逆向解题：由偏导数求解偏积分

一、题目

已知函数 $z=f(x, y)$ 满足 $\frac{\partial^{2} z}{\partial x \partial y}$ $=$ $x$ $+$ $y$, 且有 $f(x, 0)$ $=$ $x$, $f(0, y)$ $=$ $y^{2}$, 则 $f(x, y)$ $=$ $?$

难度评级：

二、解析

首先：

$$
\frac{\partial z}{\partial x}= \int(x+y) d y=x y+\frac{1}{2} y^{2}+\varphi(x) \Rightarrow
$$

$$
f(x, y) = z = \int\left(\frac{\partial z}{\partial x}\right) \mathrm{d} x \Rightarrow
$$

$$
f(x, y) = \int\left(x y+\frac{1}{2} y^{2} + \varphi(x) \right) \mathrm{d} x \Rightarrow
$$

$$
f(x, y) = \frac{1}{2} x^{2} y+\frac{1}{2} x y^{2} + \int \varphi(x) \mathrm{d} x + \varphi(y).
$$

Next

又：

$$
f(x, 0) = x \Rightarrow
$$

$$
0 + 0 + \int \varphi(x) \mathrm{d} x + 0 = x \Rightarrow
$$

$$
\int \varphi(x) \mathrm{d} x = x.
$$

Next

接着：

$$
f(0, y) = y^{2} \Rightarrow
$$

$$
0 + 0 + 0 + \varphi(y) = y^{2} \Rightarrow
$$

$$
\varphi(y) = y^{2}.
$$

Next

综上可知：

$$
f(x, y) = \frac{1}{2} x^{2} y+\frac{1}{2} x y^{2} + \int \varphi(x) \mathrm{d} x + \varphi(y) \Rightarrow
$$

$$
f(x, y) = \frac{1}{2} x^{2} y + \frac{1}{2} x y^{2} + x + y^{2}.
$$

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