13.3 习题课讲解

13.3.1 可降阶的高阶方程

解 由曲率的定义可得$$\begin{array}{}\text{(1)}& \frac{|y^{″}|}{(1+y^{\prime 2}{)}^{3/2}}=\kappa \end{array}$$ 故$|y^{″}|\ne 0$。由导函数的介质性知$y^{″}$不变号，故不妨设$y^{″}>0$。令$u=y^{\prime }$，则$$\begin{array}{}\text{(2)}& u^{\prime }=\kappa (1+u^2{)}^{3/2}⟹\frac{\mathrm{d}u}{(1+u^2{)}^{3/2}}=\kappa \mathrm{d}x\end{array}$$ 两边积分可得$$\begin{array}{}\text{(3)}& \frac{u}{\sqrt{1+u^2}}=\kappa x+C_1⟹\frac{\mathrm{d}y}{\mathrm{d}x}=u=\frac{\kappa x+C_1}{\sqrt{1-(\kappa x+C_1{)}^2}}\end{array}$$ 两边积分可得$$\begin{array}{}\text{(4)}& y=-\frac{\sqrt{1-(\kappa x+C_1{)}^2}}{\kappa }+C_2⟹{(x+\frac{C_1}{k})}^2+(y-C_2{)}^2=\frac{1}{{\kappa }^2}\end{array}$$ $◻$

解 求导可得$$\begin{array}{}\text{(6)}& 2y{y}^{(3)}=0⟹y=0\vee y^{(3)}=0⟹y(x)=A{x}^2+Bx+C\end{array}$$ 但这造成了增解，因为二阶微分方程的通解只能有两个任意常数，所以需要代入原方程验证，即$$\begin{array}{}\text{(7)}& 2(A{x}^2+Bx+C)\cdot 2A-(2Ax+B{)}^2=0⟹4AC=B^2\end{array}$$ 所以原方程的通解为$$\begin{array}{}\text{(8)}& y(x)=A{x}^2\pm 2\sqrt{AC}x+C,AC\ge 0\end{array}$$ $◻$

另解 用通常的降阶法，设$u=y^{\prime }$，则$y^{″}=\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}u}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}x}=u\frac{\mathrm{d}u}{\mathrm{d}y}$，故有$$\begin{array}{}\text{(9)}& 2yu\frac{\mathrm{d}u}{\mathrm{d}y}-u^2=0⟹u=0\vee 2\frac{\mathrm{d}u}{u}=\frac{\mathrm{d}y}{y}\end{array}$$ 解得$y$为常值函数或$u^2=C|y|$，即$$\begin{array}{}\text{(10)}& \frac{\mathrm{d}y}{\mathrm{d}x}=\pm \sqrt{C|y|}=2C_1\sqrt{|y|}⟹\sqrt{|y|}=C_{1}x+C_2\end{array}$$ 所以原方程的通解为$$\begin{array}{}\text{(11)}& y=\pm (C_{1}x+C_2{)}^2\end{array}$$ $◻$

解 因式分解可得$$\begin{array}{}\text{(13)}& (x\frac{\mathrm{d}}{\mathrm{d}x}-1)(\frac{\mathrm{d}}{\mathrm{d}x}-1)y=x^2{\mathrm{e}}^x\end{array}$$ 这等价于微分方程组$$\begin{array}{}\text{(14)}& \{\begin{array}{l}{y}^{\prime }-y=z\\ x{z}^{\prime }-z=x^2{\mathrm{e}}^x\end{array}\end{array}$$ 对于第二个方程，配凑积分因子可得$$\begin{array}{}\text{(15)}& \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{z}{x}\right)={\mathrm{e}}^x⟹z=x\int {\mathrm{e}}^x\mathrm{d}x=x{\mathrm{e}}^x+C_{1}x\end{array}$$ 对于第一个方程，配凑积分因子可得$$\begin{array}{}\text{(16)}& \frac{\mathrm{d}}{\mathrm{d}x}({\mathrm{e}}^{-x}y)=x+C_{1}x{\mathrm{e}}^{-x}⟹y={\mathrm{e}}^x\int (x+C_{1}x{\mathrm{e}}^{-x})\mathrm{d}x=\frac{1}{2}{x}^2{\mathrm{e}}^x-C_1(x+1)+C_2{\mathrm{e}}^x\end{array}$$ $◻$

