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. $$ y^{\prime}=2 x-y+\frac{3}{2} ; y(0)=3 $$

Short answer

The particular solution is \( y = 2x - \frac{1}{2} + \frac{7}{2}e^{-x} \).

Step-by-step solution

Recognize the Equation Type

The given differential equation is \( y' = 2x - y + \frac{3}{2} \), which is a first-order linear differential equation. To solve it, we can use an integrating factor.

Rearrange the Equation

Rearrange the equation to get it into the standard linear form: \( y' + y = 2x + \frac{3}{2} \). Here, \( P(x) = 1 \) and \( Q(x) = 2x + \frac{3}{2} \).

Find the Integrating Factor

The integrating factor \( \mu(x) \) is found using \( \mu(x) = e^{\int P(x) \, dx} = e^{\int 1 \, dx} = e^x \).

Multiply Through by the Integrating Factor

Multiply every term in the differential equation by the integrating factor: \[ e^x y' + e^x y = e^x (2x + \frac{3}{2}) \]

Write the Left Side as a Derivative

The left side of the equation becomes the derivative of \( e^x y \):\[ \frac{d}{dx}(e^x y) = e^x (2x + \frac{3}{2}) \]

Integrate Both Sides

Integrate both sides of the equation:\[ \int \frac{d}{dx}(e^x y) \, dx = \int e^x (2x + \frac{3}{2}) \, dx \]The left side integrates to \( e^x y \). For the right side, use integration by parts to find the integral.

Evaluate the Integral

Evaluate the integral on the right side:\[ \int e^x (2x + \frac{3}{2}) \, dx = 2 \int xe^x \, dx + \frac{3}{2} \int e^x \, dx \].Using integration by parts, \( \int xe^x \, dx = xe^x - \int e^x \, dx = xe^x - e^x \),so the entire integral becomes:\[ e^x (2x) - 2e^x + \frac{3}{2}e^x = e^x (2x - 2 + \frac{3}{2}) = e^x (2x - \frac{1}{2}) \].

Solve for \( y \)

We have:\[ e^x y = e^x (2x - \frac{1}{2}) + C \]Solve for \( y \) by dividing through by \( e^x \):\[ y = 2x - \frac{1}{2} + Ce^{-x} \]

Apply the Initial Condition

Use the initial condition \( y(0) = 3 \):\[ 3 = 2 \cdot 0 - \frac{1}{2} + Ce^{0} \]Solve for \( C \):\[ 3 = -\frac{1}{2} + C \]\[ C = 3 + \frac{1}{2} = \frac{7}{2} \]

Particular Solution

Substitute \( C = \frac{7}{2} \) into the general solution:\[ y = 2x - \frac{1}{2} + \frac{7}{2}e^{-x} \]

Plot the Slope Field and Particular Solution

To plot the slope field for \( y' = 2x - y + \frac{3}{2} \), calculate the slopes at various points \((x, y)\). Afterwards, plot the curve of the particular solution \( y = 2x - \frac{1}{2} + \frac{7}{2}e^{-x} \) on the same graph to visualize how the particular solution fits into the slope field.

Key concepts

Slope Field

A slope field, also known as a direction field, is a visual representation of a differential equation. It consists of short line segments or arrows on a graph that represent the slope of the solution curve at different points. To make a slope field, consider the differential equation given in the problem: \( y' = 2x - y + \frac{3}{2} \).

The slope at any point \((x, y)\) can be calculated by inputting the coordinates into the equation for \(y'\). This will give you the direction or slope of the solution curve at that point.

• Slope fields are helpful because they show multiple potential solutions of the differential equation at once.
• They make it easier to visualize how solutions behave, especially with nonlinear or complex equations.
• By plotting the actual solution curve against the slope field, you can see how the particular solution fits amongst many possible solutions.

Integrating Factor

An integrating factor is a function used to simplify a first-order linear differential equation, making it easier to solve. For our differential equation, \(y' + y = 2x + \frac{3}{2}\), an integrating factor \(\mu(x)\) transforms the equation.

Here's the step-by-step process:

• The integrating factor \(\mu(x)\) can be determined using \(\mu(x) = e^{\int P(x) \, dx}\), where \(P(x)\) is the coefficient of \(y\) in the equation. For this problem, \(P(x) = 1\), so \(\mu(x) = e^x\).


• Multiply the entire differential equation by \(\mu(x)\). This will convert the left side of the equation into the derivative of the product \(e^x y\).


• This simplifies the problem because you can now integrate both sides of the equation with ease.

Initial Condition

Initial conditions allow you to find a particular solution from the general solution of a differential equation. For the given problem, we have \(y(0) = 3\) as the initial condition provided.

• Initial conditions are values that specify the value of the solution at a given point.
• By plugging this condition into the general solution, one can solve for any constants of integration present.
• This transforms the general solution into a specific, particular solution that starts at the point provided by the initial condition.

Using the initial condition \(y(0) = 3\), solve by substituting \(x = 0\) and \(y = 3\) into the general solution, then identify the constant \(C\). This enables finding the specific solution that satisfies the condition at the initial point provided.

First-Order Linear Differential Equation

A first-order linear differential equation has the general form \(y' + P(x)y = Q(x)\), where \(y'\) is the derivative of \(y\) with respect to \(x\). Recognizing our given equation, \(y' = 2x - y + \frac{3}{2}\), as a first-order linear differential equation helps to understand what tools are suitable for solving it.

• These equations only involve the first derivative of the unknown function and typically can be solved using an integrating factor or separation of variables.
• They are prevalent in various fields like physics, engineering, and economics because they often model how a quantity changes over time or space.
• Solving such an equation involves manipulating it into a standard linear form and then applying the appropriate method for integration.