Integral limit function - Exponential

Question

Do the integral $\frac{d}{dt} y(t) = -a \, y(t)^p$ converge to $y(t) = e^{-at}$ for $p \to 1$? I have tried to integrate and use l'Hospital rule, without success.

Formally, $\lim\limits_{p\to 1} Y(t, p) \stackrel{?}{=} e^{-at}$.

Best regards, Bruno

Answer 1

We have $y' = -ay^p$. Integrate separately, we obtain,

$\frac{y^{-p+1}}{-p+1} = -at + C$, where $C-$ an arbitrary constant. When $p \rightarrow 1$, we've a problem. Hence use L'Hopital's to evaluate limit of the term on LHS as $p \rightarrow 1$.

$\frac{d(y^{-p+1})}{dp}=-y^{-p+1} log(y)$. Hence $\lim_{p \rightarrow1} \frac{y^{-p+1}}{-p+1}=\lim_{p \rightarrow1} y^{-p+1}log(y)=log(y)$. Terms on RHS remain the same since $p$ is independent from $t$. So, $log(y) = -at + C$ which gives $y = Ke^{-at}$, $K-$ constant