Is there a differentiable function on $[0,\infty)$ that satisfies $y'=y^n$ and $y(0)>0$? Let $n$ be an integer greater than $1$. Is there a differentiable function on $[0,\infty)$ that satisfies $y'=y^n$ and $y(0)>0$?
My attempt: We solve the differential equation noting that
$\frac{dy}{dt}=y^n$
$\int\frac{1}{y^n}=\int dt$
So we get that $y=((1-n)(t+c))^{\frac{1}{1-n}}$ for some constant $c$.
However, I notice that since $n\geq2$, we get that 
$y(0)=((1-n)c)^{\frac{1}{1-n}}$, so $c$ must be negative in order to actually evaluate this term. In this case, my answer is yes, there exists a function, but I'm not sure this is correct.
Any help appreciated!

This is the main point: whatever positive number is picked for $y(0),$ the solution of the ODE $y' = y^n$ blows up in finite time. 
If $n=2$ and $y(0) = \frac{1}{W} $ with constant $W > 0,$ then
$$ y(x) = \frac{1}{W - x}  $$
for $x < W.$ The solution does not extend to $+\infty$ as requested.
If $n=3$ and $y(0) = \frac{1}{\sqrt {2W}} $ with constant $W > 0,$ then
$$ y(x) = \frac{1}{\sqrt{2W - 2x}}  $$
for $x < W.$ The solution does not extend to $+\infty$ as requested.
