# 2016年考研数二第12题解析

## 解析

$$f(0) =$$

$$1 + 2 \int_{0}^{0} f(t)dt =$$

$$1+0=0.$$

$$f^{(1)}(0) =$$

$$0+2+2.$$

$$f^{(2)}(0) =$$

$$2+2 \times 4 = 2 + 2^{3}.$$

$$f^{(3)} (0) =$$

$$2(2 + 2^{3}) = 2^{2} + 2^{4}.$$

$$f^{(n)}(0) = 2^{n-1} + 2^{n+1}.$$

$$f(0) = 1.$$

$$f^{‘} = 2x + 2 + 2f \Rightarrow$$

$$f^{‘} = 2(x+1) + 2f \Rightarrow$$

$$f^{‘} – 2f = 2(x+1).$$

$$f = [\int 2(x+1)e^{\int -2dx}dx + C]e^{- \int -2 dx}. ①$$

$$\int 2(x+1)e^{\int -2dx}dx \Rightarrow$$

$$\int (2x + 2)e^{\int -2dx}dx \Rightarrow$$

$$\int (2x + 2)e^{\int -2 dx}dx \Rightarrow 分部积分 \Rightarrow$$

$$(e^{\int -2dx})^{‘} = (-2)e^{-2x} \Rightarrow$$

$$(- \frac{1}{2}) \int (2x + 2)d(e^{-2x}) =$$

$$(-\frac{1}{2}) [(2x + 2)e^{-2x} – (-1) \int e^{-2x}d(-2x)] =$$

$$(- \frac{1}{2}) [(2x + 2)e^{-2x} + e^{-2x}] =$$

$$(- \frac{1}{2}) [(2x + 3)e^{-2x}] =$$

$$(-x – \frac{3}{2})e^{-2x}$$

$$f = [\int 2(x+1)e^{\int -2dx}dx + C]e^{- \int -2 dx} \Rightarrow$$

$$[(-x – \frac{3}{2})e^{-2x} + C]e^{2x} \Rightarrow$$

$$(-x-\frac{3}{2}) + C e^{2x}.$$

$$f=C e^{2x} – x – \frac{3}{2}.$$

$$1= C – 0 – \frac{3}{2} \Rightarrow$$

$$C = \frac{5}{2}.$$

$$f(x) = \frac{5}{2} e^{2x} – x – \frac{3}{2}.$$

$$f^{(1)}(x) = 5e^{2x} – 1.$$

$$f^{(2)}(x) = 5 \times 2 e^{2x}.$$

$$f^{(3)}(x) = 5 \times 2 \times 2 e^{2x}.$$

$$f^{(4)}(x) = 5 \times 2 \times 2 \times 2e^{2x}.$$



$$f^{(n)}(x) = 5 \times 2^{n-1} e^{2x}.$$

$$f^{(n)}(0) = 5 \times 2^{n-1}.$$

EOF