Intervals of Validity — Question 2

Question 2 – Solution

Strategy. Keep the domain of the given equation separate from the domain of a simplified formula or a multiplied equation.

Step 1: Locate the coefficient interval. Both coefficients are continuous on $(-3,3)$, the largest coefficient interval containing $0$. The linear theorem guarantees a unique solution throughout that interval for every real $a$.

Step 2: Integrate and verify. On this interval, multiplication and the product rule yield $$((x^2-9)y)'=1,\qquad (x^2-9)y=x-9a.$$ Thus $$\boxed{y(x)=\frac{x-9a}{x^2-9}}.$$ The value at $0$ is $a$, and differentiating $(x^2-9)y=x-9a$ verifies the differential equation whenever $x\ne\pm 3$.

Step 3: Examine cancellation. At $x=3$, the numerator vanishes exactly when $a=1/3$. Then $$y=\frac 1{x+3}\quad(x\ne\pm 3),\qquad \lim_{x\to 3}y=\frac 16.$$ At $x=-3$, cancellation occurs exactly when $a=-1/3$. Then $$y=\frac 1{x-3}\quad(x\ne\pm 3),\qquad \lim_{x\to-3}y=-\frac 16.$$ No value of $a$ cancels both factors. For other parameter values the corresponding endpoint is a pole.

Step 4: Use the original domain. For every $a$, the answer remains $\boxed{I=(-3,3)}$. Even a finite removable limit does not define the original coefficients at an excluded point.

The multiplied equation $(x^2-9)y'+2xy=1$ is defined at $x=\pm 3$ and imposes algebraic conditions there. It is equivalent to the original equation only away from those points. A canceled formula can extend through one endpoint as a solution of the multiplied equation, but that is an extension for a different domain of definition.