Reduction of Order — Question 6

Question 6 – Solution

Strategy. The coefficient in the standard formula is the coefficient after division by the leading term.

Step 1: Normalize. For $e^t$, the residual is $[t-(t+1)+1]e^t=0$. Division gives $p=-1-1/t$, so $$w'+(1-1/t)w=0,\qquad w=Cte^{-t}.$$

Step 2: Integrate and fit. Since $\int te^{-t}\,dt=-(t+1)e^{-t}$, $$y=Ae^t+B(t+1).$$ At $1$ the data give $Ae+2B=0$ and $Ae+B=1$. Thus $B=-1$, $A=2/e$, and $$\boxed{y=2e^{t-1}-(t+1).}$$ Both basis functions have zero residual, and substitution at $1$ gives $0,1$.

Step 3: Diagnose the error. For any $v$, the undivided residual of $e^tv$ is $$e^t[tw'+(t-1)w].$$ The student’s $w$ satisfies $w'=(t-1)w$, giving residual $$e^t(t^2-1)e^{t^2/2-t}.$$ This is zero at $t=1$ but equals $3e^2$ at $t=2$. A single vanishing residual does not establish a differential identity.

Step 4: Examine the singular initial point. At zero the original equation requires $y'(0)=y(0)$. Both $y=0$ and $y=e^t-(t+1)$ are smooth solutions through zero with value and slope zero; the second is nonzero, for example because its second derivative at zero is $1$. The regular initial-value theorem fails here because the leading coefficient vanishes. Compatibility does not imply uniqueness.