Answer
The solution of the given initial-value problem
$$
\frac{d r}{d t}+2 t r=r , \quad \quad r(0)=5
$$
is
$$
r(t)=5 e^{t-t^{2}}
$$
Work Step by Step
The initial-value problem is:
$$
\frac{d r}{d t}+2 t r=r
$$
$$
\frac{d r}{d t}+2 t r=r \Rightarrow \frac{d r}{d t}=r-2 t r=r(1-2 t)
$$
separating and integrating both sides, we get:
$$
\int \frac{d r}{r}=\int(1-2 t) d t \Rightarrow \ln |r|=t-t^{2}+C
$$
$\Rightarrow$
$$
|r|=e^{t-t^{2}+C}=k e^{t-t^{2}}
$$
where where $ k $ is an arbitrary constant.
Since
$$
r(0)=5,5=k e^{0}=k .
$$
Thus the solution of the given initial-value problem is
$$
r(t)=5 e^{t-t^{2}}
$$
