Question

Plot a slope field for each differential equation. Use the method of separation of variables (Section 4.9) or an integrating factor (Section 7.7) to find a particular solution of the differential equation that satisfies the given initial condition, and plot the particular solution. $$ y^{\prime}=\frac{1}{2} y ; y(0)=\frac{1}{2} $$

Short answer

The particular solution is \( y(t) = \frac{1}{2}e^{\frac{1}{2}t} \).

Step-by-step solution

Identify the Differential Equation

The given differential equation is \( y' = \frac{1}{2} y \). This is a first-order linear differential equation.

Solve by Separation of Variables

To solve by separation of variables, rewrite the equation as \( \frac{dy}{y} = \frac{1}{2} dt \). Integrate both sides to obtain \( \ln|y| = \frac{1}{2}t + C \).

Solve for y

Exponentiate both sides to solve for \( y \): \( |y| = e^{\frac{1}{2}t + C} \). Therefore, \( y = Ce^{\frac{1}{2}t} \) where \( C \) is a constant.

Apply the Initial Condition

Use the initial condition \( y(0) = \frac{1}{2} \). Substitute \( t = 0 \) and \( y = \frac{1}{2} \) into the solution \( y = Ce^{\frac{1}{2}t} \) to find \( C \). This gives \( \frac{1}{2} = Ce^0 \), so \( C = \frac{1}{2} \).

Write the Particular Solution

The particular solution is \( y(t) = \frac{1}{2}e^{\frac{1}{2}t} \).

Plot the Slope Field

A slope field is a visual representation of the differential equation \( y' = \frac{1}{2}y \). It shows small line segments with slopes given by the derivative \( \frac{1}{2} y \) for various points \( (t, y) \).

Plot the Particular Solution on the Slope Field

Plot the curve \( y(t) = \frac{1}{2}e^{\frac{1}{2}t} \) on the same graph as the slope field. The curve represents the particular solution and should follow the direction of the slopes.

Key concepts

Slope Field

Slope fields are powerful visual tools in mathematics that represent the solutions of differential equations. When you have a differential equation like \( y' = \frac{1}{2}y \), the slope field helps us visualize how solutions behave without actually solving the equation.

• A slope field is made up of small line segments or arrows on a graph.
• Each segment corresponds to a point \((t, y)\).
• The direction of the segment represents the slope \(y' = \frac{1}{2}y\) at that point.

These segments indicate how a solution would move from one point to another throughout the graph. When plotting the particular solution on this field, the solution should follow the direction of these small arrows or segments.
This means it aligns with the slopes depicted, giving us an intuitive grasp of solution trajectories.

Initial Condition

An initial condition is a crucial piece of information that allows us to find a specific solution to a differential equation, known as a particular solution. In our exercise, the initial condition is given as \( y(0) = \frac{1}{2} \). Here's what you need to know:

• The initial condition specifies the value of the solution at a particular point, here when \( t = 0 \).
• By plugging this into the solution equation, \( y = Ce^{\frac{1}{2}t} \), we can determine the constant \( C \).

In this case, substituting results in:
\( \frac{1}{2} = Ce^0 \)
Which simplifies to \( C = \frac{1}{2} \).
The initial condition is vital because it determines our particular solution amidst potentially many others that would satisfy the differential equation without it.

Separation of Variables

The method of separation of variables is one of the key techniques used to solve differential equations. It is particularly effective for equations that can be expressed in the form \( y' = f(y)g(t) \). In this context, our given equation \( y' = \frac{1}{2}y \) is ideal for this method. Here's how it works:

• First, we rearrange the equation so that all \( y \)-terms are on one side and all \( t \)-terms are on the other: \( \frac{dy}{y} = \frac{1}{2} dt \).
• Next, we integrate both sides. For \( \frac{dy}{y} \), the integral is \( \ln|y| \), and for \( \frac{1}{2} dt \), it is \( \frac{1}{2}t + C \).
• This leads us to \( \ln|y| = \frac{1}{2}t + C \).

By exponentiating, we eliminate the logarithm, resulting in \( |y| = e^{\frac{1}{2}t + C} \), simplifying to \( y = Ce^{\frac{1}{2}t} \) by introducing a new constant \( C \).
The solution derived in this manner is general until further refined by the initial condition.

First-Order Linear Differential Equation

First-order linear differential equations form an essential category in the study of differential equations. They generally have the form \( y' + p(t)y = q(t) \). In our exercise, the equation \( y' = \frac{1}{2}y \) is a simplified form of this, where:

• \( p(t) = -\frac{1}{2} \) and \( q(t) = 0 \).

Such equations have solutions that tend to be exponential, just like in this case where we ended up with \( y = Ce^{\frac{1}{2}t} \). Understanding first-order linear differential equations involves:

• Recognizing their linear form.
• Applying methods such as separation of variables or integrating factors.

These equations are often simpler to solve because they can sometimes be rewritten or transformed for straightforward solution strategies, as illustrated with our solved equation.