1. **State the problem:** We want to find which graph represents the solution $y(t)$ to the initial value problem (IVP): $$y'(t) = 4\left(y(t) - \frac{1}{4}y(t)^3\right), \quad y(0) = 1.$$ 2. **Rewrite the differential equation:** $$y' = 4y - y^3.$$ 3. **Find equilibrium points:** Set $y' = 0$: $$4y - y^3 = 0 \implies y(4 - y^2) = 0 \implies y = 0, \pm 2.$$ 4. **Analyze stability of equilibria:** Compute derivative of the right side with respect to $y$: $$f'(y) = 4 - 3y^2.$$ Evaluate at equilibria: - At $y=0$: $f'(0) = 4 > 0$ (unstable) - At $y=2$: $f'(2) = 4 - 3(4) = 4 - 12 = -8 < 0$ (stable) - At $y=-2$: $f'(-2) = 4 - 12 = -8 < 0$ (stable) 5. **Interpret initial condition:** Given $y(0) = 1$, which lies between $0$ and $2$. 6. **Behavior of solution:** Since $y=0$ is unstable and $y=2$ is stable, and initial value is $1$ (between 0 and 2), the solution will increase towards $2$ as $t \to \infty$. 7. **Check the graphs:** - The solution passes through $(0,1)$. - It should rise from near $1$ at $t=0$ and approach $2$ for large $t$. From the descriptions, the **top-left graph** and **bottom-left graph** pass through $(0,1)$ and approach $2$ as $t$ increases. 8. **Distinguish between top-left and bottom-left:** - Top-left graph starts near $y=-2$ at $t=-2$, which is inconsistent with initial condition at $t=0$. - Bottom-left graph starts near $y=-2$ at $t=-2$, rises crossing $y=0$ near $t=-1$, reaches $y=1$ at $t=0$, then continues rising to flatten near $y=2$ for $t>1$. Since the IVP is defined at $t=0$ with $y(0)=1$, the solution curve must pass through $(0,1)$ and increase towards $2$ for $t>0$. The bottom-left graph matches this behavior. **Final answer:** The solution corresponds to the **bottom-left graph**.