Solutions to that $\frac{du}{dt}=u^{a}$ Show that $\frac{du}{dt}$=$u^{a}$ where $u(0)=0$ has two soultions for $0<a<1$ and one solution for $a=0,1$
I have tried integrating this by saying 
$\int\frac{du}{u^{a}}$=$\int{1}dt$
which gave me the horrible solution of 
$$\frac{u(t)}{(u(t))^{a}}=(t+c)(1-a)$$ and assuming that is correct how do I proceed to show that the solutions depend on $a$?
 A: The first step in solving the problem is to note that $u(t)=0$ is always a solution of the ODE Cauchy problem under analysis, i.e. of
$$
\begin{cases}
\dfrac{\mathrm{d}u(t)}{\mathrm{d}t}=u^a(t) \\
\\
u(0)\equiv u_0=0
\end{cases}\quad0\le a\le 1.\tag{1}\label{1}
$$
The second step is trying to find another solution of the problem \eqref{1} in the "critical exponent" case, i.e. $0< a<1$. As the already noted by the OP, the best way to do so is perhaps to use the standard Barrow's formula ([1], §1.5 p. 19), 
$$
t-t_0=\int\limits_{u_0}^{u(t)}\frac{\mathrm{d}\xi}{v(\xi)}\tag{2a}\label{2a}
$$
which, for our problem \eqref{1}, takes the form:
$$
t=\int\limits_{0}^{u(t)}\frac{\mathrm{d}\xi}{\xi^a}\tag{2b}\label{2b}
$$
Now, it's easy to see that the integral at the right side of \eqref{2b} is not defined for $a=1$ (and thus in this case it does not represent a solution for \eqref{1}), while for $a\neq 1$ we have
$$
 u(t)=
\begin{cases}
0& a=0\\
\left(\dfrac{t}{1-a}\right)^\frac{1}{1-a} &0<a<1
\end{cases}\tag{3}\label{3}
$$
Now we have reached our goal: in step 1 and step 2 we have shown that \eqref{1} has one solution $u(t)=0$ if $a=0,1$, and has at least two solutions, given by the following formula
$$
 u(t)=
\begin{cases}
0\\
\left(\dfrac{t}{1-a}\right)^\frac{1}{1-a}
\end{cases}\text{ for } 0<a<1
$$ 
Final note: for $a=0,1$ it can be proved that the shown solutions are de facto unique. The related existence and uniqueness theorem ([1], §2.2 p. 36), says that if $v:\mathbb{R}\to\mathbb{R}$ is continuously differentiable, then the solution of the given Cauchy problem exists, is unique and is given by formula \eqref{2a} if $v(u_0)\neq 0$, or by
$$
\quad u(t)=u_0=\mathrm{const.}\:\text{ if }v(u_0)=0\tag{4}\label{4}
$$
(solutions of type $\eqref{4}$ are commonly called equilibrium points). Thus, in the class of Cauchy problems to which \eqref{1} belongs, a general method to determine the non uniqueness of the solution is to check if $v(u)\notin C^1$ and, in case of affirmative answer, see if there are equilibrium points and other, non constant solutions, that have the same  equilibrium point as initial value.
[1] Vladimir Igorevic Arnol'd, "Ordinary differential equations", various editions from MIT Press and from Springer-Verlag, MR1162307 Zbl 0744.34001.
