I was reading this paper and got stuck at the transition from Equation (13) to Equation (14) (p. 16/17). We got a function of the form:
$y(t)=k(t)^{\alpha}h(t)^{\beta}$
We know it grows from zero until converging at some point $y^{*}$. We also know that:
$\frac{\partial \ln(y(t))}{\partial t} = \lambda[\ln(y^{*}) - \ln(y(t))]$
, where $\lambda$ is some constant. How exactly does it imply the following equation?
$\ln(y(t)) = (1 - e^{-\lambda t})\ln(y^{*}) + e^{-\lambda t} \ln(y(0))$
After defining: $f(t)=\ln(y(t))$, I have:
$f'(t)=\lambda e^{-\lambda t}[\ln(y^{*})-f(0)]$
What should I do next? I lack skills in solving differential equations.
First, as pointed out by enzotib, you have a mistake (probably a typo) in your equation, which should read
f'(t) = \lambda [ln(y*) -f(t)]
This is a very very easy differential equation to solve, as long as you don't worry much about using df and dt as if they were ordinary quantities (which in this sort of equation, is justified):
df/dt = λ [ln(y*) -f]
df / [ln(y*) -f] = λ dt
integral 1/[ln(y*) -f] df = integral λ dt = λ + C
ln(y*) - f = e^{-C} e^{-λt}
f(t) = ln(y*) - K e^{-λt} where we replaced one constant e^{-C} with another K.
And once you have that one-parameter solution,
ln(y(0)) = f(0) = ln(y*) - K e^{0} --> K = ln(y*) - ln(y(0))
so ln(y(t)) = f(t) = ln(y*) - [ ln(y*) - ln(y(0))] e^{-\lambda t}
Grouping y* and y(0) terms this is
ln(y(t)) = (1−e^{−λt}) ln(y∗) + e^{−λt} ln(y(0)) which is the equation they give.