Basic Concepts — Question 6

Question 6 – Solution

Strategy. Apply the linear operator to the proposed combination and track the forcing as carefully as the unknown function.

Step 1: Identify the forced and zero-forcing pieces. Direct differentiation yields $$L[t]=t,\qquad L[\sin t]=0,\qquad L[\cos t]=0.$$ Linearity then gives $L[at+b\sin t+c\cos t]=at$. Equality with $t$ for every real $t$ requires $a=1$; $b,c$ are arbitrary. Hence all such candidates are $$\boxed{y=t+b\sin t+c\cos t}.$$ The forcing fixes the coefficient of $t$, leaving the two homogeneous freedoms.

Step 2: Track sums, differences and averages. Because $L[u]=L[v]=t$, $$L[u+v]=2t,\qquad L[u-v]=0,\qquad L[(u+v)/2]=t.$$ The sum is generally not a solution of the original forced equation. The difference solves the associated homogeneous equation, and the average preserves the original forcing.

Step 3: Determine the allowed weights. For any such $u,v$, $$L[\alpha u+\beta v]=(\alpha+\beta)t.$$ This equals $t$ on $\mathbb R$ exactly when $\boxed{\alpha+\beta=1}$. The weights need not be nonnegative. For homogeneous solutions, by contrast, the right-hand side remains zero for every pair of weights. Also, the zero function is not a solution of $L[y]=t$; the forced solution set does not have all the closure properties of the homogeneous one.

Step 4: Fit and verify the initial data. For the candidate family, $y(0)=c$ and $y'(0)=1+b$. The data give $c=2$, $b=-1$, so $$\boxed{y=t-\sin t+2\cos t}.$$ Its derivative is $1-\cos t-2\sin t$, its initial value and slope are 2 and 0, and $y''+y=t$. Since $p=0$, $q=1$, $g=t$ are continuous on $\mathbb R$, the initial-value theorem proves uniqueness there. No general method for finding particular solutions was assumed; the necessary candidate was supplied.