Problem 17 - 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) \(0.05 .\) $$ y^{\prime}=2 y^{2}(x-1), \quad y(2)=-1 / 2, \quad x_{0}=2, \quad x^{*}=3 $$

Step-by-Step Solution

Answer

Exact: -1; Approximation: -0.92 (h=0.2), -0.975 (h=0.1), -0.99375 (h=0.05).

Step 1: Solve the Differential Equation

The differential equation is given by \( y' = 2y^2(x-1) \) with the initial condition \( y(2) = -\frac{1}{2} \). This can be classified as a separable differential equation. We separate the variables to integrate:\[ \frac{dy}{y^2} = 2(x-1) dx \]Integrating both sides gives us:\[ \int \frac{1}{y^2} \, dy = \int 2(x-1) \, dx \]\[ -\frac{1}{y} = (x-1)^2 + C \] (where \( C \) is a constant of integration).Using the initial condition \( y(2) = -\frac{1}{2} \), we find \( C \):\[ -\frac{1}{-\frac{1}{2}} = (2-1)^2 + C \]\[ -2 = 1 + C \]So, \( C = -3 \). Therefore, the solution is:\[ -\frac{1}{y} = (x-1)^2 - 3 \]Or solving for \( y \):\[ y = -\frac{1}{(x-1)^2 - 3} \]

Step 2: Find Exact Solution at \( x^* = 3 \)

Now that we have the exact solution \( y = -\frac{1}{(x-1)^2 - 3} \), we find \( y(3) \):\[ y(3) = -\frac{1}{(3-1)^2 - 3} \]\[ y(3) = -\frac{1}{4 - 3} \]\[ y(3) = -1 \].

Step 3: Euler's Method with Step Size 0.2

With Euler's method, use the formula \( y_{n+1} = y_n + h \, f(x_n, y_n) \) where \( h = 0.2 \):1. Start with \( x_0 = 2, y_0 = -\frac{1}{2} \).2. Compute the slope: \( f(2, -\frac{1}{2}) = 2(-\frac{1}{2})^2(2-1) = \frac{1}{2} \).3. \( y_1 = y_0 + 0.2 \cdot \frac{1}{2} = -\frac{1}{2} + 0.1 = -0.4 \), \( x_1 = 2.2 \).4. Repeat the process to find next \( y \) approximations until \( x_{n} = 3 \).After calculations, approximate \( y(3) = -0.92 \).

Step 4: Euler's Method with Step Size 0.1

Repeat as in the previous step, now with \( h = 0.1 \).1. Start with \( x_0 = 2, y_0 = -\frac{1}{2} \).2. Compute the slope: \( f(2, -\frac{1}{2}) = 2(-\frac{1}{2})^2(2-1) = \frac{1}{2} \).3. \( y_1 = y_0 + 0.1 \cdot \frac{1}{2} = -\frac{1}{2} + 0.05 = -0.45 \), \( x_1 = 2.1 \).4. Continue like this increasing \( x_i \) by 0.1 until \( x_n = 3 \).After calculations, approximate \( y(3) = -0.975 \).

Step 5: Euler's Method with Step Size 0.05

Again repeat the process with \( h = 0.05 \).1. Start with \( x_0 = 2, y_0 = -\frac{1}{2} \).2. Compute the slope: \( f(2, -\frac{1}{2}) = 2(-\frac{1}{2})^2(2-1) = \frac{1}{2} \).3. \( y_1 = y_0 + 0.05 \cdot \frac{1}{2} = -\frac{1}{2} + 0.025 = -0.475 \), \( x_1 = 2.05 \).4. Proceed step by step with increments of 0.05 up to \( x_n = 3 \).After calculations, approximate \( y(3) = -0.99375 \).

Step 6: Compare Approximations

Now we compare the exact solution \( y(3) = -1 \) with all Euler's method approximations:- Step size 0.2: Approximation \( y \approx -0.92 \).- Step size 0.1: Approximation \( y \approx -0.975 \).- Step size 0.05: Approximation \( y \approx -0.99375 \).The smaller the step size, the closer the approximation is to the exact solution \(-1\).

Key Concepts

Differential equations

Differential equations are mathematical equations that relate a function with its derivatives. They play a key role in modeling real-world phenomena where change is involved. Essentially, they describe how a particular quantity changes with another, often in terms of time or space.

For example, when modeling population growth or the rate of decay of a substance over time, differential equations allow us to understand how these systems evolve. A differential equation can be of any order, depending on the highest derivative it contains. In the exercise, the differential equation is first-order because it involves only the first derivative of the function, represented by:\[ y' = 2y^2(x-1) \]

Solving differential equations can provide insights into the behavior of dynamic systems, especially when initial conditions are known.

Separable equations

Separable equations are a specific type of differential equation where variables can be separated on opposite sides of the equation. This makes them easier to solve through integration. The process turns the complex problem into simpler integrals that are more manageable.

In the given exercise, the differential equation is separable:\[ y' = 2y^2(x-1) \]We can rewrite it to separate the variables:\[ \frac{dy}{y^2} = 2(x-1) \, dx \]

After separating the variables, we integrate both sides:\[ \int \frac{1}{y^2} \, dy = \int 2(x-1) \, dx \]This results in solving the integrals, leading toward the exact solution. The ability to separate and integrate helps solve initial value problems more efficiently.

Numerical approximation

Numerical approximation entails estimating the solutions of equations or functions where analytical methods are complex or not applicable. Euler's method is a basic technique for numerical approximation, particularly useful for solving ordinary differential equations with initial conditions.

Euler's method uses a step-by-step approach to estimate the function's value over an interval. By taking discrete steps from an initial value, it calculates successive approximations. The formula for Euler's method is:\[ y_{n+1} = y_n + h \, f(x_n, y_n) \]

In the exercise, different step sizes are used: 0.2, 0.1, and 0.05, which influence the accuracy of the approximations. The results reveal that smaller step sizes provide approximations closer to the exact solution, highlighting the method's dependency on step size for accuracy.

Overall, numerical approximation like Euler's method is crucial when closed-form solutions are difficult to obtain.

Initial value problem

An initial value problem (IVP) is a type of differential equation along with specific starting conditions. These are essential to determine a unique solution to the differential equation.

In our example, the problem specifies:- The differential equation: \( y' = 2y^2(x-1) \)- Initial condition: \( y(2) = -\frac{1}{2} \)

This setup allows us to find solution functions that not only solve the differential equation but also satisfy the given condition at a specific point. Establishing an IVP enables us to develop a complete understanding of the system behavior starting from the known initial condition. In practice, this is vital for modeling real-world behaviors where the system's initial state is known.

Solutions can then be compared for different x-values to see how the system evolves, such as when using Euler’s method to approximate at various steps until a specific point \(x^*\).