Fundamental Sets of Solutions — Question 4

Question 4 – Solution

Strategy. Check both requirements: independence and membership in the solution space of one regular equation on the stated interval.

Step 1: Prove independence. If $At^2+Bt^3=0$ for every real $t$, then $A+Bt=0$ for every nonzero $t$. Evaluating at two distinct nonzero points forces $A=B=0$.

Step 2: Reject the proposed regular equation. The nonzero function $u=t^2$ has $u(0)=u'(0)=0$. Uniqueness for continuous normalized coefficients would force it to equal the zero solution on $\mathbb R$. Equivalently, substitution at zero would require $u''(0)=2=0$. Thus these independent functions cannot be a fundamental set for any equation of the stated kind.

Step 3: Work on a valid interval. Substituting the two functions for $t>0$ gives $$2+2tp+t^2q=0,\qquad 6t+3t^2p+t^3q=0.$$ Subtracting $t$ times the first equation from the second yields $p=-4/t$, then $q=6/t^2$. Hence $$\boxed{y''-\frac 4t y'+\frac 6{t^2}y=0.}$$ Both residuals vanish, the coefficients are continuous on $(0,\infty)$, and $uv'-u'v=t^4>0$ there, so the pair is fundamental.

Step 4: Fit and interpret. The data give $A+B=0$, $2A+3B=1$, so $\boxed{y=t^3-t^2}$. Its polynomial extension has zero value and slope at zero, but the normalized equation is undefined there. A theorem requiring continuous coefficients on an interval containing zero cannot be applied to that extension.