Fundamental Sets of Solutions — Question 8

Question 8 – Solution

Strategy. Treat the two known solutions as simultaneous equations for the unknown coefficients, then check the interval where the result is regular.

Step 1: Reconstruct the equation. Substituting $u=t$ and $v=e^{-t}$ gives $$p+tq=0,\qquad 1-p+q=0.$$ For $t\ne-1$, these have the unique solution $$\boxed{p=\frac{t}{t+1},\qquad q=-\frac 1{t+1}.}$$ Thus both residuals vanish in $y''+[t/(t+1)]y'-y/(t+1)=0$ wherever these coefficients are defined.

Step 2: Find a fundamental interval. The largest connected open interval containing zero and avoiding the pole is $\boxed{(-1,\infty)}$. The initial-data determinant, or its value at any point of this interval, is $$uv'-u'v=-(t+1)e^{-t}\ne 0.$$ In particular it equals $-1$ at zero, so the two solutions form a fundamental set throughout the interval.

Step 3: Locate the obstruction. At $t=-1$, the coefficient equations would require $p-q=0$ and $1-p+q=0$, an immediate contradiction. Even finite pointwise coefficient values cannot satisfy both conditions there. Smoothness of the two functions does not guarantee a common regular normalized equation across that point.

Step 4: Solve and distinguish extensions. In $y=At+Be^{-t}$, the data give $B=2$ and $A-B=-1$, hence $$\boxed{y=t+2e^{-t}\quad(-1<t<\infty).}$$ The value and slope at zero are $2,-1$, and linearity verifies its equation. The formula is smooth for all real $t$, but the reconstructed normalized equation is undefined at $-1$. Extending a function does not remove that coefficient singularity. Multiplying by $t+1$ instead gives a different, undivided formulation with a vanishing leading coefficient there.