Basic Concepts — Question 8

Question 8 – Solution

Strategy. Insert simple known solutions to recover coefficients, then inspect what changes when the leading coefficient vanishes.

Step 1: Recover the equation on its stated interval. Inserting $y_1=1$ gives $q(t)=0$. Inserting $y_2=t^2$ then gives $2+2tp(t)=0$, so $$\boxed{p(t)=-1/t,\quad q(t)=0},\qquad \boxed{y''-\frac 1t y'=0\quad(t>0)}.$$ The constant solution has both derivatives zero, and $2-(1/t)(2t)=0$ verifies the quadratic solution.

Step 2: Rule out a regular equation through zero. At $t=0$, the function $t^2$ has value 0, first derivative 0 and second derivative 2. In any normalized equation with finite coefficients at zero it would require $$2+p(0)\cdot 0+q(0)\cdot 0=0,$$ which is impossible. Thus no continuous-coefficient normalized equation on an interval containing zero can have both supplied functions as solutions. The obstruction is already visible from the quadratic alone.

Step 3: Retain the degenerate leading coefficient. The undivided equation is $ty''-y'=0$. For $y=C_1+C_2t^2$, its left-hand side is $t(2C_2)-2C_2t=0$, including at zero. The family has $y(0)=C_1$, $y'(0)=0$, so $$\boxed{b=0:\ y=a+C_2t^2\ (C_2\in\mathbb R)},\qquad \boxed{b\ne 0:\ \text{no member}}.$$ Indeed, the undivided equation itself at zero forces $y'(0)=0$, so nonzero $b$ permits no $C^2$ solution at all.

Step 4: Explain the shared tangent. The two plotted functions share value 0 and slope 0 at zero but have different curvatures. The coefficient of $y''$ vanishes there, so the equation does not determine the second derivative from those data. Its normalized coefficient $-1/t$ is not continuous at the initial point. The regular linear theorem therefore does not apply, and the multiple zero-data solutions do not contradict it.