Question

Slope Field In Exercises 15 and \(16,(\text { a })\) sketch an approximate solution of the differential equation satisfying the given initial condition by hand on the slope field, (b) find the particular solution that satisfies the given initial condition, and (c) use a graphing utility to graph the particular solution. And (c) use a graph with the hand-drawn graph in part (a). To print an enlarged copy of the graph, go to MathGraphs.com. $$ \begin{array}{l}{\frac{d y}{d x}=e^{x}-y} \\ {(0,1)}\end{array} $$

Short answer

The particular solution to the given differential equation satisfying the initial condition (0,1) is \( y = xe^{x} + e^{x} \). The sketch of the solution on the slope field and the graphing utility should both show corresponding representations.

Step-by-step solution

Write Down and Recognize the Differential Equation

The differential equation to be solved is given as \(\frac{dy}{dx} = e^x - y\). Notice that it can be written in the form of a linear differential equation, which can be solved with integrating factors.

Solve Using Integrating Factor (IF)

First, identify the integrating factor, which is \(e^{-x}\). Multiply every term of the differential equation by the integrating factor: \(e^{-x} \frac{dy}{dx} + y e^{-x} = 1\). Simplify to get its standard form \( \frac{d (ye^{-x})}{dx}= 1\). Integrate both sides with respect to \(x\) to get \( ye^{-x} = x + C\).

Use the Given Initial Condition to Get the Particular Solution

Substitute the initial condition (0,1) into \( y = xe^{x} + Ce^{x}\) to find the value of \(C\). Thus, \( C = 1\). The particular solution of the differential equation is \(y = xe^{x} + e^{x} \).

Sketch and Verify Solution on the Slope Field

Hand-sketch the slope field for the differential equation and plot \( y = xe^{x} + e^{x}\) . Use a graphing utility to verify the sketch and the solution.

Key concepts

Slope Field

The concept of a slope field is an ingenious way to visualize solutions to differential equations without actually solving them. It's like interrogating a map to predict the direction of a hill's slope without walking it. A slope field consists of a grid of points, and at each point, a small line segment is drawn with a slope equal to the value of the differential equation at that point.

Imagine taking a step in any direction; the slope field gives an inkling of how steep your path would be. So, when you receive a differential equation like \( \frac{dy}{dx} = e^x - y \), you start equipping the graph paper with clues - the tiny line segments, which represent the rate of change or the

Integrating Factor

An integrating factor is essentially a multiplier that makes a differential equation more manageable. It's like finding the right key for a lock. Using this factor, we transform a complicated expression into something that can be integrated much more smoothly. When dealing with linear differential equations, the integrating factor is a function, often denoted as \( \mu(x) \), which, when multiplied by every term in the equation, allows the left-hand side to be written as the derivative of a product.

The formula to find the integrating factor for a first order linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \) is \( \mu(x) = e^{\int P(x) dx} \). In our given problem, the integrating factor is \( e^{-x} \) and it helps convert the equation into a perfect differential which is then easily integrated.

Initial Condition

The initial condition acts as a starting point that helps pin down one particular solution from a family of possible solutions. If a differential equation describes a family's lineage, the initial condition would be that one family member you're strictly focusing on. It is usually given as a point \( (x_0, y_0) \), where \( y_0 \) is the value of the unknown function when \( x \) equals \( x_0 \).

Initial conditions are immensely powerful - they tackle the 'C' in 'plus C', the notorious constant of integration that arises after solving a differential equation. This constant represents an infinite number of potential solutions, and the initial condition helps to nail down the exact value of 'C' to converge on the particular solution fitting the given scenario.

Particular Solution

A particular solution is your treasure chest at the end of an 'X marks the spot' map - a specific equation derived from the general solution of a differential equation, complying with the initial condition. For a gardener, if the general solution is an array of seeds, the particular solution is the one plant that grows from the specific seed chosen.

In the context of our problem, once the general solution is reached using integrating factors, the initial condition \( (0,1) \) is applied to find that one sought-after 'C' value. Hence, the resulting particular solution for the given differential equation and initial condition is \( y = xe^{x} + e^{x} \), reflecting the specific behavior that matches the provided starting point of the function.

Linear Differential Equation

A linear differential equation is like the straight-A student of the equation world: tidy, predictable, and you know exactly what to expect. It manifests as a relationship where the dependent variable and its derivative appear to the first degree and are not multiplied together. It's in the form \( P(x) \frac{dy}{dx} + Q(x)y = R(x) \), where \( P, Q, \) and \( R \) are functions of \( x \).

What makes them so welcomed in mathematics is their solvable nature. By employing techniques such as integrating factors, one can systematically find the general solution of a linear differential equation, and with the initial condition, pinpoint the particular solution. In the step-by-step solution provided, we see the linear differential equation \( \frac{dy}{dx} = e^x - y \) being tamed into a format that allows the solver to integrate and find the general solution, then use the initial condition to find the particular solution, ending up with a neat answer representing how the system behaves over time.