## Calculus: Early Transcendentals 8th Edition

The solution of the given differential equation $$\frac{d x}{d t}=1-t+x-t x$$ is $$x=-1+K e^{t-t^{2} / 2}$$ where $K$ is any nonzero constant.
$$\frac{d x}{d t}=1-t+x-t x$$ $$\begin{split} \frac{d x}{d t} &=1-t+x-t x\\ &=1(1-t)+x(1-t) \\ & = (1+x)(1-t) \end{split}$$ $\Rightarrow$ $$\frac{d x}{1+x}=(1-t) d t \quad \Rightarrow \quad \int \frac{d x}{1+x}=\int(1-t) d t$$ integrate both sides to get: $$\ln |1+x|=t-\frac{1}{2} t^{2}+C \Rightarrow |1+x|=e^{t-t^{2} / 2+C}$$ where $C$ is an arbitrary constant. this implies that $$1+x=\pm e^{t-t^{2} / 2} \cdot e^{C} \Rightarrow x=-1+K e^{t-t^{2} / 2}$$ where $K=\pm e^{C}$ is an arbitrary constant. So the solution of the differential equation $$\frac{d x}{d t}=1-t+x-t x$$ is $$x=-1+K e^{t-t^{2} / 2}$$ where $K$ is any nonzero constant.