另解 首先求解齐次方程：$$\begin{array}{}\text{(17)}& (x\frac{\mathrm{d}}{\mathrm{d}x}-1)(\frac{\mathrm{d}}{\mathrm{d}x}-1)y=0\end{array}$$ 由$y^{\prime }-y=0$解得$y=C{\mathrm{e}}^x$。利用常数变易法可设$y=C(x){\mathrm{e}}^x$，代入原方程可得$$\begin{array}{}\text{(18)}& {\mathrm{e}}^x[x{C}^{″}+(x-1)C^{\prime }]=x^2{\mathrm{e}}^x⟹x{C}^{″}+(x-1)C^{\prime }=x^2\end{array}$$ 这是关于$C^{\prime }$的一阶线性微分方程（由此实现降阶），配凑积分因子可得$$\begin{array}{}\text{(19)}& \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{{\mathrm{e}}^x{C}^{\prime }}{x}\right)=\frac{{\mathrm{e}}^x}{x^2}[x{C}^{″}+(x-1)C^{\prime }]={\mathrm{e}}^x⟹C^{\prime }=x{\mathrm{e}}^{-x}({\mathrm{e}}^x+C_1)=x+C_{1}x{\mathrm{e}}^{-x}\end{array}$$ 积分可得$$\begin{array}{}\text{(20)}& C=\int (x+C_{1}x{\mathrm{e}}^{-x})\mathrm{d}x=\frac{1}{2}{x}^2-C_1(x+1){\mathrm{e}}^{-x}+C_2\end{array}$$ 从而$$\begin{array}{}\text{(21)}& y=\frac{1}{2}{x}^2{\mathrm{e}}^x-C_1(x+1)+C_2{\mathrm{e}}^x\end{array}$$ $◻$

解 因式分解可得$$\begin{array}{}\text{(23)}& (x\frac{\mathrm{d}}{\mathrm{d}x}-1)(x\frac{\mathrm{d}}{\mathrm{d}x}-2)=2x\end{array}$$ 这等价于微分方程组$$\begin{array}{}\text{(24)}& \{\begin{array}{l}x{y}^{\prime }-y=z\\ x{z}^{\prime }-2z=2x\end{array}\end{array}$$ 对于第二个方程，配凑积分因子可得$$\begin{array}{}\text{(25)}& \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{z}{x^2}\right)=\frac{2}{x^2}⟹z=x^2\int \frac{2}{x^2}\mathrm{d}x=-2x+C_1{x}^2\end{array}$$ 对于第一个方程，配凑积分因子可得$$\begin{array}{}\text{(26)}& \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{y}{x}\right)=\frac{1}{x^2}(-2x+C_1{x}^2)⟹y=x\int \frac{1}{x^2}(-2x+C_1{x}^2)\mathrm{d}x=-2x\ln x+C_1{x}^2+C_{2}x\end{array}$$ $◻$

