Draw a direction field for the given differential equation. Based on the direction field, determine the behavior of y as t → ∞. If this behavior depends on the initial value of y at t = 0, describe this dependency. Note that the right sides of these equations depend on t as well as y; therefore, their solutions can exhibit more complicated behavior than those in the text.
- y’ = 1 /6 y3 − y – 1/ 3 t2
The Correct Answer and Explanation is :
The direction field for the differential equation ( y’ = \frac{1}{6} y^3 – y – \frac{1}{3} t^2 ) is shown above, with vectors at each point indicating the rate of change of ( y ) with respect to ( t ). The behavior of the solution ( y(t) ) as ( t \to \infty ) depends on both the equation’s structure and the initial conditions.
Key Observations from the Direction Field:
- Asymptotic Behavior: As ( t \to \infty ), the term ( -\frac{1}{3} t^2 ) dominates, causing the right-hand side of the equation to become increasingly negative for any nonzero value of ( y ). Therefore, the slope of ( y(t) ) will become negative, suggesting that ( y(t) ) decreases as ( t ) increases, regardless of the initial value of ( y ).
- Stable and Unstable Equilibria:
- The term ( \frac{1}{6} y^3 – y ) suggests that the behavior of the solution depends significantly on the value of ( y ). For large positive ( y ), the cubic term dominates, and the slope becomes very steep. For large negative ( y ), the slope tends to become more negative due to the dominance of the negative cubic term.
- There are equilibrium points where ( y’ = 0 ), which occur when ( \frac{1}{6} y^3 – y = 0 ). Solving ( y^3 – 6y = 0 ), we find that ( y = 0 ), ( y = -\sqrt{6} ), and ( y = \sqrt{6} ) are equilibrium solutions. These are points where the rate of change of ( y ) is zero.
- Dependency on Initial Conditions: The behavior of the solution is influenced by the initial value of ( y ) at ( t = 0 ). For example:
- If ( y(0) ) is close to one of the equilibrium points, the solution will tend to stabilize at that point over time.
- If ( y(0) ) is significantly different from an equilibrium point, the solution will either increase or decrease depending on the initial slope dictated by the equation, gradually tending toward lower values due to the dominant negative ( t^2 )-dependent term.
In conclusion, the solution’s behavior is a complex function of both ( t ) and the initial condition ( y(0) ). As ( t \to \infty ), the solution tends toward negative values for any initial condition, although the rate of change may vary based on the starting point.