Fundamental Sets of Solutions — Question 2

Question 2 – Solution

Strategy. Track the initial-data vectors as the two proposed basis elements approach dependence.

Step 1: Test independence. Each exponential solves $y''-y=0$, so both combinations do. The data vectors at zero are $(2,0)$ and $(1+\alpha,1-\alpha)$, with determinant $2(1-\alpha)$. The pair is fundamental exactly when $\alpha\ne 1$. At $1$, both functions equal $u$.

Step 2: Fit the data. In $y=Au+Bv_\alpha$, the data require $$2A+(1+\alpha)B=0,\qquad (1-\alpha)B=1.$$ Thus $$\boxed{A=-\frac{1+\alpha}{2(1-\alpha)},\qquad B=\frac 1{1-\alpha}.}$$

Step 3: Simplify before taking a limit. Substitution cancels the large terms exactly: $$Au+Bv_\alpha=\tfrac 12(e^t-e^{-t})=\sinh t.$$ Both coefficients grow without bound in magnitude near $\alpha=1$, while the function is independent of $\alpha$. Large coordinates in a nearly dependent pair do not imply large values of the function on a fixed interval.

Step 4: Replace the collapsing direction. Define, for $\alpha\ne 1$, $$z_\alpha=\frac{v_\alpha-(1+\alpha)u/2}{1-\alpha}=\sinh t.$$ This extends by $z_1=\sinh t$ and has data $(0,1)$. The pair $u,z_1$ remains fundamental. The original pair at $\alpha=1$ contains only one effective direction and cannot supply the requested nonzero slope.