Nonhomogeneous Differential Equations — Question 5

Question 5 – Solution

Strategy. The difference from a particular solution follows the homogeneous equation, whose growing mode determines boundedness.

Step 1: Verify and fit. Since $p''=\cos t$, $p''-p=2\cos t$. The general solution is $-\cos t+Ae^t+Be^{-t}$. The data give $A+B=a+1$, $A-B=b$, hence $$\boxed{y=-\cos t+\frac{a+1+b}{2}e^t+\frac{a+1-b}{2}e^{-t}.}$$

Step 2: Eliminate the growing mode. The bounded oscillatory and decaying terms cannot cancel a nonzero $Ae^t$ as $t\to\infty$. Thus forward boundedness holds exactly when $$\boxed{b=-a-1.}$$ In that case $y=-\cos t+(a+1)e^{-t}$, so $y-p\to 0$. This convergence is to a time-varying particular solution, not necessarily to a constant.

Step 3: Require both time directions. Boundedness as $t\to-\infty$ also forces $B=0$. Thus the unique globally bounded solution is $\boxed{y=-\cos t}$, with data $(-1,0)$. It has no limit at positive infinity: its values at $2\pi n$ and $(2n+1)\pi$ are $-1$ and $1$. The forward-bounded family also has these two limiting subsequences because its decaying correction tends to zero.

Step 4: Compare the selected data. Data $(0,-1)$ give $y_s=-\cos t+e^{-t}$, which is bounded forward. Data $(0,0)$ give $y_u=-\cos t+\cosh t$, which is unbounded forward. Both have the same bounded forcing; the second retains a growing homogeneous component. The finite plot illustrates the distinction; the mode argument proves the global claim.