题目
编号：A2016219
已知 $y_{1}(x) = \mathrm{e}^{x}$, $y_{2}(x) = u(x) \mathrm{e}^{x}$ 是二阶微分方程：
$$
(2x – 1) y^{\prime \prime} – (2x + 1) y^{\prime} + 2y = 0
$$
的两个解. 若 $u(-1) = \mathrm{e}$, $u(0) = -1$, 求 $u(x)$, 并写出该微分方程的通解.
解析
注：为了书写方便以及演算步骤的直观清晰，我们约定：
① 用 $y_{2}$ 代指题目中的 $y_{2}(x)$;
② 用 $u$ 代指题目中的 $u(x)$.
由题可得：
$$
\left\{\begin{matrix}
y_{2} = u \mathrm{e}^{x};\\
y_{2}^{\prime} = u^{\prime} \mathrm{e}^{x} + u \mathrm{e}^{x};\\
y_{2}^{\prime \prime} = u^{\prime \prime} \mathrm{e}^{x} + 2u^{\prime} \mathrm{e}^{x} + u \mathrm{e}^{x}.
\end{matrix}\right.
$$
将上式代入到 $(2x – 1)$ $y^{\prime \prime} -$ $(2x + 1) y^{\prime} +$ $2y = 0$ 中可得：
$$
(2x – 1) y^{\prime \prime} – (2x + 1) y^{\prime} + 2y = 0 \Rightarrow
$$
$$
\left\{\begin{matrix}
(2x – 1) (u^{\prime \prime} \mathrm{e}^{x} + 2u^{\prime} \mathrm{e}^{x} + u \mathrm{e}^{x}) -\\
(2x + 1) (u^{\prime} \mathrm{e}^{x} + u \mathrm{e}^{x}) + 2u \mathrm{e}^{x} = 0 \Rightarrow
\end{matrix}\right.
$$
$$
\left\{\begin{matrix}
\mathrm{e}^{x}(2x – 1) (u^{\prime \prime} + 2u^{\prime} + u) -\\
\mathrm{e}^{x}(2x + 1) (u^{\prime} + u) + 2u = 0 \Rightarrow
\end{matrix}\right.
$$
$$
等式两边同时除以 \mathrm{e}^{x} \Rightarrow
$$
$$
\left\{\begin{matrix}
(2x – 1) (u^{\prime \prime} + 2u^{\prime} + u) -\\
(2x + 1) (u^{\prime} + u) + 2u = 0 \Rightarrow
\end{matrix}\right.
$$
$$
\left\{\begin{matrix}
2xu^{\prime \prime} + 4xu^{\prime} + 2xu-u^{\prime \prime} – 2u^{\prime} – u\\
2xu^{\prime} – 2xu – u^{\prime} – u + 2u = 0 \Rightarrow
\end{matrix}\right.
$$
$$
(2x-1)u^{\prime \prime} + (2x-3)u^{\prime} = 0.
$$
接着，令 $A = u^{\prime}$, $A^{\prime} = u^{\prime \prime}$则：
$$
(2x-1)u^{\prime \prime} + (2x-3)u^{\prime} = 0 \Rightarrow
$$
$$
(2x-1)A^{\prime} + (2x-3)A = 0 \Rightarrow
$$
$$
A^{\prime} + (\frac{2x-3}{2x-1})A = 0 \Rightarrow
$$
$$
A^{\prime} + (\frac{2x-1-2}{2x-1})A = 0 \Rightarrow
$$
$$
A^{\prime} + (1-\frac{2}{2x-1})A = 0 \Rightarrow
$$
$$
一阶线性微分方程求解公式 \Rightarrow
$$
$$
A = C_{1} \mathrm{e}^{- \int (1-\frac{2}{2x-1}) \mathrm{~d} x} \Rightarrow
$$
$$
A = C_{1} \mathrm{e}^{-[\int 1 \mathrm{~d} x – \int \frac{2}{2x-1} \mathrm{~d} x]} \Rightarrow
$$
$$
A = C_{1} \mathrm{e}^{-[x – \ln(2x-1)]} \Rightarrow
$$
$$
A = C_{1} \mathrm{e}^{\ln(2x-1) + (- x)} \Rightarrow
$$
$$
A = C_{1} [\mathrm{e}^{\ln(2x-1)} \cdot \mathrm{e}^{- x}] \Rightarrow
$$
$$
A = C_{1} (2x-1) \cdot \mathrm{e}^{- x}.
$$
即：
$$
u^{\prime} = C_{1} (2x-1) \cdot \mathrm{e}^{- x}.
$$
于是：
$$
u = \int u^{\prime} \mathrm{~d} x \Rightarrow
$$
$$
u = \int C_{1} (2x-1) \cdot \mathrm{e}^{- x} \mathrm{~d} x \Rightarrow
$$
$$
u = C_{1} \int (2x-1) \cdot \mathrm{e}^{- x} \mathrm{~d} x \Rightarrow
$$
$$
u = C_{1} \int (2x \mathrm{e}^{- x} – \mathrm{e}^{- x}) \mathrm{~d} x \Rightarrow
$$
$$
u = C_{1} [2 \int x \mathrm{e}^{- x} \mathrm{~d} x – \int \mathrm{e}^{- x} \mathrm{~d} x] \Rightarrow
$$
[1]. $(\mathrm{e}^{-x})^{\prime} = – \mathrm{e}^{x}$;
[2]. $-(\mathrm{e}^{-x} + x \mathrm{e}^{-x})^{\prime} = x \mathrm{e}^{-x}$.
$$
u = C_{1} [-2(\mathrm{e}^{-x} + x \mathrm{e}^{-x}) + \mathrm{e}^{-x}] + C_{2} \Rightarrow
$$
$$
u = -C_{1}(2x+1) \mathrm{e}^{-x} + C_{2}.
$$
又：
$$
\left\{\begin{matrix}
u(-1) = \mathrm{e};\\
u(0) = -1.
\end{matrix}\right.
$$
于是：
$$
\left\{\begin{matrix}
C_{1} \mathrm{e} + C_{2} = \mathrm{e};\\
-C_{1} + C_{2} = -1.
\end{matrix}\right.
\Rightarrow
$$
$$
\left\{\begin{matrix}
C_{1}=1;\\
C_{2} = 0.
\end{matrix}\right.
$$
综上可知：
$$
u(x) = -(2x+1) \mathrm{e}^{-x}.
$$