Answer
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.
Work Step by Step
$$
\frac{d x}{d t}=1-t+x-t x
$$
$$
\begin{split}
\frac{d x}{d t} &=1-t+x-t x\\
&=1(1-t)+x(1-t)
\\
& = (1+x)(1-t)
\end{split}
$$
$ \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.
