Question

Which one of the following options is true for the initial value problem: \(\frac{dy}{dx}=2y^{1/3}\), \(y(0)=0\)?

• A. It has no solution.
• B. It has exactly one solution.
• C. It has more than one, but finite number of solutions.
• D. It has infinitely many solutions.

Hint:

Check whether \(\partial f/\partial y\) is bounded near \(y=0\); if not, uniqueness can fail even though a solution exists.

Answer 1

The Correct Option is D

Step 1: Here \(f(y)=2y^{1/3}\) is continuous everywhere, but \(\frac{\partial f}{\partial y}=\frac{2}{3}y^{-2/3}\) is unbounded as \(y\to 0\). So the Lipschitz condition fails at the initial point \((0,0)\), and Picard's uniqueness theorem does not apply.

Step 2: The trivial function \(y(x)=0\) satisfies both the ODE and \(y(0)=0\).

Step 3: Separating variables for a nontrivial solution: \(y^{-1/3}dy=2\,dx\), so \(\frac{3}{2}y^{2/3}=2x+C\). Using \(y(0)=0\) gives \(C=0\), so \(y(x)=\left(\frac{4x}{3}\right)^{3/2}\) for \(x\ge 0\).

Step 4: For any constant \(a\ge 0\), the spliced function \[y_a(x)=\begin{cases}0, & x\le a\\ \left(\frac{4(x-a)}{3}\right)^{3/2}, & x> a\end{cases}\] is differentiable everywhere (the slope is \(0\) on both sides at \(x=a\)) and satisfies the same IVP.

Step 5: Since \(a\) can be any nonnegative real number, this produces infinitely many distinct solutions.

\[\boxed{\text{Infinitely many solutions}}\]