14.1 第十次作业参考答案

14.1.1 习题8.3

解 (1) 以$p=y^{\prime }$为因变量可实现降维，同时方程可分离变量，故有$$\begin{array}{}\text{(1)}& \frac{\mathrm{d}p}{p}=\frac{2x}{1+x^2}\mathrm{d}x⟹\ln |p|=\ln (1+x^2)+C⟹y^{\prime }=p=C_1(1+x^2)\end{array}$$ 积分可得$$\begin{array}{}\text{(2)}& y=C_1(x+\frac{x^3}{3})+C_2\end{array}$$

(2) 方程不显含$x$，故可令$p=y^{\prime }$，则$y^{″}=\frac{p\mathrm{d}p}{\mathrm{d}y}$，代入原方程可得$$\begin{array}{}\text{(3)}& \frac{p\mathrm{d}p}{\mathrm{d}y}+\frac{2p^2}{1-y}=0⟹\frac{\mathrm{d}p}{p}=\frac{2\mathrm{d}y}{y-1}⟹\frac{\mathrm{d}y}{\mathrm{d}x}=p=C_1(1-y{)}^2\end{array}$$ 再次变量分离可得$$\begin{array}{}\text{(4)}& \frac{\mathrm{d}y}{(1-y{)}^2}=C_1\mathrm{d}x⟹\frac{1}{1-y}=C_{1}x+C_2⟹y=1-\frac{1}{C_{1}x+C_2}\end{array}$$

(3) 同(2)可得$$\begin{array}{}\text{(5)}& y\frac{p\mathrm{d}p}{\mathrm{d}y}-p^2=0⟹\frac{\mathrm{d}p}{p}=\frac{\mathrm{d}y}{y}⟹\frac{\mathrm{d}y}{\mathrm{d}x}=p=C_{1}y\end{array}$$ 再次变量分离可得$$\begin{array}{}\text{(6)}& \frac{\mathrm{d}y}{y}=C_1\mathrm{d}x⟹y=C_2{\mathrm{e}}^{C_{1}x}\end{array}$$

(4) 同(1)可得$$\begin{array}{}\text{(7)}& \frac{\mathrm{d}p}{p^3}=\mathrm{d}x⟹-\frac{1}{2p^2}=x+C⟹y^{\prime }=p=\pm \frac{1}{\sqrt{C_1-2x}}\end{array}$$ 积分可得$$\begin{array}{}\text{(8)}& y=\pm \sqrt{C_1-2x}+C_2\end{array}$$

(5) 注意到$$\begin{array}{}\text{(9)}& {\left(\frac{y}{x^k}\right)}^{\prime }=\frac{x{y}^{\prime }-ky}{x^{k+1}},{\left(\frac{y^{\prime }}{x^k}\right)}^{\prime }=\frac{x{y}^{″}-k{y}^{\prime }}{x^{k+1}}\end{array}$$ 为了凑出二阶和零阶项系数，尝试$$\begin{array}{}\text{(10)}& \begin{array}{rl}{(\frac{y^{\prime }}{x^k}+\frac{y^{\prime }}{x^{k-1}}-\frac{6}{k}\frac{y}{x^k})}^{\prime }& =\frac{x{y}^{″}-k{y}^{\prime }}{x^{k+1}}+\frac{x{y}^{″}-(k-1)y^{\prime }}{x^k}-\frac{6}{k}\frac{x{y}^{\prime }-ky}{x^{k+1}}\\ & =\frac{x(x+1)y^{″}-[(k-1+\frac{6}{k})x+k]y^{\prime }+6y}{x^{k+1}}\end{array}\end{array}$$ 故有$$\begin{array}{}\text{(11)}& (k-1+\frac{6}{k})x+k=4x+2⟹k=2\end{array}$$ 因此$$\begin{array}{}\text{(12)}& {\left[\frac{(1+x)y^{\prime }-3y}{x^2}\right]}^{\prime }=\frac{x(x+1)y^{″}-(4x+2)y^{\prime }+6y}{x^3}=0⟹\frac{(1+x)y^{\prime }-3y}{x^2}=C_1\end{array}$$ 利用积分因子法可得$$\begin{array}{}\text{(13)}& \frac{C_1{x}^2}{(1+x{)}^4}=\frac{(1+x)y^{\prime }-3y}{(1+x{)}^4}={\left[\frac{y}{(1+x{)}^3}\right]}^{\prime }\end{array}$$ 积分可得$$\begin{array}{}\text{(14)}& \begin{array}{rl}y& =C_2(1+x{)}^3+C_1(1+x{)}^3\int [\frac{(1+x{)}^2-2(1+x)+1}{(1+x{)}^4}\mathrm{d}(1+x)]\\ & =C_2(1+x{)}^3+C_1(1+x{)}^3[-\frac{1}{1+x}+\frac{1}{(1+x{)}^2}-\frac{1}{3(1+x{)}^3}]\\ & =C_2(1+x{)}^3-\frac{1}{3}{C}_1(3x^2+3x+1)\end{array}\end{array}$$ 重新选择$C_1,C_2$可使得$$\begin{array}{}\text{(15)}& y=C_2{x}^3+C_1(3x^2+3x+1)\end{array}$$

