《常微分方程教程》习题2.4.1,(4)

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 85   \title{\textbf{《常微分方程教程》\cite{dinglichang}习题2.4.1,(4)}}
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 87   \author{\small{叶卢庆}\\{\small{杭州师范大学理学院,学
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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 $$
143 udz+u(2x^3-2zx)dx=0.
144 $$
145 146 $$
147 \frac{du}{dx}=-2xu,
148 $$
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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posted @ 2013-11-16 02:23  叶卢庆  阅读(846)  评论(0编辑  收藏