Repeated Roots — Question 2

Question 2 – Solution

Strategy. Use the nonvanishing exponential to separate zeros of the affine multiplier from zeros of the derivative.

Step 1: Fit the repeated-root solution. The characteristic equation is $(r+2)^2=0$, so $y=(A+Bt)e^{-2t}$. The data give $A=1$, $B-2A=-3$, hence $B=-1$. Therefore $$\boxed{y=(1-t)e^{-2t}}.$$ Its derivative is $(2t-3)e^{-2t}$, with value $-3$ at zero, and substitution verifies $y''+4y'+4y=0$.

Step 2: Locate the sign change. Since $e^{-2t}>0$, the sign is that of $1-t$. The unique nonnegative zero is $\boxed{t=1}$, with positive values before and negative values after. At the zero, $y'(1)=-e^{-2}\ne 0$, so this is a crossing rather than a tangency.

Step 3: Locate the minimum and the limiting behavior. The derivative vanishes only at $t=3/2$. It is negative before and positive afterward, so the point is the unique global minimum on $[0,\infty)$: $$\boxed{y(3/2)=-\tfrac 12e^{-3}}.$$ The factor $t$ grows more slowly than $e^{2t}$, so $y\to 0$ as $t\to\infty$, approaching from below.

Step 4: Interpret the roles of the two factors. The negative repeated root ensures decay of both $e^{-2t}$ and $te^{-2t}$. It does not determine the sign of their combination or prevent a finite-time crossing. The affine multiplier supplies that sign change, while the derivative’s affine multiplier determines a later turning point. The response crosses zero at 1, reaches its minimum at $3/2$, and then rises toward zero without crossing again.