Nonhomogeneous Differential Equations — Question 2

Question 2 – Solution

Strategy. A fixed nonzero forcing is preserved by affine combinations, whose coefficients sum to one.

Step 1: Track the forcing. Each listed function has $y_j''+y_j=1$. Therefore $$L[ay_1+by_2+cy_3]=a+b+c.$$ The combination solves the same equation exactly when $\boxed{a+b+c=1}$.

Step 2: Describe the full set. Subtracting the particular solution $1$ leaves $h''+h=0$, so all solutions are $1+B\cos t+C\sin t$. The proposed combination expands to $$(a+b+c)+b\cos t+c\sin t.$$ It represents the specified solution exactly with $b=B$, $c=C$, $a=1-B-C$. Uniqueness follows from the independence of $1,\cos t,\sin t$: if their combination vanishes, adding its second derivative to itself first forces the constant coefficient to zero, then the two harmonic coefficients vanish.

Step 3: Compare operations. The sum of two solutions has forcing $2$, and their difference has forcing $0$, so neither operation stays in the set with forcing $1$. Their midpoint has forcing $(1+1)/2=1$ and does stay in the set. The zero function is absent. Thus this is an affine translate of a homogeneous solution space, rather than a vector space itself.

Step 4: Fit the data. In $1+B\cos t+C\sin t$, the data give $B=-1$, $C=2$. Hence $$\boxed{(a,b,c)=(0,-1,2),\qquad y=1-\cos t+2\sin t.}$$ The coefficients sum to one. Directly, $y''=\cos t-2\sin t$, so $y''+y=1$; the initial value and slope are $0,2$.