# 《常微分方程教程》习题2.4.1,(4)

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81 %    TITLE
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83 \begin{document}
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85   \title{\textbf{《常微分方程教程》\cite{dinglichang}习题2.4.1,(4)}}
86   % \setlength\epigraphwidth{0.7\linewidth}
87   \author{\small{叶卢庆}\\{\small{杭州师范大学理学院,学
88         号:1002011005}}\\{\small{Email:h5411167@gmail.com}}} % Institution
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114   \begin{exercise}[2.4.1,(4)]
115     求解下列微分方程:
116 $$117 y'=x^3y^3-xy. 118$$
119 \end{exercise}
120 \begin{proof}[解]
121 即为
122 $$123 \frac{dy}{dx}=x^3y^3-xy. 124$$
125 这是个 Bernoulli 方程.当 $y\neq 0$ 时,两边同时除以 $y^3$,可得
126 $$127 \frac{1}{y^3}\frac{dy}{dx}+\frac{1}{y^2}x-x^3=0. 128$$
129 令 $z=y^{-2}$,则
130 $$131 \frac{dz}{dx}=-2y^{-3}\frac{dy}{dx}, 132$$
133 因此
134 $$135 \frac{dz}{dx}-2zx+2x^3=0. 136$$
137 这是个关于 $z,x$ 的一阶线性方程.可化为
138 $$139 dz+(2x^3-2zx)dx=0. 140$$
141 乘以积分因子 $u(x)$,则
142 $$143 udz+u(2x^3-2zx)dx=0. 144$$
145 令
146 $$147 \frac{du}{dx}=-2xu, 148$$
149 不妨令 $u=e^{\int -2xdx}$.因此我们得到恰当方程
150 $$151 e^{\int -2xdx}dz+e^{\int -2xdx}(2x^3-2zx)dx=0. 152$$
153 其中两个 $e^{\int -2xdx}$ 是同一个函数.设存在二元函数 $\phi(x,y)$ 使得
154 $$155 \frac{\pa\phi}{\pa z}=e^{\int -2xdx}\ri \phi=ze^{\int -2xdx}+f(x). 156$$
157 因此
158 $$159 -2xze^{\int -2xdx}+f'(x)=e^{\int -2xdx}(2x^3-2zx). 160$$
161 可得
162 $$163 f'(x)=2x^3e^{\int -2xdx}\ri f(x)=-x^2e^{\int -2xdx}-e^{\int -2xdx}+C. 164$$
165 因此可得通积分为
166 $$167 \phi\equiv ze^{\int -2xdx}-x^2e^{\int -2xdx}-e^{\int -2xdx}+C=0. 168$$
169 其中三个 $e^{\int -2xdx}$ 都是同一个函数.将 $z=y^{-2}$ 代入,可得
170 $$171 \frac{e^{\int -2xdx}}{y^2}-x^2e^{\int -2xdx}-e^{\int -2xdx}+C=0. 172$$
173 其中三个 $e^{\int -2xdx}$ 都是同一个函数.不妨设 $\int -2xdx=-x^2+D$,因
174 此可得
175 $$176 \frac{e^{-x^2}}{y^2}-x^2e^{-x^2}-e^{-x^2}+C'=0. 177$$
178 而当 $y=0$ 时,可得曲线为 $y=0$.
179 \end{proof}
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181 %    BIBLIOGRAPHY
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posted @ 2013-11-16 02:23  叶卢庆  阅读(1095)  评论(0编辑  收藏  举报