Question

(a) Draw a direction field for the given differential equation. How do solutions appear to behave as \(t\) becomes large? Does the behavior depend on the choice of the initial value a? Let \(a_{0}\) be the value of \(a\) for which the transition from one type of behavior to another occurs. Estimate the value of \(a_{0}\). (b) Solve the initial value problem and find the critical value \(a_{0}\) cractly. (c) Describe the behavior of the solution corresponding to the initial value \(a_{0}\). $$ 2 y^{\prime}-y=e^{2 / 3}, \quad y(0)=a $$

Short answer

There is no transition point or critical value \(a_{0}\) in this case, and the behavior of the solution does not change based on the value of \(a\).

Step-by-step solution

Separate Variables

Since the differential equation is in the form of \(2y’ − y = e^{2/3}\), we can rewrite the equation as \(y’=\frac{1}{2}y+\frac{1}{2}e^{2/3}\). Now our goal is to write the equation in a separable form: \(\frac{dy}{dt}=f(y)g(t)\). We can see that the equation is already in a separable form, where \(f(y)=\frac{1}{2}y+\frac{1}{2}e^{2/3}\) and \(g(t)=1\).

Find the Integrating Factor

Since the equation is linear, we can use an integrating factor to help us solve the equation. We can rewrite the equation as \(y’ - \frac{1}{2}y = \frac{1}{2}e^{2/3}\). The integrating factor, denoted by m(t), can be found by solving \(m(t) = e^{\int -\frac{1}{2}dt}\). Therefore, \(m(t)=e^{-\frac{1}{2}t}\).

Multiply the Equation by the Integrating Factor

Now that we have found the integrating factor, we can multiply the equation by \(e^{-\frac{1}{2}t}\). By doing this, we get \((ye^{-\frac{1}{2}t})'= \frac{1}{2}e^{-\frac{1}{2}t+\frac{2}{3}}\), where our main goal is to obtain an exact differential equation.

Integrate Both Sides

Now we integrate both sides of the equation with respect to \(t\): \(\int (ye^{-\frac{1}{2}t})' dt = \int \frac{1}{2}e^{-\frac{1}{2}t+\frac{2}{3}}dt\). By doing this, we obtain \(ye^{-\frac{1}{2}t} = -\frac{2}{3}e^{-\frac{1}{2}t+\frac{2}{3}}+C\), where C is the integration constant.

Apply the Initial Condition

Apply the initial condition \(y(0) = a\) to the equation we obtained in the previous step. By doing this, we can find that the constant is \(C= \frac{2}{3}+a\). So, \(ye^{-\frac{1}{2}t}=-\frac{2}{3}e^{-\frac{1}{2}t+\frac{2}{3}}+\frac{2}{3}+a\).

Solve for y(t)

Multiply the equation we obtained in the previous step by \(e^{\frac{1}{2}t}\) to get \(y(t)= -\frac{2}{3}e^{\frac{1}{2}t}+\frac{2}{3}e^{\frac{1}{2}t}+ae^{\frac{1}{2}t}\). Notice that \(\frac{2}{3}e^{\frac{1}{2}t}-\frac{2}{3}e^{\frac{1}{2}t}=0\). Thus, the solution is \(y(t)=ae^{\frac{1}{2}t}\).

Analyze for the Critical Value and Behavior

We can analyze the equation \(y(t)=ae^{\frac{1}{2}t}\) to find the critical value \(a_{0}\) and describe the behavior of the solution based on the value of \(a\). If \(a>0\), the function grows exponentially, and if \(a<0\), the function decays exponentially. There is no transition point, so \(a_{0}\) does not exist in this case. The function does not change its behavior based on the value of \(a\). #Answer#: (b) The solution is \(y(t)=ae^{\frac{1}{2}t}\), and there is no existing \(a_{0}\) in this case. (c) Based on the initial value \(a\), if \(a>0\), the function grows exponentially, and if \(a<0\), the function decays exponentially.

Key concepts

Direction Field

Direction fields, also known as slope fields, are visual tools used to represent solutions of a differential equation without solving it explicitly. By plotting tiny line segments that indicate the slope of the solution curve at any given point \(t, y\), these fields help visualize the behavior of solutions.

Imagine you have a differential equation like \(2y' - y = e^{2/3}\). To construct a direction field, you calculate the slope \(y'\) at several grid points \(t, y\) using the equation \(y' = \frac{1}{2}y + \frac{1}{2}e^{2/3}\).

• The slopes are then marked on your graph using short line segments.
• By connecting these segments smoothly, you can see the general trends and behaviors of potential solution curves.

This allows you to predict how solutions will behave as \(t\) increases.

In our particular problem, you might observe that solution curves tend to behave exponentially, indicating either growth or decay depending on initial conditions.

Initial Value Problem

An initial value problem (IVP) involves finding a solution to a differential equation that satisfies a specific initial condition at \(t = 0\). In our exercise, we deal with the differential equation \(2y' - y = e^{2/3}\), subject to the condition \(y(0) = a\).

The goal is to find the specific solution trend \(y(t)\) that not only solves the differential equation but also respects this initial condition.

• First, solve the differential equation using appropriate methods (like using an integrating factor or separation of variables).
• Then plug in the initial condition to solve for any constants in your solution function.

Once you have the particular solution, analyze it to understand how the solution behaves over time.

The analysis assists in identifying characteristics such as growth, decay, and whether there's any transition behavior tied to specific initial values.

Separable Equations

Separable equations are a class of differential equations wherein the derivative can be expressed as a product of a function of \(y\) only and a function of \(t\) only. This means you can rewrite the differential equation so that all terms involving \(y\) are on one side and all terms involving \(t\) are on the other.

Consider the equation \(2y' - y = e^{2/3}\). By rearranging, we can express it as \(\frac{dy}{dt} = \frac{1}{2}y + \frac{1}{2}e^{2/3}\).

• This formulation immediately hints that the functions involved allow separation.
• Separate as \\frac{dy}{\left( \frac{1}{2}y + \frac{1}{2}e^{2/3} \right)} = dt\.

Once separated, you can integrate both sides to find the general solution. Even when this seems linear, the idea of separating helps lay a clearer path to solving it.

Not every differential equation can be separated, but when possible, it simplifies the integration process greatly.

Integrating Factor

The integrating factor is a powerful technique used to solve linear first-order differential equations. It involves multiplying the entire equation by a meticulously chosen function, known as the integrating factor, which makes the equation more accessible for integration.

For the equation \(y' - \frac{1}{2}y = \frac{1}{2}e^{2/3}\), identify it in the standard linear form: \( y' + p(t)y = q(t)\).

• Calculate the integrating factor \(m(t)\) using \(m(t) = e^{\int -\frac{1}{2}dt}\).
• Hence, \(m(t) = e^{-\frac{1}{2}t}\).

By multiplying the entire differential equation by \(e^{-\frac{1}{2}t}\), the left side becomes the derivative of a product: \((ye^{-\frac{1}{2}t})'\).

This transformation allows both sides to be integrated, ultimately easing the path to obtaining the solution. The integrating factor essentially streamlines finding an exact solution to seemingly complex linear equations.