Nonhomogeneous Differential Equations — Question 4

Question 4 – Solution

Strategy. Superposition combines the initial data as well as the forcing; use a homogeneous correction to achieve the desired data.

Step 1: Verify the supplied responses. We have $u''=(-\cos t+4\cos 2t)/3$, so $u''+u=\cos 2t$. Also $v'=2(1-\cos t)$ and $v''=2\sin t$, so $v''+v=2t$. At zero both values and both slopes vanish.

Step 2: Add the forcing and fit the state. The sum $u+v$ handles the combined forcing with zero initial data. Adding the homogeneous solution $\cos t-2\sin t$ gives $$\boxed{y=u+v+\cos t-2\sin t =\tfrac 43\cos t-\tfrac 13\cos 2t+2t-4\sin t.}$$ At zero the value is $4/3-1/3=1$ and the slope is $2-4=-2$. Linearity verifies the combined forcing, and regular uniqueness proves this is the initial-value solution.

Step 3: Track the full linear map. If $L[y_j]=g_j$ and the initial data are $(a_j,b_j)$, then $Ay_1+By_2$ has forcing $Ag_1+Bg_2$ and data $$(Aa_1+Ba_2,\ Ab_1+Bb_2).$$ This follows directly from linearity of $L$, evaluation and differentiation. Ignoring the data part can produce the right equation with the wrong initial state.

Step 4: Remove duplicated initial data. The naive sum $r+s$ has data $(2,0)$. Subtract one copy of the homogeneous solution $\cos t$: $$\boxed{r+s-\cos t=u+v+\cos t.}$$ The forcing is unchanged, while the data become $(1,0)$. This correction differs from Step 2 because the required initial slope here is zero, not $-2$.