12.4 讲义习题

12.4.1 线性微分方程

解 令$z=y^{1-n}$，则$n$满足的方程为$$\begin{array}{}\text{(1)}& z^{\prime }+(1-n)P(x)z=(1-n)Q(x)\end{array}$$ $◻$

解 (1) 由题可得$$\begin{array}{}\text{(2)}& \{\begin{array}{l}{y}_1^{\prime }+p(x)y_1+q(x)y_1^2-f(x)=0\\ y_2^{\prime }+p(x)y_2+q(x)y_2^2-f(x)=0\end{array}\end{array}$$ 两式相减可得$$\begin{array}{}\text{(3)}& z^{\prime }+p(x)z+q(x)z(z+2y_2)=0⟹z^{\prime }+[p(x)+2q(x)y_2(x)]z=-q(x)z^2\end{array}$$

(2) 验证略。设另一个解为$y=\frac{1}{x}+z$，则$z$满足的微分方程为$$\begin{array}{}\text{(4)}& z^{\prime }=\frac{2}{x}z+z^2⟹-\frac{z^{\prime }}{z^2}+\frac{2}{x}\frac{1}{z}+1=0\end{array}$$ 利用积分因子$x^2$可得$$\begin{array}{}\text{(5)}& {\left(\frac{x^2}{z}\right)}^{\prime }=-x^2\frac{z^{\prime }}{z}+\frac{2x}{z}=-x^2⟹\frac{x^2}{z}=-\frac{1}{3}(x^3+C_1)⟹z=-\frac{3x^2}{x^3+C_1}\end{array}$$ 故原方程的通解为$$\begin{array}{}\text{(6)}& y=\frac{1}{x}-\frac{3x^2}{x^3+C_1}\end{array}$$ $◻$

解 求导可得$$\begin{array}{}\text{(8)}& 2y{y}^{(3)}=0⟹y=0\vee y^{(3)}=0⟹y(x)=A{x}^2+Bx+C\end{array}$$ 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$$\begin{array}{}\text{(9)}& 2(A{x}^2+Bx+C)\cdot 2A-(2Ax+B{)}^2=0⟹4AC=B^2\end{array}$$ 所以原方程的通解为$$\begin{array}{}\text{(10)}& y(x)=A{x}^2\pm 2\sqrt{AC}x+C,AC\ge 0\end{array}$$ $◻$

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$$\begin{array}{}\text{(11)}& 2yu\frac{\mathrm{d}u}{\mathrm{d}y}-u^2=0⟹u=0\vee 2\frac{\mathrm{d}u}{u}=\frac{\mathrm{d}y}{y}\end{array}$$ 解得$y$为常值函数或$u^2=C|y|$，即$$\begin{array}{}\text{(12)}& \frac{\mathrm{d}y}{\mathrm{d}x}=\pm \sqrt{C|y|}=2C_1\sqrt{|y|}⟹\sqrt{|y|}=C_{1}x+C_2\end{array}$$ 所以原方程的通解为$$\begin{array}{}\text{(13)}& y=\pm (C_{1}x+C_2{)}^2\end{array}$$ $◻$

解 我们考虑更一般性的问题：设$y_1,y_2$满足二阶线性微分方程$$\begin{array}{}\text{(14)}& y^{″}+a_1(x)y^{\prime }+a_0(x)y=0\end{array}$$ 代入$y_1,y_2$可得$$\begin{array}{}\text{(15)}& \{\begin{array}{l}{a}_0(x)y_1+a_1(x)y_1^{\prime }=-y_1^{″}\\ a_0(x)y_2+a_1(x)y_2^{\prime }=-y_2^{″}\end{array}\end{array}$$ 利用Cramer法则可得$$\begin{array}{}\text{(16)}& a_1(x)=\frac{y_2{y}_1^{″}-y_1{y}_2^{″}}{W[y_1,y_2]},a_0(x)=\frac{y_1^{\prime }{y}_2^{″}-y_2^{\prime }{y}_1^{″}}{W[y_1,y_2]}\end{array}$$ 其中$W[y_1,y_2]=y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$表示Wronsky行列式。代入$y_1=\frac{1}{x}$、$y_2=\frac{1}{x-1}$可得$$\begin{array}{}\text{(17)}& y^{″}+\frac{2-4x}{x-x^2}{y}^{\prime }+\frac{2}{x^2-x}y=0\end{array}$$ $◻$

解 令$F(x)={\int }_0^{x}f(t)\mathrm{d}t$，则$F$二阶可微且满足$F^{\prime }(x)=f(x)$，故有$F$满足以下微分方程：$$\begin{array}{}\text{(19)}& F^{″}+F^{\prime }-\frac{1}{x+1}F=0,F(0)=0,F^{\prime }(0)=1\end{array}$$ 容易观察到$F$的特解为$F(x)=x+1$，代入式$(\text{???})$可得$$\begin{array}{}\text{(20)}& F(x)=C_1(x+1)+C_2(x+1){\int }_0^x\frac{{\mathrm{e}}^{-t}}{(t+1{)}^2}\mathrm{d}t\end{array}$$ 代入初值条件可得$$\begin{array}{}\text{(21)}& C_1=0,C_1+C_2=1⟹F(x)=(x+1){\int }_0^x\frac{{\mathrm{e}}^{-t}}{(t+1{)}^2}\mathrm{d}t\end{array}$$ 因此$$\begin{array}{}\text{(22)}& f(x)=F^{\prime }(x)=\frac{{\mathrm{e}}^{-x}}{x+1}+{\int }_0^x\frac{{\mathrm{e}}^{-t}}{(t+1{)}^2}\mathrm{d}t\end{array}$$ $◻$

