Nonhomogeneous Differential Equations — Question 6

Question 6 – Solution

Strategy. Verify the supplied growing response and use exact phase samples to rule out cancellation by bounded homogeneous terms.

Step 1: Verify and complete. Differentiation gives $$p'=\tfrac 12\sin t+\tfrac 12t\cos t,\qquad p''=\cos t-\tfrac 12t\sin t.$$ Thus $p''+p=\cos t$ and the general solution is $$\boxed{y=A\cos t+B\sin t+\tfrac 12t\sin t.}$$

Step 2: Fit the data. Since $p(0)=p'(0)=0$, the data give $A=a$, $B=b$. In particular the zero-data response is exactly $\boxed{y_0=\tfrac 12t\sin t}$.

Step 3: Prove both unbounded excursions. At $t_n=\pi/2+2\pi n$, $n=0,1,\ldots$, we have $y(t_n)=b+t_n/2\to\infty$. At $s_n=3\pi/2+2\pi n$, we have $y(s_n)=-b-s_n/2\to-\infty$. These values do not depend on $a$. Thus no constant homogeneous correction removes the unbounded behavior, even though the forcing is bounded.

Step 4: Distinguish bounds from extrema. On $t\ge 0$, $$\boxed{-t/2\le y_0(t)\le t/2.}$$ For $t>0$, the upper envelope is attained exactly at $t_n$ and the lower at $s_n$; both bounds also coincide with $y_0$ at zero. At a positive upper contact $y_0'=1/2$, and at a lower contact $y_0'=-1/2$. These slopes match those of the respective envelopes and are not zero, so the contacts are not stationary points of $y_0$.