Basic Concepts for nth Order Linear Equations — Question 2

Question 2 – Solution

Strategy. Apply the linear operator before deciding which combinations preserve the nonzero forcing.

Step 1: Track the right-hand side. Since $p^{(4)}=24x$ and fourth derivatives of cubics vanish, $L[p]=L[u]=L[v]=24x$. Linearity gives $$L[u+v]=48x,\qquad L[u-v]=0,\qquad L[(u+v)/2]=24x.$$ Thus the sum solves neither of the two stated equations, the difference solves the homogeneous equation, and the average solves the original one.

Step 2: Characterize every admissible combination. The proposed combination has image $24x(\alpha+\beta+\gamma)$. Because $24x$ is not identically zero on $\mathbb R$, $$\boxed{\alpha u+\beta v+\gamma p\text{ solves the equation} \iff\alpha+\beta+\gamma=1.}$$ The forcing’s isolated zero at $0$ does not remove this identity requirement.

Step 3: Describe the complete solution set. If $y$ is any solution, $(y-p)^{(4)}=0$. Four integrations show that $y-p$ is a polynomial of degree at most three. Conversely, adding any such polynomial to $p$ preserves the equation. Hence $$\boxed{y=p+c_0+c_1x+c_2x^2+c_3x^3.}$$ This is a translate of a four-dimensional vector space, hence affine. It is not itself a vector space: the zero function does not satisfy the nonzero forcing, and sums generally double the forcing.

Step 4: Fit and check the initial state. The data give $c_0=1$, $c_1=-2$, $2c_2=6$, $6c_3=0$, so $$\boxed{y=\frac{x^5}{5}+1-2x+3x^2.}$$ The added cubic has fourth derivative zero. Its value and first three derivatives at $0$ are $1,-2,6,0$; $p$ contributes zero to all four. The coefficients are uniquely fixed.