12.4.2 常系数线性微分方程（组）

解 考虑变换$u=u(x)$，使得原方程变为$y=y(u)$的常系数线性方程。注意到$$\begin{array}{}\text{(24)}& \begin{array}{rl}{y}^{\prime }& =\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}u}{\mathrm{d}x}\frac{\mathrm{d}y}{\mathrm{d}u}=u^{\prime }\frac{\mathrm{d}y}{\mathrm{d}u}\\ y^{″}& =\frac{\mathrm{d}}{\mathrm{d}x}\left(u^{\prime }\frac{\mathrm{d}y}{\mathrm{d}u}\right)=u^{″}\frac{\mathrm{d}y}{\mathrm{d}u}+u^{\prime }\frac{\mathrm{d}u}{\mathrm{d}x}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{u}^2}=u^{″}\frac{\mathrm{d}y}{\mathrm{d}u}+u^{\prime 2}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{u}^2}\end{array}\end{array}$$ 代入原方程可得$$\begin{array}{}\text{(25)}& u^{\prime 2}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{u}^2}+(u^{″}+\frac{b^{\prime }}{b}{u}^{\prime })\frac{\mathrm{d}y}{\mathrm{d}u}-\frac{a^2}{b^2}y=0\end{array}$$ 利用积分因子$\exp \int \frac{b^{\prime }}{b}\mathrm{d}x=b$可恰好将其凑成$$\begin{array}{}\text{(26)}& \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{u}^2}+\frac{(u^{\prime }b{)}^{\prime }}{u^{\prime 2}b}\frac{\mathrm{d}y}{\mathrm{d}u}-\frac{a^2}{(u^{\prime }b{)}^2}y=0\end{array}$$ 故可令$u^{\prime }b=1$，即$u(x)={\int }_{x_0}^x\frac{\mathrm{d}t}{b(t)}$，则原方程可化为$$\begin{array}{}\text{(27)}& \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{u}^2}-a^{2}y=0⟹y=C_1{\mathrm{e}}^{au(x)}+C_2{\mathrm{e}}^{-au(x)}\end{array}$$ $◻$

解 由于方程只包含$x{y}^{\prime }$和$x^2{y}^{″}$，故可尝试Euler方程的换元法：令$t=\ln |x|$，则有$$\begin{array}{}\text{(29)}& \begin{array}{rl}x{y}^{\prime }& =x\frac{\mathrm{d}t}{\mathrm{d}x}\frac{\mathrm{d}y}{\mathrm{d}t}=\frac{\mathrm{d}y}{\mathrm{d}t}\\ x^2{y}^{″}& =x^2\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{x}\frac{\mathrm{d}y}{\mathrm{d}t}\right)=x^2(-\frac{1}{x^2}\frac{\mathrm{d}y}{\mathrm{d}t}+\frac{1}{x}\frac{\mathrm{d}t}{\mathrm{d}x}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2})=\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}-\frac{\mathrm{d}y}{\mathrm{d}t}\end{array}\end{array}$$ 再令$p=\frac{\mathrm{d}y}{\mathrm{d}t}$，代入原方程可得$$\begin{array}{}\text{(30)}& 2(\frac{\mathrm{d}p}{\mathrm{d}t}-p)+a(1+p{)}^2+p-1=0⟹2\frac{\mathrm{d}p}{\mathrm{d}t}+(a-1+ap)(1+p)=0\end{array}$$ 解得$$\begin{array}{}\text{(31)}& \frac{\mathrm{d}y}{\mathrm{d}t}=p=\frac{(1-a){\mathrm{e}}^{t/2}-C_1}{a{\mathrm{e}}^{t/2}-C_1}\end{array}$$ 积分可得$$\begin{array}{}\text{(32)}& y=\{\begin{array}{ll}-t+\frac{2}{a}\ln (a{\mathrm{e}}^{t/2}+C_1)+C_2,& a\ne 0\\ -t+\frac{2}{C_1}{\mathrm{e}}^{t/2}+C_2,& a=0\end{array}\end{array}$$ 代回$t=\ln |x|$，重新选择$C_1,C_2$可得原方程的通解为$$\begin{array}{}\text{(33)}& y=\{\begin{array}{ll}-\ln |x|+\frac{2}{a}\ln (\sqrt{|x|}+C_1)+C_2,& a\ne 0\\ -\ln |x|+\frac{2}{C_1}\sqrt{|x|}+C_2,& a=0\end{array}\end{array}$$ $◻$