Question

For each of the initial-value problems use the method of successive approximations to find the first three members \(\phi_{1}, \phi_{2}, \phi_{3}\) of a sequence of functions that approaches the exact solution of the problem. \(y^{\prime}=x+y, \quad y(0)=1\).

Short answer

Using the method of successive approximations, we found the first three members of the sequence of functions as follows: \(\phi_1(x) = \frac{1}{2}x^2 + x + 1\) \(\phi_2(x) = \frac{1}{6}x^3 + \frac{3}{4}x^2 + x + 1\) \(\phi_3(x) = \frac{1}{24}x^4 + \frac{1}{4}x^3 + \frac{5}{4}x^2 + x + 1\)

Step-by-step solution

Define the initial function \(\phi_0(x)\)

For the given initial-value problem, we have \(y(0)=1\). Thus, we define the initial function \(\phi_0(x)\) as a constant function that satisfies this condition: \[\phi_0(x) = 1\]

Find the first approximation \(\phi_1(x)\)

To find the first approximation \(\phi_1(x)\), we substitute the initial function \(\phi_0(x)\) into the given differential equation: \(y' = x + \phi_0(x)\) So, we have: \(\phi_1'(x) = x + 1\) Now, integrate both sides with respect to \(x\) to find \(\phi_1(x)\): \(\phi_1(x) = \int (x + 1) dx = \frac{1}{2}x^2 + x + C\) Using the initial condition \(y(0) = 1\), we find the value of the constant \(C\): \(\phi_1(0) = \frac{1}{2}(0)^2 + (0) + C = 1\) Which gives: \(C = 1\) So we have: \(\phi_1(x) = \frac{1}{2}x^2 + x + 1\)

Find the second approximation \(\phi_2(x)\)

To find the second approximation \(\phi_2(x)\), we substitute the first approximation \(\phi_1(x)\) into the given differential equation: \(y' = x + \phi_1(x)\) So we have: \(\phi_2'(x) = x + \frac{1}{2}x^2 + x + 1\) Now, integrate both sides with respect to \(x\) to find \(\phi_2(x)\): \(\phi_2(x) = \int (x + \frac{1}{2}x^2 + x + 1) dx = \frac{1}{6}x^3 + \frac{3}{4}x^2 + x + C\) Again, using the initial condition \(y(0) = 1\), we find the value of the constant \(C\): \(\phi_2(0) = \frac{1}{6}(0)^3 + \frac{3}{4}(0)^2 + (0) + C = 1\) Which gives: \(C = 1\) So we have: \(\phi_2(x) = \frac{1}{6}x^3 + \frac{3}{4}x^2 + x + 1\)

Find the third approximation \(\phi_3(x)\)

Repeat the process to find the third approximation \(\phi_3(x)\) by substituting \(\phi_2(x)\) into the given differential equation: \(y' = x + \phi_2(x)\) So we have: \(\phi_3'(x) = x + \frac{1}{6}x^3 + \frac{3}{4}x^2 + x + 1\) Now, integrate both sides with respect to \(x\) to find \(\phi_3(x)\): \(\phi_3(x) = \int (x + \frac{1}{6}x^3 + \frac{3}{4}x^2 + x + 1) dx = \frac{1}{24}x^4 + \frac{1}{4}x^3 + \frac{5}{4}x^2 + x + C\) Finally, using the initial value \(y(0)=1\), we find the value of the constant \(C\): \(\phi_3(0) = \frac{1}{24}(0)^4 + \frac{1}{4}(0)^3 + \frac{5}{4}(0)^2 + (0) + C = 1\) Which gives: \(C = 1\) Thus, we have: \(\phi_3(x) = \frac{1}{24}x^4 + \frac{1}{4}x^3 + \frac{5}{4}x^2 + x + 1\) The first three approximations are: \[\phi_1(x) = \frac{1}{2}x^2 + x + 1\] \[\phi_2(x) = \frac{1}{6}x^3 + \frac{3}{4}x^2 + x + 1\] \[\phi_3(x) = \frac{1}{24}x^4 + \frac{1}{4}x^3 + \frac{5}{4}x^2 + x + 1\]

Key concepts

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations that involve some ordinary derivatives of a function. The function typically depends on one independent variable, often representing time, and the derivatives represent rates of change. Solving an ODE means finding a function that satisfies the equation.

For instance, the ODE given in the exercise is an example of a first-order ODE, where the term 'first-order' indicates that it involves the first derivative of the function. The equation is linear, as its terms are either constants or linear functions of the function and its derivatives.

The general form of a first-order ODE can be written as:\[\begin{equation}y' = f(x, y),\end{equation}\]where the prime symbol (') indicates a derivative with respect to the independent variable, and the function on the right-hand side, f(x, y), gives the slope of the solution curve at any point (x, y). In our example, f(x, y) = x + y.

Understanding the behavior of solutions to ODEs is critical in many fields of science and engineering as these equations commonly model physical phenomena.

Initial-Value Problem

An initial-value problem is a specific type of differential equation problem where you are given the ODE as well as initial conditions for the unknown function. These conditions specify the value of the function (and possibly some of its derivatives) at a particular point, which is typically the starting point for time-evolution problems.

In the context of the exercise provided, the initial-value problem is framed as follows:\[\begin{equation}y' = x + y, \quad\text{with}\quad y(0) = 1.\end{equation}\]The solution must not only satisfy the differential equation, but it must also pass through the point (0, 1) on the xy-plane—that is, when the independent variable x is zero, the solution y must be one. This point serves as the 'anchor' of the solution, ensuring uniqueness under most circumstances.

Initial-value problems are fundamental in applications where the state of the system is known at a specific time, and its future behavior needs to be predicted.

Approximation of Solutions

In many cases, finding an exact solution to an ODE is not possible or quite challenging. To overcome this, approximation methods such as the method of successive approximations, also known as Picard's iteration, can be employed. This method iteratively refines guesses of the solution.

The steps outlined in the provided solution demonstrate this iterative approach. Starting with a simple function that satisfies the initial condition, \[\begin{equation}\phi_0(x) = 1,\end{equation}\]each step involves substituting the last approximation into the right-hand side of the differential equation and integrating to get a new approximation.

The method of successive approximations is powerful because it not only provides a way to approximate the solution but also shows how the approximations converge towards the true solution. With each iteration, \[\begin{equation}\phi_{n+1}(x)\end{equation}\] becomes a more accurate representation of the actual solution of the ODE, under certain regularity conditions on f(x, y). Understanding this method provides insight into numerical techniques for solving ODEs, which are essential when exact solutions are unobtainable.