Question

Let \( 0<\alpha<1 \) be a real number. The number of differentiable functions \( y : [0,1] \to [0,\infty) \), having continuous derivative on \([0,1]\) and satisfying \[ y'(t) = (y(t))^{\alpha}, \ t \in [0,1], \quad y(0) = 0, \] is

• A. exactly one.
• B. exactly two.
• C. finite but more than two.
• D. infinite.

Hint:

When the exponent \( \alpha \) in \( y' = y^\alpha \) satisfies \(0<\alpha<1\), the derivative vanishes at \(y=0\), making the solution non-unique. Such cases often yield infinitely many solutions.

Answer 1

The Correct Option is D

Step 1: Solve the differential equation.
We are given \[ y'(t) = (y(t))^{\alpha}, \quad 0<\alpha<1, \quad y(0)=0. \] Separating variables, \[ \frac{dy}{(y)^{\alpha}} = dt. \] Integrating both sides, \[ \frac{y^{1-\alpha}}{1-\alpha} = t + C. \] Applying \(y(0)=0\) gives \(C=0\). Hence, \[ y(t) = ((1-\alpha)t)^{\frac{1}{1-\alpha}}. \]
Step 2: Existence of multiple solutions.
Since \(0<\alpha<1\), we have \(y'(t)=0\) when \(y=0\). Therefore, a function defined as \[ y(t) = \begin{cases} 0, & 0 \le t \le t_0,
((1-\alpha)(t-t_0))^{\frac{1}{1-\alpha}}, & t>t_0, \end{cases} \] also satisfies the differential equation for any \(t_0 \in [0,1]\).
Step 3: Conclusion.
Because \(t_0\) can take infinitely many values, there are infinitely many such differentiable functions.