# 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

## 解析

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

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

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

$$f^{‘}(u) \cdot [(\cos y) e^{x}]=$$

$$e^{x} \cos y \cdot f^{‘}(u).$$

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

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

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

$$-e^{x} \sin y \cdot f^{‘}(u).$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$e^{2x} f^{”}(u).$$

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

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

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

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

[1]. 由题知 $z=f(u)$, $u=e^{x} \cos y$.

$$f^{”}(u) = 4f(u) + u \Rightarrow$$

$$f^{”}(u) – 4f(u) = u. ①$$

$$\lambda^{2} – 4 = 0 \Rightarrow$$

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

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

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

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

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

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

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

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

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