Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

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. $$ y^{\prime}=x-y+2 ; y(0)=4 $$

Step-by-Step Solution

Answer

The particular solution is \( y = x + 1 + 3e^{-x} \).

Step 1: Write the Differential Equation

The differential equation given is \( y' = x - y + 2 \) with the initial condition \( y(0) = 4 \). We aim to both plot a slope field and find a particular solution using the initial condition.

Step 2: Rearrange the Equation for Slope Field

To plot the slope field, recognize that the slope of the solution at each point \((x, y)\) is represented by \( x - y + 2 \). For various values of \(x\) and \(y\), calculate \( x - y + 2 \) to understand the direction of the slope field.

Step 3: Introduce an Integrating Factor

Since the equation is linear, we will solve it using an integrating factor. First, rewrite the differential equation in standard linear form: \( y' + y = x + 2 \). The integrating factor is \( \, e^{\int 1 \, dx} = e^x \, \).

Step 4: Multiply by the Integrating Factor

Multiply each term of the equation by the integrating factor. This yields \( e^x y' + e^x y = e^x (x + 2) \).

Step 5: Simplify the Left Side

Notice the left side of the equation is now the derivative of \( e^x y \). Therefore, \( \frac{d}{dx} (e^x y) = e^x (x + 2) \). This allows us to write: \( e^x y = \int e^x (x + 2) \, dx \).

Step 6: Integrate the Right Side

Integrate \( \int e^x (x + 2) \, dx \) by parts. Let \( u = x + 2 \) and \( dv = e^x \, dx \), then \( du = dx \) and \( v = e^x \). The integral becomes \( e^x (x+2) - \int e^x \, dx = e^x (x + 2) - e^x + C \).

Step 7: Solve for y

Substitute back, \( e^x y = e^x (x + 1) + C \). Divide by \( e^x \) to solve for \( y \): \( y = x + 1 + Ce^{-x} \).

Step 8: Apply the Initial Condition

Use the initial condition \( y(0)=4 \) to find \( C \). Substitute into the general solution: \( 4 = 0 + 1 + C \cdot 1 \), which gives \( C = 3 \). The particular solution is \( y = x + 1 + 3e^{-x} \).

Step 9: Plot the Slope Field and Particular Solution

Plot the slope field using the expression \( x - y + 2 \) to show slope directions across a grid. Then plot the particular solution, \( y = x + 1 + 3e^{-x} \), on the same graph to visualize how it fits with the overall slope pattern.

Key Concepts

Slope Fields

Slope fields provide a visual illustration of differential equations. They allow us to see the behavior of solutions at various points in the plane. Each little line segment in the slope field shows the slope of the solution to a differential equation at that specific point.
To construct the slope field for the differential equation \(y' = x - y + 2\), we need to compute the expression \(x - y + 2\) for various pairs of \((x, y)\). This calculation gives us the direction (or slope) that the solution would "follow" at any given point.
By sketching these slopes across the coordinate plane, we get a sense of what the solutions look like without actually solving the equation completely. Observing these slope patterns helps us infer the general behavior of solutions over a range of points.

Separation of Variables

Separation of variables is a common method used to solve differential equations. It involves separating and then individually integrating the variables on each side of the equation. However, not all differential equations are directly separable.
In the given problem, the equation \(y' = x - y + 2\) is not easily separable, indicating that other solution methods, such as an integrating factor, might be better suited. This highlights the importance of recognizing an equation's form and choosing the appropriate method.
The essence of separation of variables is creating independent integrations for \(x\) and \(y\), which can simplify solving initial value problems when applicable.

Integrating Factor

For linear differential equations of the form \(y' + Py = Q\), like \(y' + y = x + 2\) in this example, an integrating factor is pivotal in finding solutions. The integrating factor method transforms a non-separable equation into a form that's straightforward to solve.
An integrating factor, usually denoted as \(e^{ \int P \, dx}\), allows the left side of the differential equation to be rewritten as the derivative of a product. For our equation, \(P = 1\), leading to an integrating factor of \(e^x\).
By multiplying the entire equation by this integrating factor, the left side simplifies to \(\frac{d}{dx}(e^x y)\). This transformation enables easy integration and simplification of the equation to find the general solution.

Initial Conditions

Initial conditions are crucial in determining specific solutions to differential equations, often called particular solutions. In our example, the initial condition \(y(0) = 4\) provides the necessary information to determine the constant \(C\) in the general solution.
With the general solution \(y = x + 1 + Ce^{-x}\), substituting \(x = 0\) and \(y = 4\) allows us to solve for \(C\). Performing this substitution: \(4 = 0 + 1 + C \cdot 1\) reveals \(C = 3\). Thus, the particular solution becomes \(y = x + 1 + 3e^{-x}\).
Applying initial conditions anchors the solution to a specific point on the curve, ensuring the particular solution reflects the real-world or theoretical situation described by the initial condition.