(5’) 由于方程是线性的，且$y^{(k)}$的系数$a_k(x)$均为次数为$k$的多项式，故我们可以猜测方程具有多项式解：$$\begin{array}{}\text{(16)}& y=a_0+a_{1}x+a_2{x}^2+a_3{x}^3\end{array}$$ 代入原方程可得$$\begin{array}{}\text{(17)}& (6a_0-2a_1)+(2a_1-2a_2)x=0⟹a_2=a_1=3a_0,a_3\in \mathbb{R}\end{array}$$ 因此$$\begin{array}{}\text{(18)}& y=C_2{x}^3+C_1(3x^2+3x+1)\end{array}$$

(6) 注意到$$\begin{array}{}\text{(19)}& x^2{y}^{″}+(2x-1)y^{\prime }+x^2{y}^{\prime }-y=(x^2{y}^{\prime }-y{)}^{\prime }-(x^2{y}^{\prime }-y)=0\end{array}$$ 由此解得$$\begin{array}{}\text{(20)}& x^2{y}^{\prime }-y=C_1{\mathrm{e}}^x\end{array}$$ 再利用常数变易法可得$$\begin{array}{}\text{(21)}& y=C_2{\mathrm{e}}^{-1/x}+C_1{\mathrm{e}}^{-1/x}{\int }_1^x\frac{{\mathrm{e}}^{t+1/t}}{t^2}\mathrm{d}t\end{array}$$

(7) 同(4)可得$$\begin{array}{}\text{(22)}& \frac{\mathrm{d}p}{p^2}=\mathrm{d}x⟹-\frac{1}{p}=x+C⟹y^{\prime }=p=-\frac{1}{x+C_1}\end{array}$$ 积分可得$$\begin{array}{}\text{(23)}& y=-\ln |x+C_1|+C_2\end{array}$$ 代入初值条件可得$C_1=-1,C_2=0$，故原方程的解为$$\begin{array}{}\text{(24)}& y=-\ln (1-x)\end{array}$$ $◻$

14.1.2 习题8.4

解 参考第12.2.6节，令$z=y^{1-n}$，则$n$满足的方程为$$\begin{array}{}\text{(25)}& z^{\prime }+(1-n)P(x)z=(1-n)Q(x)\end{array}$$ $◻$

解 (1) 由题可得$$\begin{array}{}\text{(26)}& \{\begin{array}{l}{y}_1^{\prime }+p(x)y_1+q(x)y_1^2-f(x)=0\\ y_2^{\prime }+p(x)y_2+q(x)y_2^2-f(x)=0\end{array}\end{array}$$ 两式相减可得$$\begin{array}{}\text{(27)}& z^{\prime }+p(x)z+q(x)z(z+2y_2)=0⟹z^{\prime }+[p(x)+2q(x)y_2(x)]z=-q(x)z^2\end{array}$$

(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$$\begin{array}{}\text{(28)}& z^{\prime }=\frac{2}{x}z+z^2⟹-\frac{z^{\prime }}{z^2}+\frac{2}{x}\frac{1}{z}+1=0\end{array}$$ 利用积分因子$x^2$可得$$\begin{array}{}\text{(29)}& {\left(\frac{x^2}{z}\right)}^{\prime }=-x^2\frac{z^{\prime }}{z}+\frac{2x}{z}=-x^2⟹\frac{x^2}{z}=-\frac{1}{3}(x^3+C_1)⟹z=-\frac{3x^2}{x^3+C_1}\end{array}$$ 故原方程的通解为$$\begin{array}{}\text{(30)}& y=\frac{1}{x}-\frac{3x^2}{x^3+C_1}\end{array}$$ $◻$

解 参考例13.3.2。 $◻$

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$$\begin{array}{}\text{(32)}& y^{″}+a_1(x)y^{\prime }+a_0(x)y=0\end{array}$$ 代入$y_1,y_2$可得$$\begin{array}{}\text{(33)}& \{\begin{array}{l}{a}_0(x)y_1+a_1(x)y_1^{\prime }=-y_1^{″}\\ a_0(x)y_2+a_1(x)y_2^{\prime }=-y_2^{″}\end{array}\end{array}$$ 利用Cramer法则可得$$\begin{array}{}\text{(34)}& a_1(x)=\frac{y_2{y}_1^{″}-y_1{y}_2^{″}}{W[y_1,y_2]},a_0(x)=\frac{y_1^{\prime }{y}_2^{″}-y_2^{\prime }{y}_1^{″}}{W[y_1,y_2]}\end{array}$$ 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$$\begin{array}{}\text{(35)}& y^{″}+\frac{2-4x}{x-x^2}{y}^{\prime }+\frac{2}{x^2-x}y=0\end{array}$$ $◻$

