Find the solution of the differential equation that satisfies the given initial condition:
$$ \frac{dL}{dt} = kL^2ln(t), L(1) = -8$$
The thing that's really screwing me up here is that darn k. I've separated the equation as follows:
$$ \int \frac{dL}{L^2} = \int kln(t)dt $$
And integrated:
$$ -L^{-1} = k(tln(t) – t) + C $$
But I don't know what to do with k. I'm guessing it involves substitution, but I'm not sure how.
After I've gotten rid of the k, I'm guessing it's just a matter of writing the equation as L = {whatever} and substituting 1 for t and -8 for L to find C.
Best Answer
You will not be able to get rid of $k$ without further information. The constant $C$ that you need to compute will be a function of $k$.
Added: Substituting the initial condition in $-\frac{1}{L}=k(t\ln t-t)+C$, we obtain $\frac{1}{8}=-k+C$. Thus $$-\frac{1}{L}=k(t\ln t-t)+\frac{1}{8}+k.$$ Take the negative reciprocal to find $L$ explicitly in terms of $t$ and $k$.