How can I prove that the DE $y'=y^\alpha$ has infinitely many solutions? I need to show that the DE $y'=y^{\alpha}$, where $\alpha$ is a constant with $0<\alpha<1$, has infinitely many solutions passing through the point $(0,0)$.
Also I need four of such solutions.
Thank you for your help!
Klara
 A: Here how you can proceed. First note that $y=0$ is a solution. Next, you use integration to get
$$
\int_0^y\frac{dy}{y^{\alpha}}=t+C_1.
$$
Note that on the left hand side you have an improper integral which converges if $0<\alpha<1$ -- this is the reason the solutions are not unique at the point $(0,0)$. Assuming that $0<\alpha<1$ you get
$$
y=K(t-C)^{\frac{1}{1-\alpha}}
$$ 
also a solution to your equation. Here $K$ is a constant which depends on $\alpha$.
Here is a first example of a solution
$$
y_1(t)=\begin{cases}
0&t<0,\\
Kt^{\frac{1}{1-\alpha}}&t\geq 0.
\end{cases}
$$
The only thing you need to check that it has a continuous derivative at $t=0$, which can be checked by direct calculation.
Generalizing, any function of the form
$$
y_1(t)=\begin{cases}
0&t<C,\\
K(t-C)^{\frac{1}{1-\alpha}}&t\geq C.
\end{cases}
$$
is a solution (you need to check the derivative at $t=C$) and they all pass through $(0,0)$ if $C>0$.
A: You have
$$ y' = \frac{dy}{dx} = y^{\alpha}, \ y(0)=0, \ 0<\alpha<1$$
From this we can start 
\begin{align}
 \frac{dy}{dx} &= y^{\alpha}
\\ \Leftrightarrow \frac{1}{  y^{\alpha}} dy &= dx
\\ \Leftrightarrow \int \frac{1}{  y^{\alpha}} dy &= \int 1 dx
\end{align}
Since this really does smell like homework like Siminore said, you should try from this point by your own. Remember the integration constant and fit that with the initial condition $y(0)=0$.
Edit: Okay if this isn't HW I shall continue.
Taking the above we get:
\begin{align}
  \int \frac{1}{  y^{\alpha}} dy &= \int 1 dx
\\ \Leftrightarrow \frac{y^{1-\alpha}}{1-\alpha}+c &=x \,  \text{  as } \, 0<\alpha<1
\\ \Leftrightarrow y^{1-\alpha}&=(x-c)\cdot(1-\alpha)
\\ \Leftrightarrow y &= ((x-c)\cdot(1-\alpha))^{\frac{1}{1-\alpha}}
\end{align}
Now we also have to check whether our initial condition is fulfilled.
$$ y(0) =((0-c)\cdot(1-\alpha))^{\frac{1}{1-\alpha}}=0$$
only for $c=0$, so
$$ y(x) = (x\cdot(1-\alpha))^{\frac{1}{1-\alpha}} $$
is the unique solution of the above ODE.
Edit: It is not the unique solution, since $y(t)=0 \ \forall t$ also solves the problem.
