2014年考研数二第18题解析:偏导数、二阶常系数非齐次线性微分方程

题目

设函数 $f(u)$ 二阶连续可导,$z=f(e^{x} \cos y)$ 满足 $\frac{\partial ^{2} z}{\partial x^{2}} + \frac{\partial ^{2} z}{\partial y^{2}}$ $=(4z + e^{x} \cos y)e^{2x}$, 若 $f(0)=0$, $f^{‘}(0)=0$, 求 $f(u)$ 的表达式.

解析

由题知,可令:

$$
u = e^{x} \cos y.
$$

于是,有:

$$
\frac{\partial z}{\partial x} =
$$

$$
\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} =
$$

$$
f^{‘}(u) \cdot [(\cos y) e^{x}]=
$$

$$
e^{x} \cos y \cdot f^{‘}(u).
$$

$$
\frac{\partial z}{\partial y} =
$$

$$
\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} =
$$

$$
f^{‘}(u) \cdot [(e^{x}) \cdot (- \sin y)] =
$$

$$
-e^{x} \sin y \cdot f^{‘}(u).
$$

$$
\frac{\partial ^{2} z}{\partial x ^{2}} =
$$

$$
\frac{\partial}{\partial x} (\frac{\partial z}{\partial x}) =
$$

$$
\frac{\partial}{\partial x}[e^{x} \cos y \cdot f^{‘}(u)] =
$$

$$
\cos y \cdot \frac{\partial}{\partial x}[e^{x} f^{‘}(u)] =
$$

$$
\cos y \cdot [e^{x} f^{‘}(u) + e^{x}(f^{”}(u) \cdot \frac{\partial u}{\partial x})] =
$$

$$
\cos y \cdot [e^{x} f^{‘}(u) + e^{x}(f^{”}(u) \cdot e^{x} \cos y)] =
$$

$$
e^{x} \cos y f^{‘}(u) + e^{2x} \cos^{2} y f^{”}(u).
$$

$$
\frac{\partial ^{2} z}{\partial y ^{2}} =
$$

$$
\frac{\partial}{\partial y}( \frac{\partial z}{\partial y} ) =
$$

$$
\frac{\partial}{\partial y}[-e^{x} \sin y f^{‘}(u)] =
$$

$$
(-e^{x}) \cdot \frac{\partial}{\partial y} [\sin y f^{‘}(u)] =
$$

$$
(-e^{x}) \cdot [\cos y f^{‘}(u) + \sin y f^{”}(u) \cdot \frac{\partial u}{\partial y}] =
$$

$$
(-e^{x}) \cdot [\cos y f^{‘}(u) + \sin y f^{”}(u) \cdot (-e^{x} \sin y)] =
$$

$$
-e^{x} \cos y f^{‘}(u) + e^{2x} \sin ^{2} y f^{”}(u).
$$

于是:

$$
\frac{\partial ^{2} z}{\partial x ^{2}} + \frac{\partial ^{2} z}{\partial y ^{2}} =
$$

$$
e^{x} \cos y f^{‘}(u) + e^{2x} \cos^{2} y f^{”}(u) +
$$

$$
[-e^{x} \cos y f^{‘}(u) + e^{2x} \sin ^{2} y f^{”}(u)] =
$$

$$
e^{2x} [\sin ^{2} y + \cos ^{2} y] f^{”}(u) =
$$

$$
e^{2x} f^{”}(u).
$$

又由题知:

$$
\frac{\partial ^{2} z}{\partial x^{2}} + \frac{\partial ^{2} z}{\partial y^{2}} = (4z + e^{x} \cos y)e^{2x}.
$$

于是:

$$
\frac{\partial ^{2} z}{\partial x^{2}} + \frac{\partial ^{2} z}{\partial y^{2}} = e^{2x} f^{”}(u).
$$

即:

$$
e^{2x} f^{”}(u) = (4z + e^{x} \cos y)e^{2x} \Rightarrow
$$

$$
f^{”}(u) = 4z + e^{x} \cos y \Rightarrow
$$

注:

[1]. 由题知 $z=f(u)$, $u=e^{x} \cos y$.

$$
f^{”}(u) = 4f(u) + u \Rightarrow
$$

$$
f^{”}(u) – 4f(u) = u. ①
$$

观察可知,$①$ 式就是一个二阶常系数非齐次线性微分方程,接下来可以使用适用于二阶常系数非齐次线性微分方程的方式求解 $f(u)$.

首先,求解出 $f^{”}(u) – 4f(u) = u$ 对应的齐次线性方程 $f^{”}(u) – 4f(u) = 0$ 的通解:

$$
\lambda^{2} – 4 = 0 \Rightarrow
$$

注:

[1]. 这里要注意,$f^{”}(u) – 4f(u) = 0$ 对应的特征方程【不是】 $\lambda ^{2} – 4 \lambda = 0$.

$$
\lambda_{1} = 2, \lambda_{2} = -2.
$$

于是,$f^{”}(u) – 4f(u) = 0$ 的通解可设为:

$$
C_{1}e^{2u} + C_{2}e^{-2u}.
$$

又根据 $f^{”}(u) – 4f(u) = u$ 中的右端项 $u$ 可以将 $f^{”}(u) – 4f(u) = u$ 的特解设为 $au+b$.

接着,将 $au+b$ 代入到 $f^{”}(u) – 4f(u) = u$ 可得:

$$
a=\frac{-1}{4}, b=0.
$$

于是:

$$
f(u) = C_{1}e^{2u} + C_{2}e^{-2u} – \frac{1}{4}u; ②
$$

$$
f^{‘}(u) = 2C_{1}e^{2u} – 2C_{2}e^{-2u} – \frac{1}{4}. ③
$$

将 $f(0)=0$, $f^{‘}(0)=0$ 分别代入到 $②$ 式和 $③$ 式可得:

$$
\left\{\begin{matrix}
C_{1} + C_{2} = 0;\\
2C_{1} – 2C_{2} – \frac{1}{4} = 0
\end{matrix}\right.
\Rightarrow
$$

$$
C_{1} = \frac{1}{16}, C_{2} = \frac{-1}{16}.
$$

于是,可知:

$$
f(u) = \frac{1}{16} e^{2u} – \frac{1}{16} e^{-2u} – \frac{1}{4}u.
$$