Problem 51 - Calculus Early Transcendentals

Question

A differential equation of the form \(y^{\prime}(t)=F(y)\) is said to be autonomous (the function \(F\) depends only on \(y\) ). The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(F\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0,\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal line segments in the direction field. Note also that for autonomous equations, the direction field is independent of \(t\). Consider the following equations. a. Find all equilibrium solutions. b. Sketch the direction field on either side of the equilibrium solutions for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=y^{2}$$

Step-by-Step Solution

Answer

Question: Sketch the direction field and solution curve for the autonomous differential equation \(y'(t) = y^2\) with the initial condition \(y(0) = 1\). Answer: For the given differential equation, the direction field consists of arrows with positive slopes as \(y'(t) > 0\) for all \(y\). The only equilibrium solution is \(y=0\), so there will be horizontal line segments along the \(t\)-axis representing the constant solution. The solution curve corresponding to the initial condition \(y(0) = 1\) starts at the point \((0,1)\) and increases as \(t\) increases, with the curve being asymptotic to the line \(y=0\).

Step 1: Find the equilibrium solutions

To find the equilibrium solutions, we'll set \(F(y)=y^2=0\) and solve for \(y\). We get: $$y=\sqrt{0}=0$$ So \(y=0\) is the only equilibrium solution.

Step 2: Determine the direction field

To sketch the direction field, we analyze the sign of \(y'(t)\): - If \(y>0:\) then \(y^2>0\) and \(y'(t)>0,\) which means the slope is positive. - If \(y<0:\) then \(y^2>0\) and \(y'(t)>0,\) which means the slope is also positive. Based on these observations, we can conclude that the slope of the direction field is always positive, irrespective of the value of \(y\).

Step 3: Sketch the direction field on either side of the equilibrium solution

Using the information gathered in Step 2, we can now sketch the direction field. It consists of arrows with positive slopes as \(y'(t)>0\) for all \(y\). Since the equilibrium solution is \(y=0,\) there will be horizontal line segments along the \(t\)-axis representing the constant solution.

Step 4: Sketch the solution curve for the initial condition \(y(0) = 1\)

Since we know the direction field consists of arrows with positive slopes and the equilibrium solution is \(y=0,\) we can sketch a solution curve that starts at the point \((0,1)\) and increases as \(t\) increases. The curve should be on the right side of the \(y\)-axis as \(t \geq 0\). The solution curve should be asymptotic to the line \(y=0\), approaching it but never actually touching it.

Key Concepts

Equilibrium Solutions

In autonomous differential equations, like the one given in the exercise, equilibrium solutions are crucial in understanding the behavior of solutions over time. An equilibrium solution is a constant solution, which means that the derivative is zero, making the function unchanging over time. To find the equilibrium solutions for a differential equation of the form \( y'(t) = F(y) \), we need to solve the equation \( F(y) = 0 \).
For the provided example, \( y'(t) = y^2 \), we set \( y^2 = 0 \). Solving this, the only solution is \( y = 0 \). This means that \( y = 0 \) is an equilibrium solution.
Equilibrium solutions are often depicted as horizontal lines in a direction field. They suggest stability in the system, where small perturbations do not drive the solution away from the equilibrium. This fundamental concept helps anticipate how other solutions behave in the vicinity of this constant solution.

Direction Field

A direction field, also known as a slope field, provides a graphical representation of a differential equation without needing to solve it analytically. For an autonomous equation \( y'(t) = F(y) \), the direction field depends solely on \( y \) and not on \( t \).
To construct a direction field, we evaluate the sign and magnitude of \( F(y) \) for various values of \( y \). In the specific case of \( y'(t) = y^2 \), we find that:

• When \( y > 0 \): \( y^2 > 0 \) yields a positive slope.
• When \( y < 0 \): \( y^2 > 0 \) also gives a positive slope.

This creates a pattern of field lines that have positive slopes regardless of whether \( y \) is positive or negative. In the sketch of such a direction field, you'll encounter uniformly upward pointing arrows. Examining these arrows allows us to visualize how solutions evolve over time.

Initial Value Problem

An initial value problem involves finding a specific solution curve that passes through a given point, often serving as the starting condition for our differential equation. In this exercise, the initial condition provided is \( y(0) = 1 \).
The presence of the direction field makes it easier to predict the behavior of this solution. Starting at the point \( (0,1) \), the solution curve follows the pattern of arrows dictated by the direction field, moving upwards as time \( t \) increases. The slope of these arrows aligns with the derivative \( y'(t) = y^2 \), ensuring the curve ascends as it continues.
A significant trait of this solution curve is how it behaves relative to the equilibrium solution of \( y = 0 \). It is asymptotic to the line \( y = 0 \), meaning it approaches zero as \( t \) becomes large but never converges to it. Thus, the illustration of the initial value problem provides a tangible sense of the dynamic evolution of solutions from specific initial conditions.