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1. 求解下列方程的通解:
(1). $\dps{\frac{\p u}{\p x}+\frac{\p u}{\p y}+\frac{\p u}{\p z}=0}$.
解答: 特征方程为 $$\bex \frac{\rd x}{1}=\frac{\rd y}{1}=\frac{\rd z}{1}, \eex$$ 而有两个相互独立的首次积分 (本书中叫初积分) $$\bex x-y=C_1,\quad y-z=C_2. \eex$$ 于是原方程的通解为 $$\bex u=\varPhi(x-y,y-z). \eex$$
(2). $\dps{(x-y^3)\frac{\p z}{\p x} +y\frac{\p z}{\p y}=0}$.
解答: 特征方程为 $$\bex \frac{\rd x}{x-y^3}=\frac{\rd y}{y}, \eex$$ 求解之. 先化为 $$\bex \frac{\rd x}{\rd y}=\frac{x}{y}-y^2. \eex$$ 对应的齐次方程的通解为 $x=cy$, 而可设上述非齐次方程的通解为 $x=c(y)y$, 代入即得 $$\bex \frac{\rd c}{\rd y}y=-y^2\ra c(y)=-\frac{y^2}{2}+C\ra x=-\frac{y^3}{2}+Cy. \eex$$ 而有首次积分 $$\bex \frac{x+\frac{y^3}{2}}{y}=C. \eex$$ 于是原方程的通解为 $$\bex u=\varPhi\sex{\frac{2x+y^3}{2y}}. \eex$$
(3). $\dps{xz\frac{\p u}{\p x} +yz\frac{\p u}{\p y}-(x^2+y^2)\frac{\p u}{\p z}=0}$.
解答: 特征方程为 $$\bex \frac{\rd x}{xz} =\frac{\rd y}{yz}=\frac{\rd z}{-(x^2+y^2)}, \eex$$ 而 $$\beex \bea \frac{\rd x}{xz}=\frac{\rd y}{yz}&\ra \frac{\rd x}{x}=\frac{\rd y}{y}\ra y=C_1x,\\ \frac{\rd(x^2)}{2x^2z}=\frac{\rd (y^2)}{2y^2z} =\frac{\rd z}{-(x^2+y^2)} &\ra \frac{\rd (x^2+y^2)}{2(x^2+y^2)z}=\frac{\rd z}{-(x^2+y^2)}\\ &\ra \rd (x^2+y^2)=-2z\rd z\ra x^2+y^2+z^2=C_2. \eea \eeex$$ 于是原方程的通解为 $$\bex u=\varPhi\sex{\frac{y}{x},x^2+y^2+z^2}. \eex$$
(4). $\dps{y\frac{\p u}{\p x} +(xy^\frac{1}{2}-xy^2)\frac{\p u}{\p y} +yz \frac{\p u}{\p z}=0}$.
解答: 特征方程为 $$\bex \frac{\rd x}{y}=\frac{\rd y}{xy^\frac{1}{2}-xy^2}=\frac{\rd z}{yz}, \eex$$ 而 $$\bex \rd x=\frac{\rd y}{xy^{-\frac{1}{2}-xy}}=\frac{\rd z}{z}. \eex$$ 由 $$\beex \bea x\rd x=\frac{\rd y}{y^{-\frac{1}{2}}-y}=\frac{y^\frac{1}{2}\rd y}{1-y^\frac{3}{2}} =-\frac{2}{3}\rd \ln |1-y^\frac{2}{3}|&\ra e^{\frac{x^2}{2}}(1-y^\frac{3}{2})^\frac{2}{3}=C_1,\\ \rd x=\frac{\rd z}{z}&\ra \frac{e^x}{z}=C_2 \eea \eeex$$ 知原方程的通解为 $$\bex u=\varPhi\sex{e^{\frac{x^2}{2}}(1-y^\frac{3}{2})^\frac{2}{3},\frac{e^x}{z}}. \eex$$
(5). $\dps{(mz-ny)\frac{\p z}{\p x} +(nx-lz)\frac{\p u}{\p y} +(ly-mx)\frac{\p u}{\p z}=0}$ ($m,n,l$ 是常数).
解答: 特征方程为 $$\bex \frac{\rd x}{m z-ny}=\frac{\rd y}{nx-lz}=\frac{\rd z}{l y-m x}. \eex$$ 由 $$\beex \bea \frac{l\rd x}{l(mz-ny)}&=\frac{m\rd y}{m(nx-lz)} =\frac{\rd (lx+my)}{-n(ly-mx)}=\frac{\rd z}{ly-mx}\\ &\ra \rd (lx+my)+n\rd z=0 \ra lx+my+nz=C_1,\\ \frac{x\rd x}{x(mz-ny)} &=\frac{y\rd y}{y(nx-lz)} =\frac{x\rd x+y\rd y}{-z(ly-mx)}=\frac{\rd z}{ly-mx} \ra x^2+y^2+z^2=C_2 \eea \eeex$$ 知原方程的通解为 $$\bex u=\varPhi\sex{lx+my+nz,x^2+y^2+z^2}. \eex$$
(6). $\dps{\sqrt{x_1}\frac{\p u}{\p x_1} +\cdots+\sqrt{x_n}\frac{\p u}{\p x_n}=0}$.
