The correct answer is:
(C) finite but more than two.
Explanation:
To solve this differential equation, we can use separation of variables.
Given: $y'(t) = (y(t))^{\alpha}$
Separating variables:
$\frac{dy}{(y(t))^{\alpha}} = dt$
Integrating both sides:
$\int \frac{1}{(y(t))^{\alpha}} dy = \int dt$
Now, let's integrate the left-hand side:
$\int (y(t))^{\alpha-1} dy = \frac{1}{\alpha} (y(t))^{\alpha} + C_1$
where $C_1$ is the constant of integration.
Now, the right-hand side of the equation is simply:
$\int dt = t + C_2$
where $C_2$ is another constant of integration.
Putting it all together:
$\frac{1}{\alpha} (y(t))^{\alpha} + C_1 = t + C_2$
Since the function $y(t)$ is positive for all $t \in [0, 1]$ (given that $y : [0, 1] \rightarrow [0, \infty)$), we can ignore the constant of integration $C_1$ and rewrite the equation as:
$\frac{1}{\alpha} (y(t))^{\alpha} = t + C$
where $C$ is a constant.
Now, solve for $y(t)$:
$y(t) = \left(\frac{\alpha}{t + C}\right)^{\frac{1}{\alpha}}$
Since $y(0) = 0$, we have:
$0 = \left(\frac{\alpha}{C}\right)^{\frac{1}{\alpha}}$
This implies that $C = \alpha$.
So, the function becomes:
$y(t) = \left(\frac{\alpha}{t + \alpha}\right)^{\frac{1}{\alpha}}$
Now, let's check the conditions given in the problem:
$0 < \alpha < 1$ (given)
$y(t)$ is differentiable for $t \in [0, 1]$ (satisfied by the function)
$y'(t) = (y(t))^{\alpha}$ (satisfied by the function)
$y(0) = 0$ (satisfied by the function)
Therefore, there is a unique solution for $y(t)$, which is:
$y(t) = \left(\frac{\alpha}{t + \alpha}\right)^{\frac{1}{\alpha}}$
Hence, the number of differentiable functions satisfying the given conditions is exactly one.
Note: This solution is AI-generated but verified by an expert.