Intervals of Validity — Question 4

Question 4 – Solution

Strategy. Track the forbidden value $y=0$ and the slope, rather than checking only whether the solution stays bounded.

Step 1: Select the branch. Multiplying by $y$ and integrating gives $$yy'=-x,\qquad x^2+y^2=4.$$ Continuity and $y(0)=2$ select $$\boxed{y=\sqrt{4-x^2}}.$$ Its positive branch has $y'=-x/\sqrt{4-x^2}$ and directly satisfies the IVP for $-2<x<2$.

Step 2: Test both endpoints. Thus $\boxed{I=(-2,2)}$, and $$\begin{array}{c|cc} &x\downarrow-2&x\uparrow 2\\ \hline y(x)&0^+&0^+\\ y'(x)&+\infty&-\infty \end{array}$$ At either endpoint the original right-hand side is undefined, and a continuously differentiable extension would also require a finite derivative.

Step 3: Exclude branch switching. The negative square root fails the initial condition. Joining the two semicircles at an endpoint still passes through $y=0$, and the circle has no real points with $|x|>2$. Following the circle as a parametrized curve is not an extension as a differentiable graph $y(x)$ on a larger interval.

Step 4: Identify the obstruction. Although $0<y\le 2$, the graph approaches the boundary of the equation’s domain. Boundedness alone does not keep it inside a region where the differential equation is defined and regular. The open circles indicate excluded endpoints, not additional solution values.