《常微分方程教程》习题2.4.1,(4)

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  	itle{	extbf{《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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  egin{exercise}[2.4.1,(4)]
    求解下列微分方程:
$$
y'=x^3y^3-xy.
$$
end{exercise}
egin{proof}[解]
即为
$$
frac{dy}{dx}=x^3y^3-xy.
$$
这是个 Bernoulli 方程.当 $y
eq 0$ 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
$$
frac{paphi}{pa z}=e^{int -2xdx}
i phi=ze^{int -2xdx}+f(x).
$$
因此
$$
-2xze^{int -2xdx}+f'(x)=e^{int -2xdx}(2x^3-2zx).
$$
可得
$$
f'(x)=2x^3e^{int -2xdx}
i f(x)=-x^2e^{int -2xdx}-e^{int -2xdx}+C.
$$
因此可得通积分为
$$
phiequiv ze^{int -2xdx}-x^2e^{int -2xdx}-e^{int -2xdx}+C=0.
$$
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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