I have inconsistent problem that involves $~ax'=ab-kx~$. If $~k=0~$ in the first place, I get the expected
$$x=x_0+by$$
However, suppose there is $~k~$, then the solution becomes
$$\int dy=\int\frac{a}{ab-kx}dx$$
$$y=-\frac{a}{k}\ln(\frac{ab-kx}{ab-kx_0})$$
$$x=\frac{ab}{k}+(x_0-\frac{ab}{k})e^{-\frac{k}{a}y}$$
now, if $~k~$ is zero, then I don't recover $~x=x_0+by~$ this baffles me!
I was sure I made the correct integration but the result does not seem to add up resulting to inconsistent answer. Where might probably be the mistake? I want to recover the old solution when I set $~k=0~$.
Write the last line as $$ x=x_0-ab\frac{e^{-k\frac ya}-1}k $$ then in the limit $k\to 0$ the last term is to be seen as difference quotient that converges to the derivative $-ab(-\frac ya)=by$ of $-ab\,e^{-k\frac ya}$ at $k=0$.