Nonhomogeneous Differential Equations — Question 7

Question 7 – Solution

Strategy. The forcing can impose endpoint behavior that no homogeneous correction can cancel.

Step 1: Verify and complete. Since $p'=\ln t$ and $p''=1/t$, it is a particular solution. Adding all affine homogeneous functions gives $$\boxed{y=t\ln t-t+At+B\qquad(t>0).}$$

Step 2: Fit and analyze. The slope at $1$ gives $A=0$, and the value then gives $B=1$. Thus $$\boxed{y=t\ln t-t+1,\qquad y'=\ln t,\quad y''=1/t>0.}$$ The function decreases on $(0,1)$, increases on $(1,\infty)$, and is strictly convex. Its unique global minimum is $y(1)=0$.

Step 3: Test extension of the selected function. Since $t\ln t\to 0$ as $t\downarrow 0$, we have $y(t)\to 1$. Setting $y(0)=1$ and, for example, $y(t)=1$ for $t<0$ gives a continuous extension. A finite derivative at zero is impossible because $$\frac{y(t)-1}{t}=\ln t-1\longrightarrow-\infty.$$ Hence there is no $C^1$ extension and therefore no $C^2$ extension. The limiting point in the graph is excluded from the original domain.

Step 4: Test every correction. For general $A,B$, continuity would require $y(0)=B$, but the difference quotient is $\ln t-1+A$, again tending to $-\infty$. No finite homogeneous constants cure this. Although $1,t$ extend smoothly, the forcing $1/t$ is undefined at zero. The maximal regular interval containing the initial point is $(0,\infty)$, not all of $\mathbb R$.