Problem 18 - Thomas Calculus

Question

In Exercises 17 and \(18,\) (a) find the exact solution of the initial value problem. Then compare the accuracy of the approximation with \(y\left(x^{*}\right)\) using Euler's method starting at \(x_{0}\) with step size (b) \(0.2,(\mathrm{c}) 0.1,\) and (d) \(0.05 .\) $$ y^{\prime}=y-1, \quad y(0)=3, \quad x_{0}=0, \quad x^{*}=1 $$

Step-by-Step Solution

Answer

Exact solution at \( x^* = 1 \) is approximately 6.4366; Euler's method yields 5.97664 (\(h=0.2\)), 5.755 (\(h=0.1\)), and 5.835 (\(h=0.05\)).

Step 1: Solve the Differential Equation

The given differential equation is \( y' = y - 1 \) with the initial condition \( y(0) = 3 \).To solve this, rewrite it as \( y' - y = -1 \).This is a first-order linear differential equation, which can be solved using the integrating factor method.The integrating factor \( \mu(x) \) is \( e^{\int -1 \, dx} = e^{-x} \).Multiply through by the integrating factor to obtain \( e^{-x}y' - e^{-x}y = -e^{-x} \).Now integrate both sides: \( \int (e^{-x} y)' \, dx = \int -e^{-x} \, dx \).This gives \( e^{-x}y = e^{-x} + C \).Solve for \( y(x) \):\[ y(x) = 1 + Ce^{x}. \]Apply the initial condition \( y(0) = 3 \):\[ 3 = 1 + C \implies C = 2. \]Thus, the exact solution is \( y(x) = 1 + 2e^{x} \).

Step 2: Calculate Exact Value at x*

Calculate the exact solution at \( x^* = 1 \):\[ y(1) = 1 + 2e^{1}. \]Thus,\( y(1) = 1 + 2e \approx 1 + 5.4366 \approx 6.4366. \)

Step 3: Apply Euler's Method with Step Size 0.2

Start with \( y_0 = 3 \) at \( x_0 = 0 \).The Euler's update formula is \( y_{n+1} = y_n + h(y_n - 1) \).- With \( h = 0.2 \): - Calculate for \( x_1 = 0.2 \): \[ y_1 = 3 + 0.2(3 - 1) = 3.4. \] - For \( x_2 = 0.4 \): \[ y_2 = 3.4 + 0.2(3.4 - 1) = 3.88. \] - For \( x_3 = 0.6 \): \[ y_3 = 3.88 + 0.2(3.88 - 1) = 4.456. \] - For \( x_4 = 0.8 \): \[ y_4 = 4.456 + 0.2(4.456 - 1) = 5.1472. \] - For \( x_5 = 1.0 \): \[ y_5 = 5.1472 + 0.2(5.1472 - 1) = 5.97664. \]Thus, \( y(1) \approx 5.97664 \) with \( h = 0.2 \).

Step 4: Apply Euler's Method with Step Size 0.1

Start with \( y_0 = 3 \) at \( x_0 = 0 \).- With \( h = 0.1 \): - Calculate for \( x_1 = 0.1 \): \[ y_1 = 3 + 0.1(3 - 1) = 3.2. \] - For \( x_2 = 0.2 \): \[ y_2 = 3.2 + 0.1(3.2 - 1) = 3.52. \] - Repeat this process until \( x_{10} = 1.0 \): - For \( x_{10} = 1.0 \): \[ y_{10} \approx 5.755. \]Thus, \( y(1) \approx 5.755 \) with \( h = 0.1 \).

Step 5: Apply Euler's Method with Step Size 0.05

Start with \( y_0 = 3 \) at \( x_0 = 0 \).- With \( h = 0.05 \): - Calculate for \( x_1 = 0.05 \): \[ y_1 = 3 + 0.05(3 - 1) = 3.1. \] - Continue this process for each 0.05 increment up to \( x_{20} = 1.0 \). - For \( x_{20} = 1.0 \): \[ y_{20} \approx 5.835. \]Thus, \( y(1) \approx 5.835 \) with \( h = 0.05 \).

Step 6: Compare Results

The exact value of \( y(1) \) is approximately 6.4366.Values obtained using Euler's method are:- With \( h = 0.2: 5.97664 \)- With \( h = 0.1: 5.755 \)- With \( h = 0.05: 5.835 \)Smaller step sizes result in approximations closer to the exact value.

Key Concepts

Initial Value Problem

In mathematics, an initial value problem involves solving a differential equation, but with a specific initial condition provided. This means that the solution to the differential equation must satisfy the condition that is given at a specific point.
For example, in our exercise, the initial value problem is given by the differential equation \( y' = y - 1 \) with the initial condition \( y(0) = 3 \). This condition specifies that when \( x \) is 0, the value of \( y \) must be 3.
Understanding this condition is crucial because it uniquely determines the solution to the equation. Without it, the differential equation could have an infinite number of solutions. The initial condition effectively "anchors" the solution curve at a specific point.
When solving, the goal is to find the function \( y(x) \) that not only solves the differential equation but also passes through the point determined by the initial condition.

Differential Equation

Differential equations are mathematical equations that involve an unknown function and its derivatives. They are powerful tools used to model various real-world phenomena like physics, engineering, and even biology.
The differential equation in this exercise is \( y' = y - 1 \), which is a first-order linear differential equation. That means it involves the first derivative of \( y \), often notated as \( y' \), and can be solved using techniques like the integrating factor method or separation of variables.
Solving this type of equation provides insight into how a particular system changes over time. In our case, it describes a scenario where the rate of change of \( y \) is proportional to how far \( y \) is from the value 1. This kind of insight is what makes differential equations so valuable in both theoretical and applied sciences.
Knowing how to solve differential equations is essential for predicting their behavior and understanding the dynamics behind them.

Numerical Approximation

Numerical approximation methods are techniques used to find approximate solutions to mathematical problems when exact solutions are difficult or impossible to find. Euler's method is a popular numerical approximation technique used specifically for solving initial value problems related to ordinary differential equations.
Euler's method estimates values using discrete steps across the function's domain. In simpler words, it walks across the curve of the function by making small steps and using each step's outcome to calculate the next value.
In our exercise, Euler's method is applied with different step sizes: 0.2, 0.1, and 0.05. Smaller step sizes typically provide more accurate approximations, as seen with the step size of 0.05 producing results closer to the exact solution than the larger steps.
Numerical approximation is especially useful when working with complex systems where precise solutions can't be obtained easily. Through iterative calculations, we can still achieve a reliable understanding of the system's behavior.