Consider the differential equation $y' = y^\alpha$ with initial condition $y(0) = c_0 > 0$. Let $\alpha$ vary in a neighborhood of $1 \in \mathbb{R}$. Of course, for $\alpha = 1$ we get the solution $y_{c_0, \alpha =1}(x) = c_0 e^x$. On the other hand, when $\alpha \neq 1$ we get a solution of the form $$y_{c_0, \alpha \neq 1}(x) = (x+c_0^{1-\alpha})^{\frac{1}{1-\alpha}} . $$ Now consider the family of solutions $y_{c_0, \alpha}(x)$ resulting from letting the triple $(x, c_0, \alpha)$ vary in a neighbourhood of $(0,1,1)$. I'd like two show that these vary continuously.
I have the following idea. Rearranging the $\alpha \neq 1$ solution gives $$ y_{c_0, \alpha \neq 1}(x) = c_0(1+c_0^{\alpha-1} x)^{\frac{1}{1-\alpha}} . $$ Recall also that we can rewrite the $\alpha = 1$ solution by using the limit definition of the exponential function: $$ y_{c_0, \alpha =1}(x) = c_0 \lim_{n \to \infty} (1 + \frac{x}{n})^n . $$ If I could show that $$ \lim_{\alpha \to 1} y_{c_0, \alpha \neq 1}(x) = y_{c_0, \alpha =1}(x) $$ then I will have shown that the solutions vary continuously in $\alpha$ around $1 \in \mathbb{R}$.
To this end, recall $y_{c_0, \alpha \neq 1}(x) = c_0(1+c_0^{\alpha-1} x)^{\frac{1}{1-\alpha}}$. As $\alpha \to 1 $, then $\frac{1}{1-\alpha} \to \infty$ so it seems like the behaviour of the power in the limit $\lim_{\alpha \to 1} y_{c_0, \alpha \neq 1}(x)$ behaves like the power in the limit definition of $e^x$. However, in the limit definition of $e^x$ there is a $\frac{1}{n}$ term whereas I'm not sure how to extract this from $y_{c_0, \alpha \neq 1}(x)$. Instead of $\frac1n$ we have $c_0^{\alpha - 1}$. It seems like instead of the behaviour $1/n \to 0$ as $n \to \infty$, we are getting $c_0^{\alpha - 1} \to 1$ as $\alpha \to 1$ which seems like not what we want?
Is this the right approach or am I missing something?
Also, for continuity in the other variables $x$ and $c_0$, is this just something like Picard-Lindelof?
Thanks.