Is there a unique solution to this simple differential equation? I am trying to establish uniqueness for a solution to a bigger problem, and it boils down to whether or not the following differential equation has a unique solution:
$$f'(t)⋅(f(t)-t)=K$$
Clearly, one solution to this differential equation is $f(t)=t+K$. Are there other solutions to this differential equation?
 A: Let $g(t)=f(t)-t$ then $f$ is a solution if and only if $g$ solves the autonomous differential equation $$g'(t)=A(g(t)),\qquad A:x\mapsto(K-x)/x,$$ for every $x\ne0$. Studying the sign of $A$, one sees that every initial condition $g(0)=x_0$ yields:


*

*a decreasing solution $g$ with limit $-\infty$ if $x_0\lt0$,

*an increasing solution $g$ with limit $K$ if $0\lt x_0\lt K$,

*the constant solution $g=K$ if $x_0=K$,

*a decreasing solution $g$ with limit $K$ if $x_0\gt K$. 


Thus, there are lots of solutions... Solving the ODE in $g$ yields, for every $t$, $$(f(t)-t-K)\cdot\exp(f(t)/K)=(f(0)-K)\cdot\exp(f(0)/K).$$
A: Short answer; Yes.
Long answer;
Let $$v(t)  :=  f(t)-t, v'(t)= f'(t)-1$$
$$(v'(t)+1) v(t) = k \iff
v'(t) = \frac{k-v(t)}{v(t)} \iff
\frac{v'(t) v(t)}{k-v(t)}  =  1 \iff
\int \frac{v'(t) v(t)}{k-v(t)} \mathrm{d}t  =   \int 1 \mathrm{d}t \iff
-(k \log(-k+v(t)))-v(t)  =  t+c_1 \iff 
v(t) = k \left (W \left (\frac{1}{k e^{\frac{t+c_1}{k}}}\right )+1 \right ) \iff f(t) = k+t+k W\left (\frac{1}{k e^{\frac{t+c_1}{k}}} \right )$$
