Bernoulli Differential Equations — Question 3

Question 3 – Solution

Strategy. The linearized equation applies only on positive pieces; restore zero intervals and enforce the sign of the square root.

Step 1: Solve a positive piece. Writing $y=w^2$ with $w>0$ gives $2ww'+2w^2=2w$, hence $w'+w=1$. Thus $w=1+Ce^{-t}$. If positivity starts at $b$ with $w(b)=0$, then $w=1-e^{-(t-b)}$, positive precisely for $t>b$.

Step 2: Restore the zero intervals. For each finite $b\ge 0$, define $$\boxed{y_b(t)=\begin{cases}0,&0\le t\le b,\\ (1-e^{-(t-b)})^2,&t\ge b. \end{cases}}$$ Also include $y\equiv 0$, denoted by $b=\infty$. On the positive piece let $r=e^{-(t-b)}$; then $y'=2r(1-r)$ and $y'+2y=2(1-r)=2\sqrt y$. At a positive joining time, both one-sided derivatives are $0$; if $b=0$, the right derivative is $0$. Thus the solution is $C^1$ on its stated domain and satisfies the equation at the join.

Step 3: Prove completeness and reject a false extension. Any positive component must start from zero at its finite left endpoint, including $b=0$, and the linear formula above fixes that component uniquely. It stays positive forever after $b$, so it cannot end at another zero. Hence there is at most one positive component and the list is exhaustive.

For $t<b$, the same linear expression has $w<0$. If it is squared, $\sqrt{w^2}=|w|=-w$, while the computed left side is $2w$. Thus it fails the original equation there. Squaring cannot remove this sign restriction.