Solve differential equation, $x'=x^2-2t^{-2}$ Solve differential equation:
$x'=x^2- \frac{2}{t^2}$
Maybe is it sth connected with homogeneous equation?
I have no idea how to solve it.
 A: $$(tx)'=tx'+x=tx^2-\frac{2}{t}+x=\frac{(tx-1)(tx+2)}{t}$$
$$\Rightarrow\frac{(tx)'}{tx-1}-\frac{(tx)'}{tx+2}=\frac{3}{t}$$
$$\Rightarrow\ln|\frac{tx-1}{tx+2}|=3\ln|t|+\ln|C|$$
$$\Rightarrow x=\frac{1+2Ct^3}{t(1-Ct^3)}$$
A: let $t = s^{\alpha}$
thus
$$
\frac{d}{dt} = \frac{ds}{dt}\frac{d}{ds} = \frac{1}{\alpha}s^{1-\alpha}{d}{ds} 
$$
therefore
$$
x' = \frac{1}{\alpha}s^{1-\alpha}{dx}{ds}  = x^2 - 2s^{-2\alpha}
$$
this leads to
$$
\frac{dx}{ds} = \alpha x^2s^{\alpha-1} - 2s^{-(\alpha+1)}
$$
lets try $\alpha = -1$
we obtain
$$
\frac{dx}{ds} = -x^2s^{-2} - 2
$$
let $x = vs$
we obtain
$$
sv' + v = - v^2 - 2
$$
we obtain 
$$
\dfrac{dv}{ds} = -\frac{v^2+v+2}{s}
$$
A: $x'=x^2- \frac{2}{t^2}$  is a Riccati ODE which can be solved thanks to the usual method for these kind of equations.
A more direct way is possible if we see that $x=\frac{1}{t}$ is obviously a particular solution. This draw us to try the change of function $x(t)=\frac{y(t)}{t}$ which leads to a separable ODE.

A: Let $x=\dfrac{1}{u}$ ,
Then $x'=-\dfrac{u'}{u^2}$
$\therefore-\dfrac{u'}{u^2}=\dfrac{1}{u^2}-\dfrac{2}{t^2}$
$u'=\dfrac{2u^2}{t^2}-1$
Let $v=\dfrac{u}{t}$ ,
Then $u=tv$
$u'=tv'+v$
$\therefore tv'+v=2v^2-1$
$tv'=2v^2-v-1$
$t\dfrac{dv}{dt}=(2v+1)(v-1)$
$\dfrac{dv}{(2v+1)(v-1)}=\dfrac{dt}{t}$
$\int\dfrac{dv}{(2v+1)(v-1)}=\int\dfrac{dt}{t}$
$\int\left(\dfrac{1}{3(v-1)}-\dfrac{2}{3(2v+1)}\right)dv=\int\dfrac{dt}{t}$
$\dfrac{\ln(v-1)-\ln(2v+1)}{3}=\ln t+c_1$
$\ln\dfrac{v-1}{2v+1}=3\ln t+c_2$
$\dfrac{v-1}{2v+1}=Ct^3$
$v=\dfrac{1+Ct^3}{1-2Ct^3}$
$\dfrac{1}{xt}=\dfrac{1+Ct^3}{1-2Ct^3}$
$x=\dfrac{1-2Ct^3}{t(1+Ct^3)}$
A: Hint: 
$$t^2x'=(tx)^2-2$$
with $tx=:y$, you get a separable equation
$$tx'+x=y'\qquad\Rightarrow\qquad t^2x'=ty'-y$$
$$ty'-y=y^2-2\qquad\Rightarrow\qquad \frac{dy}{dt}=\frac{y^2+y-2}{t}$$
