How do you find a solution of this differential equation on both of its intervals of existence?

Is this correct?

$x'=\frac{k}{k+t}x \implies x'-\frac{k}{k+t}x=0$

$\implies (|k+t|^{-k}x)'=0 \implies x(t)=c|k+t|^{k}$

x(t)=\begin{cases} c(k+t)^{k} & k>-t \\ -c(k+t)^k & k<-t \\ \end{cases}


This is mostly right, but there is one issue. Because of the singularity at $t = -k$, you have two separate problems on $t > -k$ and $t < -k$. In terms of your solution, this would mean you can have two different constants $c$ on these two domains.

  • $\begingroup$ So it would be correct if I renamed the constant $-c$ to $c_1$? $\endgroup$ – user707991 Sep 26 '19 at 23:08
  • $\begingroup$ Yes, that would be enough. $\endgroup$ – Zarrax Sep 26 '19 at 23:28

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