# Finding a Differential Equation that Satisfies an Initial Condition

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.

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 $k$.
• Substituting say in your displayed equation, we get $\frac{1}{8}=-k +C$, so $C=k+\frac{1}{8}$. Replace $C$ by $k+\frac{1}{8}$, and then, since it can be done, solve for $L$ in terms of $t$. Apr 1 '14 at 5:11
• Ah, I think I get it. So: $$L = \frac{-1}{k(tln(t)-t)} + \frac {1}{k} +8$$ should be my final answer? Apr 1 '14 at 5:22
• I get $L=\dfrac{-1}{k(t\ln t -t)+k+\frac{1}{8}}$. This can be manipulated in various ways, but the one in the comment above is not one of them. Apr 1 '14 at 5:33