Intervals of Validity — Question 10

Question 10 – Solution

Strategy. For a quadratic-growth equation, the reciprocal solution is a quadratic polynomial. Put its zeros at the desired endpoints.

Step 1: Determine the reciprocal polynomial. While $y\ne 0$, $$u'=-(Ax+B),\qquad u(0)=1,\qquad u=1-Bx-\frac A2x^2.$$ Blow-up at both prescribed endpoints requires $u(-2)=u(3)=0$. A quadratic with these roots and value $1$ at $0$ must be $$u=\frac{(x+2)(3-x)}6=1+\frac x6-\frac{x^2}6.$$ Thus $\boxed{A=1/3,\ B=-1/6}$ and $$\boxed{y=\frac 6{(x+2)(3-x)},\qquad I=(-2,3)}.$$

Step 2: Check the interval and uniqueness of the design. Both denominator factors are positive on $I$. At either endpoint the denominator tends to $0^+$, so $y\to+\infty$ and no finite continuous extension exists.

Also $y'=-u'/u^2=((2x-1)/6)y^2$ and $y(0)=1$. For any proposed affine pair, a solution through $1$ cannot cross the zero solution, by local uniqueness for the smooth right-hand side. Hence its reciprocal on the proposed interval has exactly the polynomial form above. The two endpoint zeros and $u(0)=1$ determine it uniquely, proving uniqueness of the coefficient pair.

Step 3: Locate the minimum. The denominator is $25/4-(x-1/2)^2$, so it is largest at $x=1/2$. Therefore $\boxed{y_{\min}=24/25}$ there. Equivalently, $y'$ changes from negative to positive at $1/2$.

Step 4: Interpret the construction. The coefficients and the right-hand side are smooth everywhere, but the solution grows without bound in both finite endpoint directions. Smoothness ensures local solvability and uniqueness, not global existence for this nonlinear equation.