1991 年考研数二真题解析

五、解答题 (本题满分 9 分)

求微分方程 $y^{\prime \prime }+y=x+\cos x$ 的通解.

齐通：

$${\lambda }^2+1=0\Rightarrow \lambda =0\pm i\cdot 1\Rightarrow$$

$$y^{\ast }=e^{\alpha x}(C_1\cos \beta x+C_2\sin \beta x)\Rightarrow$$

$$y^{\ast }=C_1\cos x+C_2\sin x$$

非奇特：

先拆分：

$$y^{\prime \prime }+y=x+\cos x\{\begin{array}{l}{y}^{\prime \prime }+y=x\\ y^{\prime \prime }+y=\cos x\end{array}$$

对 $y^{\prime \prime }+y=x$, 有：

$$Y_1^{\ast }=x^k(Ax+B)e^{0x}\Rightarrow 0\ne \lambda \Rightarrow k=0\Rightarrow$$

$$Y_1^{\ast }=Ax+B\Rightarrow {\left(Y_1^{\ast }\right)}^{\prime }=A{\left(Y_1^{\ast }\right)}^{\prime \prime }=0\Rightarrow$$

$${\left(Y_1^{\ast }\right)}^{\prime \prime }+Y^{\ast }=X\Rightarrow A=1,B=0$$

$$Y_1^{\ast }=x$$

对 $y^{\prime \prime }+y=\cos x$, 有：

$$Y_2^{\ast }=x^k{e}^{\alpha x}[Q_n(x)\cos \beta x+{\omega }_n(x)\sin \beta x]\Rightarrow$$

$$\alpha =0,\beta =1\Rightarrow \alpha \pm i\beta ={\lambda }_1={\lambda }_2\Rightarrow k=1\Rightarrow$$

$$Y_2^{\ast }=x[C\cdot \cos x+D\cdot \sin x]\Rightarrow$$

$${\left(Y_2^{\ast }\right)}^{\prime }=(C\cdot \cos x+D\sin x)+x(-\cos x+D\cos x)\Rightarrow$$

$${\left(Y_2^{\ast }\right)}^{\prime \prime }=(-C\cdot \sin x+D\cos x)+(-C\sin x+$$

$$D\cos x)+x(-C\cdot \cos x-D\cdot \sin x)\Rightarrow$$

$${\left(Y_2^{\ast }\right)}^{\prime \prime }+Y_2^{\ast }=\cos x\Rightarrow$$

$$-2C\cdot \sin x+2D\cos x=\cos x\Rightarrow$$

$$C=0,D=\frac{1}{2}\Rightarrow$$

$$Y_2^{\ast }=\frac{1}{2}x\sin x\Rightarrow$$

于是：

$$Y=y^{\ast }+Y_1^{\ast }+Y_2^{\ast }\Rightarrow$$

$$Y=C_1\cos x+C_2\sin x+x+\frac{1}{2}x\sin x$$