Hello.

Problem: Find the solution of the differential equation that satisfies the given intial condition.

$\displaystyle \frac{dL}{dt}=kL^2ln(t)$

The intial condition:$\displaystyle L(1)=-1$.

My solution:

Well, I noticed its a separable equation,

The problem here, is the question did not identify the K!

so I assumed its a constant.

then I rewrite the D.E. as:

$\displaystyle \frac{dL}{L^2}=k ln(t) dt$

By integrating both sides and using integration by parts, I obtained:

$\displaystyle \frac{-1}{L}=k ( tln(t)-t+C)$

$\displaystyle \frac{-1}{L}=k ( tln(t)-t)+A$ where $\displaystyle A=Ck$.

Solving for L,I got:

$\displaystyle L=\frac{-1}{k(tln(t)-t)+A}$.

since $\displaystyle L(1)=-1$:

$\displaystyle L(1)=\frac{-1}{-k+A}=-1$.

so:

$\displaystyle A-k=1$, i.e. $\displaystyle A=1+k$.

Then ????

I stopped here!