Bernoulli Differential Equations — Question 5

Question 5 – Solution

Strategy. A linear denominator exposes the threshold, while the sign of the original solution must be carried separately.

Step 1: Solve the transformed IVP. For $a\ne 0$, $v=y^{-2}$ gives $v'-2v=-2$, so $$v=1+(a^{-2}-1)e^{2t},\qquad \boxed{y=\frac{a}{\sqrt{a^2+(1-a^2)e^{2t}}}}.$$ The denominator is the positive square root on the interval through $0$. For $a=0$, the unique solution is $y\equiv 0$; the smooth right-hand side $y^3-y$ ensures local uniqueness.

Step 2: Classify the thresholds. If $|a|<1$, the denominator is positive for every $t\ge 0$ and $y\to 0$. This includes $a=0$. If $a=\pm 1$, the formula is constant: $\boxed{y\equiv\pm 1}$. If $|a|>1$, the denominator first vanishes at $$\boxed{T=\frac 12\ln\frac{a^2}{a^2-1}}.$$ The maximal forward lifetime is $0\le t<T$. As $t\uparrow T$, $y\to+\infty$ for $a>1$ and $y\to-\infty$ for $a<-1$.

Step 3: Verify and interpret. For nonzero branches, $v'=-2y^{-3}y'=2v-2$ implies $y'=-y+y^3$, and the formula gives $y(0)=a$. The three constants $0,1,-1$ also check directly. At $a=\pm 1$ the exponential term in $v$ is exactly zero; these solutions neither decay nor blow up. A threshold case must be substituted into the equation, not inferred from neighboring values.