另解 仿照上例，借助常数变易法。 $◻$

13.3.2 高阶线性微分方程

解 设常数$C_1,C_2,\cdots ,C_6$使得$$\begin{array}{}\text{(28)}& y:=C_1{\mathrm{e}}^{\alpha x}+C_2{\mathrm{e}}^{\alpha x}x+C_3{\mathrm{e}}^{\alpha x}\frac{x^2}{2}+C_4{\mathrm{e}}^{\beta x}+C_5{\mathrm{e}}^{\beta x}x+C_6{\mathrm{e}}^{\beta x}\frac{x^2}{2}=0\end{array}$$ 注意到$$\begin{array}{}\text{(29)}& (\frac{\mathrm{d}}{\mathrm{d}x}-\alpha ){\mathrm{e}}^{\alpha x}\frac{x^n}{n!}={\mathrm{e}}^{\alpha x}\frac{x^{n-1}}{(n-1)!},(\frac{\mathrm{d}}{\mathrm{d}x}-\alpha ){\mathrm{e}}^{\beta x}=(\beta -\alpha ){\mathrm{e}}^{\beta x}\end{array}$$ 逐次求导可得$$\begin{array}{}\text{(30)}& \begin{array}{rl}0& ={(\frac{\mathrm{d}}{\mathrm{d}x}-\alpha )}^3{(\frac{\mathrm{d}}{\mathrm{d}x}-\beta )}^{2}y={(\frac{\mathrm{d}}{\mathrm{d}x}-\alpha )}^3{C}_6{\mathrm{e}}^{\beta x}=C_6(\beta -\alpha {)}^3{\mathrm{e}}^{\beta x}\\ 0& ={(\frac{\mathrm{d}}{\mathrm{d}x}-\alpha )}^2{(\frac{\mathrm{d}}{\mathrm{d}x}-\beta )}^{3}y={(\frac{\mathrm{d}}{\mathrm{d}x}-\beta )}^3{C}_3{\mathrm{e}}^{\alpha x}=C_3(\alpha -\beta {)}^3{\mathrm{e}}^{\alpha x}\end{array}\end{array}$$ 故$C_3=C_6=0$。继续求导可得$$\begin{array}{}\text{(31)}& \begin{array}{rl}0& ={(\frac{\mathrm{d}}{\mathrm{d}x}-\alpha )}^2(\frac{\mathrm{d}}{\mathrm{d}x}-\beta )y={(\frac{\mathrm{d}}{\mathrm{d}x}-\alpha )}^2{C}_5{\mathrm{e}}^{\beta x}=C_5(\beta -\alpha {)}^2{\mathrm{e}}^{\beta x}\\ 0& =(\frac{\mathrm{d}}{\mathrm{d}x}-\alpha ){(\frac{\mathrm{d}}{\mathrm{d}x}-\beta )}^{2}y={(\frac{\mathrm{d}}{\mathrm{d}x}-\beta )}^2{C}_2{\mathrm{e}}^{\alpha x}=C_2(\alpha -\beta {)}^2{\mathrm{e}}^{\alpha x}\end{array}\end{array}$$ 故$C_2=C_5=0$。继续求导可得$$\begin{array}{}\text{(32)}& \begin{array}{rl}0& =(\frac{\mathrm{d}}{\mathrm{d}x}-\alpha )y=(\frac{\mathrm{d}}{\mathrm{d}x}-\alpha )C_4{\mathrm{e}}^{\beta x}=C_4(\beta -\alpha ){\mathrm{e}}^{\beta x}\\ 0& =(\frac{\mathrm{d}}{\mathrm{d}x}-\beta )y=(\frac{\mathrm{d}}{\mathrm{d}x}-\beta )C_1{\mathrm{e}}^{\alpha x}=C_1(\alpha -\beta ){\mathrm{e}}^{\alpha x}\end{array}\end{array}$$ 故$C_1=C_4=0$。综上所述，这些函数线性无关。 $◻$

解 先考虑齐次方程，特征多项式为$$\begin{array}{}\text{(34)}& {\lambda }^4-{\lambda }^3-3{\lambda }^2+\lambda +2=(\lambda -2)(\lambda -1)(\lambda +1{)}^2=0\end{array}$$ 因此齐次方程解空间的基为${\mathrm{e}}^{2x},{\mathrm{e}}^x,{\mathrm{e}}^{-x},x{\mathrm{e}}^{-x}$，即齐次方程的通解为$$\begin{array}{}\text{(35)}& y_{\mathrm{h}}=C_1{\mathrm{e}}^{2x}+C_2{\mathrm{e}}^x+C_3{\mathrm{e}}^{-x}+C_{4}x{\mathrm{e}}^{-x}\end{array}$$ 接下来寻找非齐次方程的特解。设$y_{\mathrm{p}}=Ax+B$，代入原方程可得$$\begin{array}{}\text{(36)}& A+2(Ax+B)=3x+4⟹A=\frac{3}{2},B=\frac{5}{4}\end{array}$$ 故非齐次方程的通解为$$\begin{array}{}\text{(37)}& y=y_{\mathrm{h}}+y_{\mathrm{p}}=\frac{3}{2}x+\frac{5}{4}+C_1{\mathrm{e}}^{2x}+C_2{\mathrm{e}}^x+C_3{\mathrm{e}}^{-x}+C_{4}x{\mathrm{e}}^{-x}\end{array}$$ $◻$

