Real & Distinct Roots — Question 1

Question 1 – Solution

Strategy. Keep the leading coefficient in the characteristic polynomial and choose a shifted exponential basis to simplify the data.

Step 1: Derive, factor and check the polynomial. Substitution gives $(6r^2+r-2)e^{rt}=0$. Since $e^{rt}\ne 0$, $$6r^2+r-2=(2r-1)(3r+2)=0, \qquad \boxed{r_1=1/2,\quad r_2=-2/3}.$$ Direct evaluation gives $6(1/2)^2+1/2-2=0$ and $6(-2/3)^2-2/3-2=0$.

Step 2: Use the actual initial time. A shifted basis only rescales the arbitrary constants, so write $$y=Ae^{(t-1)/2}+Be^{-2(t-1)/3}.$$ The data give $A+B=7$ and $A/2-2B/3=0$. Thus $A=4$, $B=3$, and $$\boxed{y=4e^{(t-1)/2}+3e^{-2(t-1)/3}}.$$

Step 3: Verify the equation and uniqueness. Each exponential contributes its coefficient times $6r^2+r-2$ to the residual, so their sum solves the original equation. At 1, its value is $4+3=7$ and its derivative is $2-2=0$. After division by 6, all coefficients are continuous on $\mathbb R$. The linear initial-value theorem therefore gives uniqueness there.

Step 4: Interpret the initial balance of modes. Differentiating twice gives $$y''=e^{(t-1)/2}+\frac 43e^{-2(t-1)/3}>0.$$ Thus $y'$ is strictly increasing, crosses zero at 1, and $\boxed{y(1)=7}$ is the unique global minimum. The positive-root mode gives $y\to+\infty$ as $t\to\infty$; the negative-root mode gives $y\to+\infty$ as $t\to-\infty$.

Zero slope at one instant is a balance between two changing modes. Indeed the equation at 1 gives $y''(1)=7/3$, not zero. A constant solution would have to satisfy $-2y=0$ and could not have value 7.