Separable Equations — Question 1

Question 1 – Solution

Strategy. Separate using the everywhere-positive factor $1+y^2$, then respect the range of the inverse tangent.

Step 1: Integrate and select the branch. $$\frac{dy}{1+y^2}=2x\,dx,\qquad \arctan y=x^2+C,\qquad C=\frac\pi 4.$$ Thus $\boxed{y(x)=\tan(x^2+\pi/4)}$ near $0$. No constant solution was lost, since $1+y^2$ never vanishes for real $y$.

Step 2: Identify the interval and shape. For the branch through $0$, we need $-\pi/2<x^2+\pi/4<\pi/2$. Hence $$\boxed{I=(-\sqrt{\pi}/2,\sqrt{\pi}/2)}.$$ The derivative has the sign of $x$: the solution decreases to its minimum $y(0)=1$ and then increases. At either endpoint approached from inside $I$, the angle tends to $\pi/2$ from below, so $y\to+\infty$.

Step 3: Verify and explain the obstruction. Differentiation gives $y'=2x\sec^2(x^2+\pi/4)=2x(1+y^2)$, and $y(0)=1$. A real differentiable extension through an endpoint would have a finite, continuous value there, contradicting blow-up. Other tangent branches lie on separate intervals; periodicity does not connect them across a pole.