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：$$\begin{array}{}\text{(37)}& F^{″}+F^{\prime }-\frac{1}{x+1}F=0,F(0)=0,F^{\prime }(0)=1\end{array}$$ 容易观察到$F$的特解为$F(x)=x+1$，代入式$(\text{???})$可得$$\begin{array}{}\text{(38)}& F(x)=C_1(x+1)+C_2(x+1){\int }_0^x\frac{{\mathrm{e}}^{-t}}{(t+1{)}^2}\mathrm{d}t\end{array}$$ 代入初值条件可得$$\begin{array}{}\text{(39)}& C_1=0,C_1+C_2=1⟹F(x)=(x+1){\int }_0^x\frac{{\mathrm{e}}^{-t}}{(t+1{)}^2}\mathrm{d}t\end{array}$$ 因此$$\begin{array}{}\text{(40)}& f(x)=F^{\prime }(x)=\frac{{\mathrm{e}}^{-x}}{x+1}+{\int }_0^x\frac{{\mathrm{e}}^{-t}}{(t+1{)}^2}\mathrm{d}t\end{array}$$ $◻$

14.1.3 习题8.6

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$$\begin{array}{}\text{(42)}& \begin{array}{rl}{y}^{\prime }& =\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}u}{\mathrm{d}x}\frac{\mathrm{d}y}{\mathrm{d}u}=u^{\prime }\frac{\mathrm{d}y}{\mathrm{d}u}\\ y^{″}& =\frac{\mathrm{d}}{\mathrm{d}x}\left(u^{\prime }\frac{\mathrm{d}y}{\mathrm{d}u}\right)=u^{″}\frac{\mathrm{d}y}{\mathrm{d}u}+u^{\prime }\frac{\mathrm{d}u}{\mathrm{d}x}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{u}^2}=u^{″}\frac{\mathrm{d}y}{\mathrm{d}u}+u^{\prime 2}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{u}^2}\end{array}\end{array}$$ 代入原方程可得$$\begin{array}{}\text{(43)}& u^{\prime 2}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{u}^2}+(u^{″}+\frac{b^{\prime }}{b}{u}^{\prime })\frac{\mathrm{d}y}{\mathrm{d}u}-\frac{a^2}{b^2}y=0\end{array}$$ 利用积分因子$\exp \int \frac{b^{\prime }}{b}\mathrm{d}x=b$可恰好将其凑成$$\begin{array}{}\text{(44)}& \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{u}^2}+\frac{(u^{\prime }b{)}^{\prime }}{u^{\prime 2}b}\frac{\mathrm{d}y}{\mathrm{d}u}-\frac{a^2}{(u^{\prime }b{)}^2}y=0\end{array}$$ 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$$\begin{array}{}\text{(45)}& \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{u}^2}-a^{2}y=0⟹y=C_1{\mathrm{e}}^{au(x)}+C_2{\mathrm{e}}^{-au(x)}\end{array}$$ $◻$

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$$\begin{array}{}\text{(47)}& \begin{array}{rl}x{y}^{\prime }& =x\frac{\mathrm{d}t}{\mathrm{d}x}\frac{\mathrm{d}y}{\mathrm{d}t}=\frac{\mathrm{d}y}{\mathrm{d}t}\\ x^2{y}^{″}& =x^2\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{x}\frac{\mathrm{d}y}{\mathrm{d}t}\right)=x^2(-\frac{1}{x^2}\frac{\mathrm{d}y}{\mathrm{d}t}+\frac{1}{x}\frac{\mathrm{d}t}{\mathrm{d}x}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2})=\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}-\frac{\mathrm{d}y}{\mathrm{d}t}\end{array}\end{array}$$ 再令$p=\frac{\mathrm{d}y}{\mathrm{d}t}$，代入原方程可得$$\begin{array}{}\text{(48)}& 2(\frac{\mathrm{d}p}{\mathrm{d}t}-p)+a(1+p{)}^2+p-1=0⟹2\frac{\mathrm{d}p}{\mathrm{d}t}+(a-1+ap)(1+p)=0\end{array}$$ 解得$$\begin{array}{}\text{(49)}& \frac{\mathrm{d}y}{\mathrm{d}t}=p=\frac{(1-a){\mathrm{e}}^{t/2}-C_1}{a{\mathrm{e}}^{t/2}-C_1}\end{array}$$ 积分可得$$\begin{array}{}\text{(50)}& y=\{\begin{array}{ll}-t+\frac{2}{a}\ln (a{\mathrm{e}}^{t/2}+C_1)+C_2,& a\ne 0\\ -t+\frac{2}{C_1}{\mathrm{e}}^{t/2}+C_2,& a=0\end{array}\end{array}$$ 代回$t=\ln |x|$，重新选择$C_1,C_2$可得原方程的通解为$$\begin{array}{}\text{(51)}& y=\{\begin{array}{ll}-\ln |x|+\frac{2}{a}\ln (\sqrt{|x|}+C_1)+C_2,& a\ne 0\\ -\ln |x|+\frac{2}{C_1}\sqrt{|x|}+C_2,& a=0\end{array}\end{array}$$ $◻$