Problem 19 - Calculus Early Transcendentals

Question

(a) Use Euler's Method with a step size of \(\Delta x=0.2\) to approximate the solution of the initial-value problem $$ y^{\prime}=x+y, \quad y(0)=1 $$ over the interval \(0 \leq x \leq 1\) (b) Solve the initial-value problem exactly, and calculate the error and the percentage error in each of the approximations in part (a). (c) Sketch the exact solution and the approximate solution together.

Step-by-Step Solution

Answer

Euler’s method gives approximate values, exact solution is \( y = -x - 1 + 2e^x \); errors calculated at discrete \( x \) points.

Step 1: Understand Euler's Method

Euler's method is an iterative way to approximate solutions to ordinary differential equations. Given the differential equation \( y' = f(x, y) \) with an initial condition \( y(x_0) = y_0 \), the formula for Euler's method is \( y_{n+1} = y_n + \Delta x \cdot f(x_n, y_n) \).

Step 2: Apply Euler's Method

We start with \( x_0 = 0 \) and \( y_0 = 1 \). Using \( \Delta x = 0.2 \) and \( y' = x + y \), compute subsequent points up to \( x = 1 \). 1. At \( x_0 = 0, y_0 = 1 \): \( y_1 = 1 + 0.2 \times (0 + 1) = 1.2 \).2. At \( x_1 = 0.2, y_1 = 1.2 \): \( y_2 = 1.2 + 0.2 \times (0.2 + 1.2) = 1.44 \).3. At \( x_2 = 0.4, y_2 = 1.44 \): \( y_3 = 1.44 + 0.2 \times (0.4 + 1.44) = 1.728 \).4. At \( x_3 = 0.6, y_3 = 1.728 \): \( y_4 = 1.728 + 0.2 \times (0.6 + 1.728) = 2.0736 \).5. At \( x_4 = 0.8, y_4 = 2.0736 \): \( y_5 = 2.0736 + 0.2 \times (0.8 + 2.0736) = 2.48832 \).6. At \( x_5 = 1.0, y_5 = 2.48832 \).The values at each step are \( y_1 = 1.2, y_2 = 1.44, y_3 = 1.728, y_4 = 2.0736, y_5 = 2.48832 \).

Step 3: Solve the Differential Equation Exactly

To solve \( y' = x + y \), use an integrating factor. Rewrite it as \( y' - y = x \). The integrating factor is \( e^{-x} \). Multiply through by this factor: \( e^{-x}y' - e^{-x}y = x e^{-x} \) becomes the derivative \( \frac{d}{dx}(e^{-x}y) = x e^{-x} \).Integrate both sides: \[ e^{-x}y = \int x e^{-x} dx \].Using integration by parts, find \( \int x e^{-x} dx = -xe^{-x} - e^{-x} + C \).Thus, \( e^{-x}y = -xe^{-x} - e^{-x} + C \), or \( y = -x - 1 + Ce^x \). Using the initial condition \( y(0) = 1 \), solve for \( C \). Substitute \( x = 0, y = 1 \): \( 1 = -0 - 1 + C \), so \( C = 2 \).The exact solution is \( y = -x - 1 + 2e^x \).

Step 4: Calculate the Errors

Compute the exact solution at step points and compare with the Euler approximated values:- \( x = 0.2 \): Exact \( y(0.2) = -0.2 - 1 + 2e^{0.2} \approx 1.224 \); Euler's \( y_1 = 1.2 \). Error = |1.224 - 1.2| = 0.024; Percentage error = |0.024/1.224| * 100%.- Continue this for each \( x = 0.4, 0.6, 0.8, 1.0 \), calculating exact \( y \) values and errors.

Step 5: Graph the Solutions

Sketch the graph for both the exact solution \( y = -x - 1 + 2e^x \) and the approximate points \((0, 1), (0.2, 1.2), (0.4, 1.44), (0.6, 1.728), (0.8, 2.0736), (1, 2.48832)\). The exact solution graph will be a smooth curve, and the Euler's points will form a piecewise linear path approximating this curve.

Key Concepts

Initial-value problem

An initial-value problem is a type of differential equation accompanied by an initial condition. This initial condition specifies the value of the unknown function at a given point. In our exercise, the differential equation is given as \( y' = x + y \), and the initial condition is specified as \( y(0) = 1 \).

The purpose of including an initial condition is to allow us to identify a unique solution among the many possibilities a differential equation might have. Without these conditions, we could have a broad range of solutions that meet the differential equation, but not necessarily the specific scenario we are examining. It is the way to 'anchor' a solution to a known physical, geometrical, or other condition.

When solving an initial-value problem, the process generally involves two steps:

• Solve the differential equation to find the general solution.
• Use the initial condition(s) to find the specific values for constants involved in the general solution.

This two-step process ensures that the solution is both analytic where possible and specific to the problem at hand.

Ordinary differential equations

Ordinary differential equations (ODEs) involve functions of a single variable and their derivatives. These equations form the backbone of various scientific fields, allowing us to predict system behaviors such as motion, heat, electric circuits, or population dynamics. In our case, the ODE provided is \( y' = x + y \). It involves two variables - \( x \) and \( y \) - but is ordinary because it does not involve partial derivatives of more than one variable.

To approach solving ODEs, you can consider methods like:

• **Separation of Variables:** Used when variables in equations can be separated to opposite sides of the equation.
• **Integrating Factors:** As in our example, this method is useful for linear first-order ODEs. It involves multiplying the entire equation by a specific function to allow the left-hand side to be rewritten as a derivative of a product.
• **Homogeneous Equations:** Equations where the degree of all terms is the same can be simplified using substitutions.

Understanding these techniques helps in analyzing ODEs comprehensively, which is crucial for getting the correct specific solution given an initial condition.

Integration by parts

Integration by parts is a technique derived from the product rule of differentiation. It is used to integrate products of functions and is especially helpful when facing integrals that do not easily fit basic integration formulas.

The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]

In the context of solving our differential equation, integration by parts is utilized to solve the integral \( \int x e^{-x} dx \), which is part of finding the exact solution. Here's how it typically works:

• Select which part of the integral to differentiate (\( u \)) and which to integrate (\( dv \)).
• Differentiating \( u \) gives \( du \); integrating \( dv \) gives \( v \).
• Apply the integration by parts formula to find \( \int u \, dv \).

By selecting \( u = x \) and \( dv = e^{-x} dx \) in our problem, differentiation and integration lead us to the solution. This method can simplify seemingly complex integrals, making it a powerful tool in calculus, especially within ODE contexts.