这是一份kcl伦敦大学学院 5CCM131B作业代写的成功案
This can be separated to give
$$
\frac{1-3 y}{y} \mathrm{~d} y=\frac{5 x-2}{x} \mathrm{~d} x
$$
Integrating both sides we have
$$
\int\left(\frac{1}{y}-3\right) \mathrm{d} y=\int\left(5-\frac{2}{x}\right) \mathrm{d} x
$$
taking $x$ and $y$ positive, because they represent the size of a population, we have no need of modulus signs in the logarithms arising from the integration,
$$
\ln y-3 y=5 x-2 \ln x+c .
$$
We can do no better than this implicit solution relating $x$ and $y$. However, we can represent the curves defined by
$$
F(x, y)=\ln y+2 \ln x-3 y-5 x=\text { constant }
$$
5CCM131B COURSE NOTES ：
For this equation $x(t)=\mathrm{e}^{k t}$ is a solution if $k$ satisfies
$$
k^{2}+2 k+5=0
$$
which has roots
$$
k=\frac{-2 \pm \sqrt{4-20}}{2}=-1 \pm \sqrt{-4}=-1 \pm 2 \mathrm{i}
$$
So the general solution of $(12.10)$ is
$$
x(t)=\mathrm{e}^{-t}(A \cos 2 t+B \sin 2 t),
$$
showing that the origin is stable. Since
$$
\dot{x}(t)=e^{-t}((2 B-A) \cos 2 t-(2 A+B) \sin 2 t)
$$
the initial conditions pick out the solution with
$$
A=1 \quad 2 B-A=0,
$$
i.e. $A=1$ and $B=\frac{1}{2}$, so that
$$
x(t)=\mathrm{e}^{-t}\left(\cos 2 t+\frac{1}{2} \sin 2 t\right) .
$$