Reduction of Order — Question 7

Question 7 – Solution

Strategy. Divide the solution family by its positive seed so that zeros become level crossings of a monotone function.

Step 1: Construct the companion. The seed residual is $2(1+t^2)-2(1+t^2)=0$. Since $p=0$, $$I(t)=\int_0^t\frac{ds}{(1+s^2)^2} =\frac{t}{2(1+t^2)}+\frac 12\arctan t.$$ The integral follows by setting $s=\tan u$ and integrating $\cos^2u$. Hence $$\boxed{y_2=\tfrac 12t+\tfrac 12(1+t^2)\arctan t.}$$

Step 2: Verify. We obtain $y_2'=1+t\arctan t$ and $y_2''=\arctan t+t/(1+t^2)=2y_2/(1+t^2)$. The equation holds and the data are $0,1$.

Step 3: Determine the ratio range. Since $I'=1/(1+t^2)^2>0$, the ratio is strictly increasing. Its limits are $-\pi/4$ and $\pi/4$; neither is attained. Its range is exactly $(-\pi/4,\pi/4)$ by continuity.

Step 4: Classify all parameters. The sign of $y_C$ is the sign of $C-I(t)$. If $|C|<\pi/4$, there is exactly one finite zero, with positive values before it and negative values after it. If $C\ge\pi/4$, the solution is strictly positive everywhere; if $C\le-\pi/4$, it is strictly negative everywhere. Equality in these last two cases does not create a finite zero because the limiting ratio levels are excluded.