Calculus: Early Transcendentals, Chapter 9, 9.3, Section 9.3, Problem 18

This differential equation can be solved by the method of separating variables.
Multiply both sides by dt and divide both sides of the equation by L^2 . This will bring all terms depending on t to the right side and all terms depending on L to the left side:
(dL)/L^2 = kln(t)dt
Now both sides can be integrated. On the left side, we have an integral of a power function:
int (dL)/L^2 = int L^(-2) dL = -1L^(-1) = -1/L
(The constant of integration can be omitted for now.)
On the right side, the integral of ln(t) equals
int ln(t)dt = tln(t) - t
(This can be verified by taking the derivative using the product rule and the fact that (ln(t))' = 1/t : (tln(t) - t)' = ln(t) + t/t - 1 = ln(t))
So the after the integration the equation becomes
-1/L = k(tln(t) - t) + C
The constant of integration C is added to the right side here.
Since the initial condition is given, L(1) = -1, C can be found now:
for t = 1, ln(t) = ln(1) = 0, and
-1/(-1) = k(1*0 - 1) + C = -k + C
1 = -k + C, so from here C = 1 + k
Putting C back into the equation, we get
-1/L = k(tln(t)-t) + 1 + k
Multiplying both sides by -1 and opening parenthesis on the right side results in
1/L = -ktln(t) + kt - 1 - k
From here,
L(t) = 1/(kt - ktln(t) - 1 - k)
So, the solution of the given equation satisfying initial condition L(1) = -1 is
L(t) = 1/(kt-kln(t)-1-k)

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