Separable Equations — Question 2

Question 2 – Solution

Strategy. Preserve the two equilibrium solutions, then use partial fractions on a nonconstant branch.

Step 1: Separate without losing constants. The constant solutions are $\boxed{y\equiv 0,\ y\equiv 3}$. Away from those values, $$\frac 1{y(3-y)}=\frac 1{3y}+\frac 1{3(3-y)},\qquad \frac 13\ln\left|\frac{y}{3-y}\right|=x+C.$$ The nonconstant family can be written $y=3/(1+A e^{-3x})$, $A\ne 0$, on intervals where the denominator is nonzero. The value $A=0$ restores $y\equiv 3$; $y\equiv 0$ must still be listed separately.

Step 2: Apply the data and determine the domain. From $4=3/(1+A)$, $A=-1/4$, so $$\boxed{y(x)=\frac 3{1-\frac 14e^{-3x}},\qquad I=(-\ln 4/3,\infty)}.$$ Here $y>3$, so $y'<0$. The limits are $+\infty$ at the left endpoint and $3$ as $x\to\infty$. Writing $D=1-\tfrac 14e^{-3x}$ gives $y'=-9e^{-3x}/(4D^2)=y(3-y)$, verifying the equation.

Step 3: Compute the passage time. The equation $7/2=3/(1-\tfrac 14e^{-3x})$ gives $\boxed{x=\tfrac 13\ln(7/4)}$. At any finite $x\in I$, the exponential is positive, so $y-3>0$: the limiting level is never attained or crossed.