Reduction of Order — Question 2

Question 2 – Solution

Strategy. Normalize the equation before reduction, then use exact derivatives to study the singular endpoint.

Step 1: Find the missing solution. For $t^2$, the residual is $(2-6+4)t^2=0$. With $y=t^2v$, the divided equation reduces to $$v''+\frac 1t v'=0,\qquad v'=D/t,\qquad \boxed{y=t^2(C+D\ln t).}$$

Step 2: Impose the data. At $1$, $y=C$ and $y'=2C+D$. Thus $C=D=1$ and $$\boxed{y=t^2(1+\ln t)},\qquad y'=t(3+2\ln t),\quad y''=5+2\ln t.$$ These give $y(1)=1$ and $y'(1)=3$.

Step 3: Locate the features. The sole zero is $t=e^{-1}$. The derivative changes from negative to positive at $t=e^{-3/2}$, giving the unique global minimum $$(t,y)=(e^{-3/2},-\tfrac 12 e^{-3}).$$

Step 4: Test endpoint regularity. As $t\downarrow 0$, both $y$ and $y'$ tend to zero, but $y''\to-\infty$. Setting $y=0$ for $t\le 0$ gives a $C^1$ extension (the derivative at zero is also $\lim y(t)/t=0$). No $C^2$ extension is possible. The leading coefficient vanishes at zero, so the regular normalized theorem does not apply there.