Intervals of Validity — Question 5

Question 5 – Solution

Strategy. Regard $x$ as a function of $y$. Its turning points identify where the selected inverse branch stops being a regular graph.

Step 1: Integrate without solving the cubic. The chain rule gives $$\frac{d}{dx}(y^3-y)=(3y^2-1)y'=1, \qquad \boxed{x=y^3-y}.$$ The initial value sets the integration constant to zero. Put $h(y)=y^3-y$. On $-1/\sqrt 3<y<1/\sqrt 3$, $h'(y)<0$, so $h$ is strictly decreasing and has a differentiable inverse there.

Step 2: Map the branch endpoints. Let $c=2/(3\sqrt 3)$. Then $h(1/\sqrt 3)=-c$ and $h(-1/\sqrt 3)=c$. Hence the inverse through $(0,0)$ exists exactly on $$\boxed{I=(-c,c)=\left(-\frac{2}{3\sqrt 3},\frac{2}{3\sqrt 3}\right)}.$$ Differentiating its defining relation recovers $y'=1/(3y^2-1)$ and verifies the original equation on this branch.

Step 3: Analyze the folds. As $x\downarrow-c$, $y\to 1/\sqrt 3$; as $x\uparrow c$, $y\to-1/\sqrt 3$. In both cases $3y^2-1\to 0^-$, so $y'\to-\infty$. Both finite limiting values are excluded by the original equation.

Step 4: Reject a change of root. The dashed pieces show other parts of the algebraic curve. A continuous switch at a fold must include an excluded value of $y$ with unbounded slope; a switch to a different root at that $x$ would be discontinuous. Neither gives a classical extension. The cubic relation alone does not specify the IVP’s interval or branch.