Question

Use i) Euler's Method and ii) the Midpoint Method to determine the given value of \(y\) for the following problems: a. \(\frac{d y}{d x}=2 y, y(0)=2\). Find \(y(1)\) with \(h=0.1\). b. \(\frac{d y}{d x}=x-y, y(0)=1\). Find \(y(2)\) with \(h=0.2\). c. \(\frac{d y}{d x}=x \sqrt{1-y^{2}}, y(1)=0 .\) Find \(y(2)\) with \(h=0.2\).

Short answer

Numerically computed values of \(\ y(1)\), \(\ y(2)\) and \(\ y(2)\) for problems a, b, and c must be calculated through the implementation of both Euler's and Midpoint methods. The exact results will depend significantly on your iteration accuracy but may differ from the true values as both the Euler's and Midpoint methods are approximation methods.

Step-by-step solution

Problem a: Using Euler's Method

Start with the initial condition \(y(0)=2\). The Euler's method iteratively calculates the next \(y\) value by adding \(h\) times the derivative at the point to the current \(y\) value. Do this 10 times with \(h=0.1\) to reach \(x=1\).

Problem a: Applying the Midpoint Method

Start again with the initial condition \(y(0)=2\). The Midpoint Method improves upon Euler's Method by taking the slope at the midpoint of the interval rather than at the beginning. Iteratively calculate the midpoint, evaluate the derivative at the midpoint, and use that to calculate the next \(y\) value. Repeat this 10 times to reach \(x=1\).

Problem b: Using Euler's Method

Start with the initial condition \(y(0)=1\). Using Euler's method, iteratively calculate the next \(y\) value by adding the product of \(h\) and the derivative at each point to the current \(y\) value. Repeat this 10 times with \(h=0.2\) to reach \(x=2\).

Problem b: Applying the Midpoint Method

Start again with the initial condition \(y(0)=1\). Apply the Midpoint Method 10 times to reach \(x=2\).

Problem c: Using Euler's Method

Start with the initial condition \(y(1)=0\). Using Euler's method, iteratively calculate the next \(y\) value by adding the product of \(h\) and the derivative at each point to the current \(y\) value. Repeat this 5 times with \(h=0.2\) to reach \(x=2\).

Problem c: Applying the Midpoint Method

Start again with the initial condition \(y(1)=0\). Apply the Midpoint Method 5 times to reach \(x=2\).

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They describe the rate at which quantities change and are fundamental in expressing the laws of nature, such as physics, engineering, and biology. In our exercise, we deal with first-order differential equations of the form \( \frac{dy}{dx} = f(x, y) \), where \( \frac{dy}{dx} \) represents the rate of change of the variable \( y \) with respect to \( x \) and \( f(x, y) \) is a given function. Solving a differential equation involves finding the function \( y \) that satisfies this equation for given initial conditions.
For example, in problem a, \( \frac{dy}{dx} = 2y \) is an exponential growth equation, suggesting that the rate of change of \( y \) is proportional to \( y \) itself. Whereas, problem b's \( \frac{dy}{dx} = x - y \) indicates a relationship where the change in \( y \) depends on both \( y \) and \( x \) linearly.

Numerical Methods

Numerical methods are techniques used to approximate solutions for mathematical problems that cannot be solved analytically or would be very difficult to solve exactly. These methods provide approximate answers to varying degrees of precision and are a vital tool, especially when dealing with differential equations.
In our context, the methods in question are Euler's Method and the Midpoint Method. Both are used to approximate the solutions of initial value problems in differential equations. Euler's Method is a very straightforward approach that progresses step by step, while the Midpoint Method is a refinement that offers better accuracy by considering the trend at the midpoint of each step.
Numerical methods like these are essential when dealing with real-world problems because they allow us to handle complex equations that represent systems where exact solutions are not feasible to obtain.

Initial Value Problem

An initial value problem is a type of differential equation along with a specific condition, called an initial condition, which specifies the value of the unknown function at a given point. It's essentially stating 'We're starting here, now tell me where I'll end up'. These problems are crucial since they provide a starting point for the solution to the differential equation to be computed.
The initial conditions in our exercise are \( y(0) = 2 \) for the first problem, \( y(0) = 1 \) for the second, and \( y(1) = 0 \) for the third. These give us the necessary starting values from which we can apply numerical methods to predict the behavior of the system described by the differential equation as it evolves over time or space.

Derivative Calculation

Derivative calculation plays a fundamental role in differential equations, as it determines the rate of change of quantities. The derivative of a function at a point gives the slope of the tangent line at that point, which geometrically represents how fast the function is changing.
In Euler's Method, derivative calculation at each step guides the prediction of the function's next value. Meanwhile, the Midpoint Method also relies on derivative calculation, but at the midpoint between the current and next points for a more accurate prediction. These calculations hinge on analytical expressions in simple cases or on numerical differentiation in complex scenarios. For instance, in problem a, the derivative calculation is straightforward since \( \frac{dy}{dx} = 2y \) implies that the slope of \( y \) at any point is two times its current value, guiding us towards the exponential nature of the solution.