Answer
$\displaystyle r = \frac{1}{\ln t}(e^t + C)$
Work Step by Step
$\displaystyle t \ln t \frac{dr}{dt}+r=te^t$
$\displaystyle \frac{dr}{dt}+\frac{1}{t \ln t}r=\frac{e^t}{\ln t}$
Find the integrating factor $I(t)$:
$\displaystyle I(t) = e^{\int \frac{1}{t \ln t} dt}$
Substitute using $u= \ln t$, and $du=\frac{1}{t}dt$.
$\displaystyle I(t) = e^{\ln \ln t} = \ln t$
Now, use the integrating factor $I(t)$ to solve the first-order linear differential equation:
$\displaystyle \frac{d(r\ln t)}{dt} = \frac{e^t \ln t}{\ln t} = e^t$
Integrate both sides:
$\displaystyle r\ln t = \int e^t dt$
$\displaystyle r = \frac{1}{\ln t}(e^t + C)$
