这是一份ucl伦敦大学学院 math0046作业代写的成功案
$$
x\left(1-x y^{2}\right) \frac{d y}{d x}+2 y=0
$$
whence
$$
\frac{d y}{d x}=-\frac{2 y}{x\left(1-x y^{2}\right)} .
$$
This would seem to be difficult to solve but, by inverting, we obtain
$$
\frac{d x}{d y}=-\frac{x\left(1-x y^{2}\right)}{2 y}=-\frac{x}{2 y}+\frac{x^{2} y}{2}
$$
and so
$$
\frac{d x}{d y}+\frac{x}{2 y}=\frac{y x^{2}}{2} .
$$
We see that (3.73) is an equation of Bernoulli type when viewed as an equation for $x$ in terms of $y$. Letting $z=1 / x$, then
$$
\frac{d z}{d y}-\frac{z}{2 y}=-\frac{y}{2},
$$
for which the integrating factor is $y^{-\frac{1}{3}}$. Hence
$$
z y^{-\frac{1}{2}}=-\int \frac{y}{2} y^{-\frac{1}{2}} d y+C,
$$
MATH0046 COURSE NOTES ：
The form of this equation is
$$
\frac{d^{2} y}{d x^{2}}+\frac{2}{x} \frac{d y}{d x}+y^{\alpha}=0
$$
where $\alpha$ is a constant. Solutions which satisfy the boundary conditions
$$
y=1, \quad d y / d x=0 \text { at } x=0,
$$
are called Lane-Emden functions. Exact solutions are known when $\alpha=0,1$ and 5 . In the cases $\alpha=0$ and $\alpha=1$, the equation is linear and the solutions are easily found. When $\alpha=5$, let
$$
x=\mathrm{e}^{-t}, \quad y=\frac{1}{\sqrt{2}} \mathrm{e}^{/ / 2} u(t)
$$