解 令$z^{\prime }=y$，把上述二阶方程改写为一阶方程组$$\begin{array}{}\text{(39)}& \frac{\mathrm{d}}{\mathrm{d}x}\left(\begin{array}{c}y\\ z\end{array}\right)=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\left(\begin{array}{c}y\\ z\end{array}\right)+\left(\begin{array}{c}0\\ \sin x\end{array}\right)\end{array}$$ 齐次部分的一对线性无关解为$$\begin{array}{}\text{(40)}& \left(\begin{array}{c}\cos x\\ -\sin x\end{array}\right),\left(\begin{array}{c}\sin x\\ \cos x\end{array}\right)\end{array}$$ 它们给出可逆矩阵$$\begin{array}{}\text{(41)}& U(x)=\left(\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right),U^{\prime }(x)=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)U(x)\end{array}$$ 设$y$是非齐次方程的解，且满足$$\begin{array}{}\text{(42)}& \left(\begin{array}{c}y\\ y^{\prime }\end{array}\right)=U(x)\left(\begin{array}{c}{C}_1(x)\\ C_2(x)\end{array}\right)\end{array}$$ 则$$\begin{array}{}\text{(43)}& \frac{\mathrm{d}}{\mathrm{d}x}\left(\begin{array}{c}y\\ y^{\prime }\end{array}\right)-\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\left(\begin{array}{c}y\\ y^{\prime }\end{array}\right)=U(x)\left(\begin{array}{c}{C}_1^{\prime }(x)\\ C_2^{\prime }(x)\end{array}\right)=\left(\begin{array}{c}0\\ \sin x\end{array}\right)\end{array}$$ 所以$$\begin{array}{}\text{(44)}& \left(\begin{array}{c}{C}_1^{\prime }(x)\\ C_2^{\prime }(x)\end{array}\right)=U^{-1}(x)\left(\begin{array}{c}0\\ \sin x\end{array}\right)=\left(\begin{array}{c}-{\sin }^{2}x\\ \sin x\cos x\end{array}\right)⟹\{\begin{array}{l}{C}_1(x)=\frac{1}{4}\sin 2x-\frac{x}{2}+A\\ C_2(x)=-\frac{1}{4}\cos 2x+B\end{array}\end{array}$$ 因此$$\begin{array}{}\text{(45)}& y=C_1(x)\cos x+C_2(x)\sin x=\cancel{\frac{1}{4}\sin x}-\frac{x}{2}\cos x+A\cos x+B\sin x\end{array}$$ $◻$

解 令$t=\ln |x|$，则$$\begin{array}{}\text{(47)}& \begin{array}{rl}{y}^{\prime }& =\frac{\mathrm{d}y}{\mathrm{d}t}\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{x}\frac{\mathrm{d}y}{\mathrm{d}t}\\ y^{″}& =\frac{\mathrm{d}{y}^{\prime }}{\mathrm{d}t}\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{x}(-\frac{1}{x^2}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{1}{x}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2})=\frac{1}{x^2}(\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}-\frac{\mathrm{d}y}{\mathrm{d}t})\end{array}\end{array}$$ 代入原方程可得$$\begin{array}{}\text{(48)}& \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}+2\frac{\mathrm{d}y}{\mathrm{d}t}+y=0\end{array}$$ 解得$$\begin{array}{}\text{(49)}& y=({\tilde{C}}_1+{\tilde{C}}_{2}t){\mathrm{e}}^{-t}=({\tilde{C}}_1+{\tilde{C}}_2\ln |x|)\frac{1}{|x|}=\frac{C_1+C_2\ln |x|}{x}\end{array}$$ $◻$