Basic Concepts — Question 5

Question 5 – Solution

Strategy. Locate coefficient obstructions before applying the theorem, then compare its guaranteed interval with the behavior of actual solutions.

Step 1: Separate domain from degeneracy. The logarithm requires $t>-3$ and the forcing excludes $t=4$. Thus the original coefficient domain is $(-3,4)\cup(4,\infty)$. Normalization additionally excludes $t=0,2$, where the leading coefficient vanishes. For other allowed $t$, $$p=\frac{t+1}{t(t-2)},\qquad q=\frac{\ln(t+3)}{t(t-2)},\qquad g=\frac 1{(t-4)t(t-2)}.$$

Step 2: Choose the connected interval containing the data. The regular intervals are $(-3,0)$, $(0,2)$, $(2,4)$ and $(4,\infty)$. Therefore the largest applicable intervals are $$\boxed{t_0=1:\ (0,2)},\qquad \boxed{t_0=3:\ (2,4)}.$$ The initial values do not change these coefficient-based guarantees.

Step 3: State the limitation accurately. The theorem guarantees existence and uniqueness within the selected interval. It does not classify the behavior at a singular endpoint or prove that every solution becomes unbounded there. Whether a particular solution extends requires examination of that solution and the original equation. A coefficient-domain boundary is also distinct from a pole in the solution itself.

Step 4: Solve the comparison equation completely. Since $(ty')'=ty''+y'$, integration on $t>0$ gives $ty'=C$, then $y=C\ln t+D$. The data at 1 imply $C=b$, $D=a$, so $$\boxed{y=a+b\ln t\quad(t>0)}.$$ If $b\ne 0$, $y$ is unbounded as $t\downarrow 0$, so no continuous, hence no $C^2$, extension exists. If $b=0$, the constant solution $y=a$ satisfies the undivided equation on all of $\mathbb R$.

Thus extension occurs exactly when $\boxed{b=0}$. In fact it is unique among $C^2$ extensions: on the negative side every solution has form $C_-\ln|t|+D_-$; continuity forces $C_-=0$, $D_-=a$. The normalized theorem alone did not provide that endpoint conclusion.