Reduction of Order — Question 9

Question 9 – Solution

Strategy. Solve on each side of the singular coefficient, then compare one-sided derivatives at the join.

Step 1: Reduce on the positive side. The seed has residual $t^2(6t)-6t^3=0$. In normalized form $p=0$, so $$v'=Ct^{-6},\quad v=D-\tfrac 15Ct^{-5},\quad \boxed{y=At^3+Bt^{-2}\quad(t>0).}$$ These two powers each satisfy the original equation by direct substitution.

Step 2: Exclude the divergent mode. If $B\ne 0$, then $y$ diverges at zero, so even a continuous extension is impossible. If $B=0$, the limits of $y,y',y''$ are $0,0,0$. These are the required values for a $C^2$ extension.

Step 3: Classify continuations. On $t<0$ the same calculation gives $a t^3+b t^{-2}$; continuity forces $b=0$. Thus all classical continuations of the right branch $At^3$ are $$\boxed{y(t)=\begin{cases}a t^3,&t<0,\\0,&t=0,\\At^3,&t>0,\end{cases} \qquad a\in\mathbb R.}$$ The first and second derivatives match at zero for every $a$. At zero the undivided residual is $0-6(0)=0$. Therefore these are sufficient as well as necessary, and infinitely many continuations share $y(0)=y'(0)=0$ even with the right branch fixed.

Step 4: Require one more derivative. The one-sided third derivatives are $6a$ and $6A$; a $C^3$ join occurs exactly when $a=A$. In that case the single polynomial $At^3$ holds on both sides. The leading coefficient is zero at the joining point, and the normalized coefficient $-6/t^2$ is undefined there, so the regular uniqueness theorem has no applicable interval containing zero.