first orderr non linear ODE I came along this first order non linear ODE, and cannot solve it.
$$\frac{dv}{dt}=\frac{-b}{(vt)^2}+k$$ (where b and k are constants)
The question asked to express v as a function of t. 
Thank you very much, any help is welcome!
 A: Assume $b\neq0$ and $k\neq0$ for the key case:
Hint:
Approach $1$:
$\dfrac{dv}{dt}=-\dfrac{b}{(vt)^2}+k$
$\dfrac{dv}{dt}-k=-\dfrac{b}{v^2t^2}$
$\dfrac{d(v-kt)}{dt}=-\dfrac{b}{v^2t^2}$
Let $x=v-kt$ ,
Then $\dfrac{dx}{dt}=-\dfrac{b}{(x+kt)^2t^2}$
$\dfrac{dt}{dx}=-\dfrac{k^2t^4+2kxt^3+x^2t^2}{b}$
This belongs to a "Chini-like" equation.
Approach $2$:
Let $x=vt$ ,
Then $v=\dfrac{x}{t}$
$\dfrac{dv}{dt}=\dfrac{1}{t}\dfrac{dx}{dt}-\dfrac{x}{t^2}$
$\therefore\dfrac{1}{t}\dfrac{dx}{dt}-\dfrac{x}{t^2}=-\dfrac{b}{x^2}+k$
$\dfrac{1}{t}\dfrac{dx}{dt}=k-\dfrac{b}{x^2}+\dfrac{x}{t^2}$
$\dfrac{1}{t}\dfrac{dx}{dt}=\dfrac{(kx^2-b)t^2+x^3}{x^2t^2}$
$((kx^2-b)t^2+x^3)\dfrac{dt}{dx}=x^2t$
Let $u=\dfrac{1}{t^2}$ ,
Then $\dfrac{du}{dx}=-\dfrac{2}{t^3}\dfrac{dt}{dx}$
$\therefore-\dfrac{((kx^2-b)t^2+x^3)t^3}{2}\dfrac{du}{dx}=x^2t$
$\left(\dfrac{1}{t^2}+\dfrac{k}{x}-\dfrac{b}{x^3}\right)\dfrac{du}{dx}=-\dfrac{2}{xt^4}$
$\left(u+\dfrac{k}{x}-\dfrac{b}{x^3}\right)\dfrac{du}{dx}=-\dfrac{2u^2}{x}$
This belongs to an Abel equation of the second kind.
Let $v=u+\dfrac{k}{x}-\dfrac{b}{x^3}$ ,
Then $u=v-\dfrac{k}{x}+\dfrac{b}{x^3}$
$\dfrac{du}{dx}=\dfrac{dv}{dx}+\dfrac{k}{x^2}-\dfrac{3b}{x^4}$
$\therefore v\left(\dfrac{dv}{dx}+\dfrac{k}{x^2}-\dfrac{3b}{x^4}\right)=-\dfrac{2}{x}\left(v-\dfrac{k}{x}+\dfrac{b}{x^3}\right)^2$
$v\dfrac{dv}{dx}+\left(\dfrac{k}{x^2}-\dfrac{3b}{x^4}\right)v=-\dfrac{2}{x}\left(v^2-\left(\dfrac{2k}{x}-\dfrac{2b}{x^3}\right)v+\dfrac{(kx^2-b)^2}{x^6}\right)$
$v\dfrac{dv}{dx}+\left(\dfrac{k}{x^2}-\dfrac{3b}{x^4}\right)v=-\dfrac{2v^2}{x}+\left(\dfrac{4k}{x^2}-\dfrac{4b}{x^4}\right)v-\dfrac{2(kx^2-b)^2}{x^7}$
$v\dfrac{dv}{dx}=-\dfrac{2v^2}{x}+\left(\dfrac{3k}{x^2}-\dfrac{b}{x^4}\right)v-\dfrac{2(kx^2-b)^2}{x^7}$
Let $v=\dfrac{w}{x^2}$ ,
Then $\dfrac{dv}{dx}=\dfrac{1}{x^2}\dfrac{dw}{dx}-\dfrac{2w}{x^3}$
$\therefore\dfrac{w}{x^2}\left(\dfrac{1}{x^2}\dfrac{dw}{dx}-\dfrac{2w}{x^3}\right)=-\dfrac{2w^2}{x^5}+\left(\dfrac{3k}{x^2}-\dfrac{b}{x^4}\right)\dfrac{w}{x^2}-\dfrac{2(kx^2-b)^2}{x^7}$
$\dfrac{w}{x^4}\dfrac{dw}{dx}-\dfrac{2w^2}{x^5}=-\dfrac{2w^2}{x^5}+\left(\dfrac{3k}{x^2}-\dfrac{b}{x^4}\right)\dfrac{w}{x^2}-\dfrac{2(kx^2-b)^2}{x^7}$
$\dfrac{w}{x^4}\dfrac{dw}{dx}=\left(\dfrac{3k}{x^2}-\dfrac{b}{x^4}\right)\dfrac{w}{x^2}-\dfrac{2(kx^2-b)^2}{x^7}$
$w\dfrac{dw}{dx}=\left(3k-\dfrac{b}{x^2}\right)w-\dfrac{2(kx^2-b)^2}{x^3}$
