Answer
$$
\frac{dL}{dt} =K L^{2} \ln t , \quad L(1)=-1.
$$
The solution of the given differential equation is
$$
L= \frac{1}{K t-K t \ln t -(1+K)}
$$
Work Step by Step
$$
\frac{dL}{dt} =K L^{2} \ln t , \quad L(1)=-1.
$$
We write the equation in terms of differentials:
$$
\frac{dL}{L^{2}} =K \ln t dt
$$
and integrate both sides:
$$
\int \frac{dL}{L^{2}} =\int K \ln t dt
$$
use integration by parts with
$$
\quad\quad\quad \left[\begin{array}{c}{u=\ln t, \quad\quad dv= dt } \\ {d u= \frac{dt}{t} , \quad\quad v= t }\end{array}\right] , \text { then }\\
$$
$ \Rightarrow$
$$
-\frac{1}{L}= K t \ln t - K \int dt +C
$$
$\Rightarrow$
$$
-\frac{1}{L}= K t \ln t - K t +C
$$
$\Rightarrow$
$$
L= \frac{1}{K t-K t \ln t -C}
$$
where $C $ is an arbitrary constant.
Now, to satisfy the condition $ L(1)=-1, $ we have
$$
-1= \frac{1}{K -K \ln (1) -C} \quad \Rightarrow C-K=1 \quad \Rightarrow C=1+K
$$
Thus, the solution of the given differential equation is
$$
L= \frac{1}{K t-K t \ln t -(1+K)}
$$
