$$(-1) \cdot 30k = \ln{10^{-1}} \Rightarrow$$

$$(-1) \cdot 30k = (-1) \cdot \ln{10} \Rightarrow$$

$$30k = \ln{10} \Rightarrow$$

$$k = \frac{\ln{10}}{30}, k > 0.$$

$$T(t) = 100 e^{\frac{-\ln{10}}{30} \cdot t} + 20.$$

$$T(t) = 21 = 100 e^{\frac{-\ln{10}}{30} \cdot t} + 20 \Rightarrow$$

$$100 e^{\frac{-\ln{10}}{30} \cdot t} = 1 \Rightarrow$$

$$e^{\frac{-\ln{10}}{30} \cdot t} = \frac{1}{100} \Rightarrow$$

$$\ln 100^{-1} = \frac{-\ln{10}}{30} \cdot t \Rightarrow$$

$$(-1) \cdot \ln 100 = \frac{-\ln{10}}{30} \cdot t \Rightarrow$$

$$\ln 100 = \frac{\ln{10}}{30} \cdot t \Rightarrow$$

$$t = \ln 100 \cdot \frac{30}{\ln 10} \Rightarrow$$

$$t = 30 \cdot \log_{10}^{100} \Rightarrow$$

$$t = 30 \cdot 2 = 60.$$