Real & Distinct Roots — Question 6

Question 6 – Solution

Strategy. Use the strict inequality of real exponentials to solve the endpoint problem and control the number of zeros.

Step 1: Invert the endpoint constraints. For $y=Ae^{r_1t}+Be^{r_2t}$, the conditions give $A+B=a$ and $Ae^{r_1L}+Be^{r_2L}=b$. Therefore $$B=\frac{b-ae^{r_1L}}{e^{r_2L}-e^{r_1L}},\qquad A=a-B.$$ Because $r_2>r_1$ and $L>0$, the denominator is strictly positive. There is exactly one coefficient pair for all endpoint data, unlike some oscillatory endpoint problems.

Step 2: Fit the concrete equation. Here the roots are 1 and 2. At $L=\ln 2$, the equations are $A+B=1$ and $2A+4B=1$. Hence $$\boxed{y=\frac 32e^t-\frac 12e^{2t}}.$$ Both endpoints have value 1, and each mode satisfies the original equation.

Step 3: Bound the number of zeros and deduce positivity. Factor the general solution as $e^{r_1t}(A+Be^{(r_2-r_1)t})$. If $B\ne 0$, the bracket is strictly monotone, so it has at most one zero. If $B=0$, a nonzero solution is a nonvanishing pure exponential; the case $A=0$ is likewise nonvanishing.

If a solution with positive endpoints became negative inside, continuity would force two distinct zeros. If it merely touched zero inside while staying nonnegative, that interior minimum would have $y=y'=0$; initial-value uniqueness would force the identically zero solution, contradicting the endpoints. Thus it is strictly positive throughout the interval.

Step 4: Distinguish positivity from an upper bound. For the concrete solution, $y'=e^t(3/2-e^t)$ changes from positive to negative at $t=\ln(3/2)$, which lies between 0 and $\ln 2$. Its value there is $$\boxed{y_{\max}=\frac 32\frac 32-\frac 12\frac 94=\frac 98>1}.$$ The curve stays positive but exceeds both endpoint values. A restriction on zeros is not a maximum principle bounding the response by its endpoints.