Separable Equations — Question 6

Question 6 – Solution

Strategy. Separate only where $y\ne 0$, then use the sign of $y'$ and check any joins at zero directly.

Step 1: Reach zero. While $y>0$, integration gives $3y^{1/3}=-3t+C$. The data yield $y=(2-t)^3$ for $0\le t<2$, so the first zero is $\boxed{t=2}$.

Step 2: Enforce nonnegativity. The equation always gives $y'\le 0$. Once a nonnegative solution reaches zero it cannot increase or decrease without violating that constraint. Thus the unique nonnegative continuation is $(2-t)^3$ up to $2$, followed by $0$ forever.

Step 3: Classify real continuations. For any finite $b\ge 2$, a real continuation is $$\boxed{y_b(t)=\begin{cases} (2-t)^3,&0\le t\le 2,\\ 0,&2\le t\le b,\\ -(t-b)^3,&t\ge b. \end{cases}}$$ Also allow $b=\infty$, meaning the solution stays zero forever. These exhaust the possibilities: monotonicity makes the zero set after $2$ an interval; once negative, separation gives $\sqrt[3]{y}=b-t$, where continuity fixes the constant at the departure time $b$.

Step 4: Verify the joins and the lost solutions. Each cubic piece has $y'=-3(\sqrt[3]{y})^2$. At $t=2$ and any finite $b$, both one-sided derivatives are $0$, matching the equation at $y=0$. The pieces are therefore $C^1$. Division excludes zero itself, so it misses the constant zero solution and all zero waiting intervals; using one integration constant through zero is unjustified.