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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?

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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 $t$, $$(f(t)-t-K)\cdot\exp(f(t)/K)=(f(0)-K)\cdot\exp(f(0)/K).$$

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  • $\begingroup$ Thanks. This is super helpful. It looks like the solution is monotonic regardless of the initial condition. If I insist on the solution (to the transformed problem of solving for g(t)) being bounded, and t is allowed to range from -∞ to ∞, is the constant solution the only solution? (i.e., is the constant solution the only bounded solution if the domain is R?) $\endgroup$ – Mike Sep 10 '14 at 22:26
  • $\begingroup$ Yes. $ $ $ $ $ $ $\endgroup$ – Did Sep 10 '14 at 22:43
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Short answer; Yes. Long answer;

Let $$v(t) := f(t)-t, v'(t)= f'(t)-1$$

$$(v'(t)+1) v(t) = k \iff v'(t) = \frac{k-v(t)}{v(t)} \iff \frac{v'(t) v(t)}{k-v(t)} = 1 \iff \int \frac{v'(t) v(t)}{k-v(t)} \mathrm{d}t = \int 1 \mathrm{d}t \iff -(k \log(-k+v(t)))-v(t) = t+c_1 \iff v(t) = k \left (W \left (\frac{1}{k e^{\frac{t+c_1}{k}}}\right )+1 \right ) \iff f(t) = k+t+k W\left (\frac{1}{k e^{\frac{t+c_1}{k}}} \right )$$

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