解答: 特征方程为 $$\bex \frac{\rd x_1}{\sqrt{x_1}}=\cdots=\frac{\rd x_n}{\sqrt{x_n}}, \eex$$ 容易看出有 $n-1$ 个相互独立的首次积分 $$\bex \sqrt{x_2}-\sqrt{x_1}=C_1,\quad\cdots,\quad \sqrt{x_n}-\sqrt{x_1}=C_{n-1}. \eex$$ 故原方程的通解为 $$\bex u=\varPhi(\sqrt{x_2}-\sqrt{x_1},\cdots, \sqrt{x_n}-\sqrt{x_1}). \eex$$
(7). $\dps{(y+z)\frac{\p u}{\p x} +(z+x)\frac{\p u}{\p y} +(x+y)\frac{\p u}{\p z}=0}$.
解答: 特征方程为 $$\bex \frac{\rd x}{y+z} =\frac{\rd y}{z+x} =\frac{\rd z}{x+y}. \eex$$ 由 $$\beex \bea \frac{\rd (x-y)}{y-x}=\frac{\rd (y-z)}{z-y} &\ra \frac{x-y}{y-z}=C_1,\\ \frac{\rd (x+y+z)}{2(x+y+z)}=\frac{\rd (x-y)}{y-x} &\ra (x+y+z)(x-y)^2=C_2 \eea \eeex$$ 知原方程的通解为 $$\bex u=\varPhi\sex{\frac{x-y}{y-z},(x+y+z)(x-y)^2}. \eex$$
2. 求解下列初值问题:
(1). $\dps{\sedd{\ba{ll} \cfrac{\p z}{\p x}-2x\cfrac{\p z}{\p y}=0\\ z|_{x=1}=y^2 \ea}}$.
解答: 特征方程为 $$\bex \frac{\rd x}{1}=\frac{\rd y}{-2x}, \eex$$ 而有首次积分 $$\bex x^2+y=C. \eex$$ 故原方程的通解为 $$\bex z=\varPhi(x^2+y). \eex$$ 又 $$\bex y^2=z|_{x=1}=\varPhi(1+y)\ra \varPhi(s)=(s-1)^2, \eex$$ 我们有 $$\bex z=(x^2+y-1)^2. \eex$$
(2). $\dps{\sedd{\ba{ll} (x^2-y^2)\cfrac{\p z}{\p x}+2xy \cfrac{\p z}{\p y}=0\\ z|_{x=0}=1+\sqrt{y} \ea}}$.
解答: 特征方程为 $$\bex \frac{\rd x}{x^2-y^2}=\frac{\rd y}{2xy}. \eex$$ 求解之, $$\beex \bea &\quad \frac{\rd x}{\rd y}=\frac{x^2-y^2}{2xy}=\frac{\sex{\frac{x}{y}}^2-1}{2\frac{x}{y}}\\ &\ra s+y\frac{\rd s}{\rd y}=\frac{s^2-1}{2s}\\ &\ra y\frac{\rd s}{\rd y}=-\frac{s^2+1}{2s}\\ &\ra \frac{2s}{s^2+1}\rd s+\frac{\rd y}{y}=0\\ &\ra (s^2+1)y=C\\ &\ra \frac{x^2+y^2}{y}=C. \eea \eeex$$ 故原方程的通解为 $$\bex z=\varPhi\sex{\frac{x^2+y^2}{y}}. \eex$$ 又 $$\bex 1+\sqrt{y}=z|_{x=0}=\varPhi(y), \eex$$ 我们有 $$\bex z=1+\sqrt{\frac{x^2+y^2}{y}}. \eex$$
(3). $\dps{\sedd{\ba{ll} (y+z)\cfrac{\p u}{\p x} +(z+x)\cfrac{\p u}{\p y} +(x+y)\cfrac{\p u}{\p x}=0\\ u|_{z=0}=y^2 \ea}}$.
解答: 有第 1 题第 (7) 小题, 原方程的通解为 $$\bex u=\varPhi\sex{\frac{x-y}{y-z},(x+y+z)(x-y)^2}. \eex$$ 又 $$\bex y^2=u|_{z=0}=\varPhi\sex{\frac{x-y}{y},(x+y)(x-y)^2}\ra \varPhi(s,t)=\sez{\frac{t}{s^2(s+2)}}^\frac{2}{3}, \eex$$ 我们有 $$\bex u=\frac{(y-z)^2(x+y+z)^\frac{2}{3}}{(x+y-2z)^\frac{2}{3}}. \eex$$
(4). $\dps{\sedd{\ba{ll} (x^2+y^2)\cfrac{\p u}{\p x} +2xy\cfrac{\p u}{\p y}=0\\ u|_{x=2y}=y^2 \ea}}$.
解答: 特征方程为 $$\bex \frac{\rd x}{x^2+y^2}=\frac{\rd y}{2xy}, \eex$$ 而 $$\bex \frac{\rd (x+y)}{(x+y)^2}=\frac{\rd (x-y)}{(x-y)^2}\ra \frac{1}{x+y}-\frac{1}{x-y}=C. \eex$$ 故原方程的通解为 $$\bex u=\varPhi\sex{\frac{1}{x+y}-\frac{1}{x-y}}. \eex$$ 又 $$\bex y^2=u|_{x=2y}=\varPhi\sex{-\frac{2}{3y}} \ra \varPhi(s)=\frac{4}{9s^2}, \eex$$ 我们有 $$\bex u=\frac{(x^2-y^2)^2}{9y^2}. \eex$$
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原文地址：http://www.cnblogs.com/zhangzujin/p/4211241.html
