Intervals of Validity — Question 8

Question 8 – Solution

Strategy. The original equation remains meaningful at the vanishing leading coefficient. Solve on the two half-lines and test the join in that original equation.

Step 1: Solve where division is allowed. For $x\ne 0$, the equation gives $(y/x^2)'=0$. The initial condition therefore forces $y=x^2$ throughout $x>0$. On $x<0$, the general solution is $y=Cx^2$, with an independent real constant $C$.

The divided equation $y'=2y/x$ has no meaning at $0$, but the original equation does: it requires $y(0)=0$. A singularity introduced by division is not automatically an obstruction for the original problem.

Step 2: List all global solutions. Every global candidate must have the form $$\boxed{y_C(x)= \begin{cases} Cx^2,&x<0,\\ 0,&x=0,\\ x^2,&x>0, \end{cases}\qquad C\in\mathbb R.}$$ The half-line solutions and the equation at $0$ leave no other possible global classical solutions.

Step 3: Verify the join. For every finite $C$, $y_C$ is continuous at $0$. The difference quotient there is $Ch$ for $h<0$ and $h$ for $h>0$, so $y_C'(0)=0$. Moreover, $$y_C'(x)=2Cx\ (x<0),\qquad y_C'(x)=2x\ (x>0)$$ tends to $0$ from both sides. Thus $y_C$ is continuously differentiable. The identity $xy_C'=2y_C$ holds on each half-line and reduces to $0=0$ at $0$. Also $y_C(1)=1$.

Step 4: Resolve maximality and uniqueness. Each $y_C$ has maximal interval $\boxed{\mathbb R}$. The unique positive-half-line solution extends through $0$ in infinitely many ways because the original equation loses its derivative term there. The usual normal-form uniqueness theorem cannot cross that degeneracy.

If the problem had instead specified $y'=2y/x$ only for $x\ne 0$, its IVP interval containing $1$ would be $(0,\infty)$, even for the polynomial formula. Which equation and domain were actually given matters.