Reduction of Order — Question 8

Question 8 – Solution

Strategy. A constant seed reduces the equation directly to a first-order equation for the slope.

Step 1: Integrate the slope. Set $w=y'$. An integrating factor is $1+t^2$, so $$((1+t^2)w)'=0,\qquad w=B/(1+t^2),\qquad y=A+B\arctan t.$$

Step 2: Fit the two limits. The equations $A-B\pi/2=-1$ and $A+B\pi/2=3$ give $$\boxed{y=1+\frac 4\pi\arctan t.}$$ This is the unique choice of the two constants. Its derivatives $4/[\pi(1+t^2)]$ and $-8t/[\pi(1+t^2)^2]$ cancel in the original equation.

Step 3: Describe the trajectory. The value and slope at zero are $1$ and $4/\pi$. The slope is positive everywhere; continuity and the limiting values give range $(-1,3)$. Every member $A+B\arctan t$ is bounded, since $|\arctan t|<\pi/2$.

Step 4: Determine the tail rate. For $t>0$, $$3-y(t)=\frac 4\pi\arctan(1/t),\qquad \boxed{\lim_{t\to\infty}t[3-y(t)]=\frac 4\pi.}$$ Here $\arctan u/u\to 1$ as $u\to 0$. An exponential bound would force this positive limit to be at most $\lim Mt e^{-ct}=0$, a contradiction. The approach to the limit has an algebraic $1/t$ tail.