《常微分方程教程》习题2.4.1,(4)

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  \title{\textbf{《常微分方程教程》\cite{dinglichang}习题2.4.1,(4)}}
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  \author{\small{叶卢庆}\\{\small{杭州师范大学理学院,学
        号:1002011005}}\\{\small{Email:h5411167@gmail.com}}} % Institution
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  \begin{exercise}[2.4.1,(4)]
    求解下列微分方程:
$$
y'=x^3y^3-xy.
$$
\end{exercise}
\begin{proof}[解]
即为
$$
\frac{dy}{dx}=x^3y^3-xy.
$$
这是个 Bernoulli 方程.当 $y\neq 0$ 时,两边同时除以 $y^3$,可得
$$
\frac{1}{y^3}\frac{dy}{dx}+\frac{1}{y^2}x-x^3=0.
$$
令 $z=y^{-2}$,则
$$
\frac{dz}{dx}=-2y^{-3}\frac{dy}{dx},
$$
因此
$$
\frac{dz}{dx}-2zx+2x^3=0.
$$
这是个关于 $z,x$ 的一阶线性方程.可化为
$$
dz+(2x^3-2zx)dx=0.
$$
乘以积分因子 $u(x)$,则
$$
udz+u(2x^3-2zx)dx=0.
$$
令 
$$
\frac{du}{dx}=-2xu,
$$
不妨令 $u=e^{\int -2xdx}$.因此我们得到恰当方程
$$
e^{\int -2xdx}dz+e^{\int -2xdx}(2x^3-2zx)dx=0.
$$
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
$$
\frac{\pa\phi}{\pa z}=e^{\int -2xdx}\ri \phi=ze^{\int -2xdx}+f(x).
$$
因此
$$
-2xze^{\int -2xdx}+f'(x)=e^{\int -2xdx}(2x^3-2zx).
$$
可得
$$
f'(x)=2x^3e^{\int -2xdx}\ri f(x)=-x^2e^{\int -2xdx}-e^{\int -2xdx}+C.
$$
因此可得通积分为
$$
\phi\equiv ze^{\int -2xdx}-x^2e^{\int -2xdx}-e^{\int -2xdx}+C=0.
$$
其中三个 $e^{\int -2xdx}$ 都是同一个函数.将 $z=y^{-2}$ 代入,可得
$$
\frac{e^{\int -2xdx}}{y^2}-x^2e^{\int -2xdx}-e^{\int -2xdx}+C=0.
$$
其中三个 $e^{\int -2xdx}$ 都是同一个函数.不妨设 $\int -2xdx=-x^2+D$,因
此可得
$$
\frac{e^{-x^2}}{y^2}-x^2e^{-x^2}-e^{-x^2}+C'=0.
$$
而当 $y=0$ 时,可得曲线为 $y=0$.
\end{proof}
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