For the ODE with negative initial condition, say for example \begin{align*} \dot{x}= rx\left(1-\dfrac{x}{K}\right),x(0)=-K \end{align*}
From the phase portrait, $x(t)$ should converge to $-\infty$ as $t\to \infty$. However, this is my exact solution for it: \begin{align*} &\dot{x} =rx\left(1-\dfrac{x}{K}\right) \\ \Leftrightarrow &\left[\dfrac{1}{x\left(1-\frac{x}{K}\right)}\right]dx = rdt \\ \Leftrightarrow & \left[\dfrac{1-\frac{x}{K}+\frac{x}{K}}{x\left(1-\frac{x}{K}\right)}\right]dx = rdt\\ \Leftrightarrow&\left(\dfrac{1}{x}+\dfrac{1}{K-x}\right)dx = rdt.\\ \end{align*} Integrate both sides, we claim that \begin{align*} \ln|x|-\ln|K-x|=rt+\ln C, \end{align*} where $C>0$. Therefore we get \begin{align*} &\left|\dfrac{x}{K-x}\right|=Ce^{rt}\\ \Leftrightarrow&\dfrac{x}{K-x}=\pm Ce^{rt}. \end{align*} Since $C$ can be any positive number, then $\pm C$ can take on any non-zero value, so that we can introduce the solution of the form \begin{align} \dfrac{x}{K-x}=Ce^{rt}, \end{align} where $C \in \mathbb{R}, C\ne 0$ so \begin{align*} x(t) = \dfrac{CKe^{rt}}{1+Ce^{rt}}. \end{align*} For the initial condition $x_0=-K$. So, we get \begin{align*} -K=C\left(K+K\right) \Leftrightarrow C=-\dfrac{1}{2}. \end{align*} So the solution in this case is \begin{align*} x(t) = \dfrac{-1/2Ke^{rt}}{1-1/2 e^{rt}}=\dfrac{-1/2K}{e^{-rt}-1/2} \to K, \end{align*} as $t \to \infty$ (since $r>0$ then $0<e^{-r}<1$ so $e^{-rt}\to 0$ as $t \to \infty$), which is not true due to the phase portrait. Can someone please show me where I get It wrong please? Thank you!