下册 7.3 微分方程的验证及变量替换 第5题
📝 题目
5．设 $F(u, v)$ 是具有连续偏导数的二元函数．
（1）验证 $w=F(x y, y z)$ 满足 $\displaystyle x \frac{\partial w}{\partial x}+z \frac{\partial w}{\partial z}=y \frac{\partial w}{\partial y}$ ，这时 $F(x y, y z)$ 称为以上偏微分方程的完全积分．
（2）利用猜试法求以下偏微分方程的完全积分：$\displaystyle x_{1} \frac{\partial w}{\partial x_{1}}+\cdots+x_{m} \frac{\partial w}{\partial x_{m}}=y \frac{\partial w}{\partial y}$ ．
💡 答案解析
\section*{解题过程：}
$\displaystyle \frac{\partial w}{\partial x}=y \frac{\partial F}{\partial u}, \frac{\partial w}{\partial z}=y \frac{\partial F}{\partial v}, \frac{\partial w}{\partial y}=x \frac{\partial F}{\partial u}+z \frac{\partial F}{\partial v}$ ．于是 $\displaystyle x \frac{\partial w}{\partial x}+z \frac{\partial w}{\partial z}=y \frac{\partial w}{\partial y}$ ．
如果 $F\left(u_{1}, \cdots, u_{m}\right)$ 是具有连续偏导数的多元函数，则原偏微分方程的完全积分为 $w=F\left(x_{1} y, x_{2} y, \cdots, x_{m} y\right)$ 。则
$$
\frac{\partial w}{\partial x_{i}}=y \frac{\partial F}{\partial u_{i}}, i=1,2, \cdots, m, \frac{\partial w}{\partial y}=\sum_{i=1}^{\infty} x_{i} \frac{\partial F}{\partial y} .
$$
于是
$$
x_{1} \frac{\partial w}{\partial x_{1}}+\cdots+x_{m} \frac{\partial w}{\partial x_{m}}=y \frac{\partial w}{\partial y} .
$$
📋 详细解题步骤
步骤 1/5
目标：验证第一部分：计算偏导数
设 $w = F(xy, yz)$，其中 $F(u,v)$ 具有连续偏导数。令 $u = xy$, $v = yz$。由链式法则：
$$\frac{\partial w}{\partial x} = \frac{\partial F}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \cdot \frac{\partial v}{\partial x} = y \frac{\partial F}{\partial u} + 0 = y F_u,$$
$$\frac{\partial w}{\partial z} = \frac{\partial F}{\partial u} \cdot \frac{\partial u}{\partial z} + \frac{\partial F}{\partial v} \cdot \frac{\partial v}{\partial z} = 0 + y \frac{\partial F}{\partial v} = y F_v,$$
$$\frac{\partial w}{\partial y} = \frac{\partial F}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial F}{\partial v} \cdot \frac{\partial v}{\partial y} = x \frac{\partial F}{\partial u} + z \frac{\partial F}{\partial v} = x F_u + z F_v.$$
公式：链式法则：$\frac{\partial w}{\partial x_i} = \sum_j \frac{\partial F}{\partial u_j} \frac{\partial u_j}{\partial x_i}$
提示：注意区分 $F$ 对中间变量 $u,v$ 的偏导数和 $w$ 对自变量的偏导数。
步骤 2/5
目标：验证第一部分：代入方程验证
计算 $x \frac{\partial w}{\partial x} + z \frac{\partial w}{\partial z}$：
$$x \frac{\partial w}{\partial x} + z \frac{\partial w}{\partial z} = x (y F_u) + z (y F_v) = xy F_u + yz F_v.$$
计算 $y \frac{\partial w}{\partial y}$：
$$y \frac{\partial w}{\partial y} = y (x F_u + z F_v) = xy F_u + yz F_v.$$
两者相等，故 $x \frac{\partial w}{\partial x} + z \frac{\partial w}{\partial z} = y \frac{\partial w}{\partial y}$ 成立。
提示：注意等式两边形式一致，避免计算错误。
步骤 3/5
目标：推广到多维：猜测完全积分形式
对于方程 $x_1 \frac{\partial w}{\partial x_1} + \cdots + x_m \frac{\partial w}{\partial x_m} = y \frac{\partial w}{\partial y}$，类比第一问，猜测完全积分为 $w = F(x_1 y, x_2 y, \cdots, x_m y)$，其中 $F(u_1, \dots, u_m)$ 具有连续偏导数，$u_i = x_i y$。
提示：注意变量对应关系：第一问中 $u=xy$, $v=yz$，这里每个 $u_i = x_i y$。
步骤 4/5
目标：计算多维情况下的偏导数
对 $i=1,\dots,m$，有
$$\frac{\partial w}{\partial x_i} = \frac{\partial F}{\partial u_i} \cdot \frac{\partial u_i}{\partial x_i} = y \frac{\partial F}{\partial u_i},$$
$$\frac{\partial w}{\partial y} = \sum_{i=1}^m \frac{\partial F}{\partial u_i} \cdot \frac{\partial u_i}{\partial y} = \sum_{i=1}^m x_i \frac{\partial F}{\partial u_i}.$$
公式：链式法则
提示：注意 $\frac{\partial w}{\partial y}$ 是对所有 $u_i$ 求和。
步骤 5/5
目标：代入多维方程验证
计算左边：
$$\sum_{i=1}^m x_i \frac{\partial w}{\partial x_i} = \sum_{i=1}^m x_i \left( y \frac{\partial F}{\partial u_i} \right) = y \sum_{i=1}^m x_i \frac{\partial F}{\partial u_i}.$$
右边：
$$y \frac{\partial w}{\partial y} = y \sum_{i=1}^m x_i \frac{\partial F}{\partial u_i}.$$
两者相等，故 $w = F(x_1 y, \dots, x_m y)$ 满足方程，即为完全积分。
提示：注意求和符号的运用，避免遗漏项。
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