Separable Equations — Question 9

Question 9 – Solution

Strategy. Separate from the left endpoint, then check the denominator on the whole prescribed interval before accepting the right endpoint.

Step 1: Obtain the candidate. For a nonzero solution, $$-\frac 1y=x^2+C,\qquad \frac 1y=\frac 1A+1-x^2, \qquad \boxed{y=\frac{A}{1+A(1-x^2)}}.$$ For $A=0$, $y\equiv 0$ is a solution. No solution can leave or cross zero: the right-hand side $2xy^2$ has a continuous $y$-derivative, $4xy$, so the usual local uniqueness theorem applies at every finite point. Consequently the separated formula applies to any solution with $A\ne 0$ throughout its interval.

Step 2: Test the entire interval. On $[-1,1]$, the quantity $1-x^2$ ranges over $[0,1]$. If $A\ge 0$, the denominator is at least $1$. If $-1<A<0$, it is at least $1+A>0$. For $A=-1$ it vanishes at $x=0$; for $A<-1$ it vanishes at $$x=\pm\sqrt{1+1/A}\in(-1,1).$$ The nonzero numerator cannot cancel these zeros. Any admissible formula gives $y(1)=A$, so the necessary conditions are $$\boxed{B=A\quad\text{and}\quad A>-1}.$$ They are also sufficient: the displayed formula, including $A=0$, is smooth on a neighborhood of $[-1,1]$. Its derivative is $2xA^2/[1+A(1-x^2)]^2=2xy^2$, and it gives both specified endpoint values.

Step 3: Expose the false endpoint test. For $A=B=-2$, the candidate is $y=-2/(2x^2-1)$. It has the desired values at $x=\pm 1$ but poles at $x=\pm 1/\sqrt 2$. Hence it is not a solution on the entire interval.

Step 4: Give the center value and uniqueness. For every admissible pair, $\boxed{y(0)=A/(1+A)}$. The initial value at $-1$ already fixes the candidate uniquely; for $A=0$, local uniqueness also forbids any departure from zero. The second endpoint is a compatibility condition, not an extra free integration constant.