Chapter 8: Problem 4
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. \(y^{\prime}=1+x y^{2}, \quad y(0)=0\).
Short Answer
Expert verified
The first three members of the sequence of functions that approach the exact solution of this initial-value problem are:
\(\phi_1(x) = x\),
\(\phi_2(x) = x + \frac{1}{4}x^4\),
\(\phi_3(x) = x + \frac{1}{4}x^4 + \frac{1}{16}x^8 + \frac{1}{192}x^{12}\).
Step by step solution
01
Understand the method of successive approximations
The method of successive approximations involves iterating multiple times to find an approximation to the solution of the differential equation. In each iteration, we plug the function from the previous step into the \(y\) and \(y'\) parts of the differential equation to form a new function. We repeat this process until we have \(\phi_1\), \(\phi_2\), and \(\phi_3\).
02
Choose the initial function \(\phi_0\)
We first choose an initial function \(\phi_0(x)\) such that it satisfies the given initial condition, i.e., \(\phi_0(0)=0\). A simple function that fulfills this requirement is \(\phi_0(x) = 0\).
03
Compute \(\phi_1\)
Now we will compute \(\phi_1\) by substituting \(\phi_0(x)\) into the \(y\) and \(y'\) parts of the differential equation:
\(y' = 1 + xy^2\)
Here, \(y = \phi_0(x) = 0\), so we will plug in this value:
\(\phi_1'(x) = 1 + x(0)^2\)
\(\phi_1'(x) = 1\)
Now, integrate both sides with respect to \(x\):
\(\phi_1(x) = \int \phi_1'(x) dx = \int 1 dx = x + C_1\)
As per the given initial condition \(y(0) = 0\), we substitute \(\phi_1(0) = 0\) and find the constant \(C_1\):
\(0 = 0 + C_1\)
\(C_1 = 0\)
So, \(\phi_1(x) = x\).
04
Compute \(\phi_2\)
Now, using the same method, we will compute \(\phi_2\):
\(y' = 1 + xy^2\)
Here, \(y = \phi_1(x) = x\), so we will plug in this value:
\(\phi_2'(x) = 1 + x(x^2)\)
\(\phi_2'(x) = 1 + x^3\)
Now, integrate both sides with respect to \(x\):
\(\phi_2(x) = \int \phi_2'(x) dx = \int (1 + x^3) dx = x + \frac{1}{4}x^4 + C_2\)
Again, using the initial condition \(y(0) = 0\), we find the constant \(C_2\):
\(0 = 0 + 0 + C_2\)
\(C_2 = 0\)
So, \(\phi_2(x) = x + \frac{1}{4}x^4\).
05
Compute \(\phi_3\)
Finally, we compute \(\phi_3\) using the same method:
\(y' = 1 + xy^2\)
Here, \(y = \phi_2(x) = x + \frac{1}{4}x^4\), so we plug in this value:
\(\phi_3'(x) = 1 + x \left(x + \frac{1}{4}x^4\right)^2\)
\(\phi_3'(x) = 1 + x^3 + \frac{1}{2}x^7 + \frac{1}{16}x^{11}\)
Now, integrate both sides with respect to \(x\):
\(\phi_3(x) = \int \phi_3'(x) dx = \int \left(1 + x^3 + \frac{1}{2}x^7 + \frac{1}{16}x^{11}\right) dx = x + \frac{1}{4}x^4 + \frac{1}{16}x^8 + \frac{1}{192}x^{12} + C_3\)
Again, using the initial condition \(y(0) = 0\), we find the constant \(C_3\):
\(0 = 0 + 0 + 0 + 0 + C_3\)
\(C_3 = 0\)
So, \(\phi_3(x) = x + \frac{1}{4}x^4 + \frac{1}{16}x^8 + \frac{1}{192}x^{12}\).
The first three members of the sequence of functions that approach the exact solution of this initial-value problem are:
\(\phi_1(x) = x\),
\(\phi_2(x) = x + \frac{1}{4}x^4\),
\(\phi_3(x) = x + \frac{1}{4}x^4 + \frac{1}{16}x^8 + \frac{1}{192}x^{12}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial-Value Problems
Initial-value problems are a specific type of problem in differential equations where the solution is required to pass through a given point on the graph. This involves solving a differential equation with one or more initial conditions specified. In this type of problem, the differential equation describes how a certain function behaves, and the initial condition gives a specific starting point that the function must go through. This ensures that there is a unique solution that fits the criteria given by both the equation and the initial condition. Each initial condition acts like a boundary that must be satisfied and ensures that the sequence of functions, such as our \(\phi_i(x)\) series, begins at the correct starting point. Examples of these problems often arise in physics and engineering, where the initial state of the system is known, and the goal is to determine its future behavior.
Successive Approximations
The method of successive approximations, also known as Picard iteration, is a technique used to iteratively find an approximate solution to a differential equation. It starts with an initial guess that fulfills the given initial conditions and uses this guess to calculate further approximations. In each step, the previous approximation is substituted back into the equation. Through this iterative process, each approximation becomes increasingly closer to the true solution. This method is particularly useful for ensuring that solutions adhere to the required initial conditions and can provide a series of approximations that gradually converge to the true solution. It is often used when direct integration of the differential equation is complex or impossible.
Ordinary Differential Equations
Ordinary Differential Equations (ODEs) are equations that involve functions and their derivatives. They are called "ordinary" because they contain one independent variable, typically time or space, which differentiates them from partial differential equations that involve several independent variables. The general purpose of solving an ODE is to find a function or a set of functions that satisfy the equation throughout its domain. ODEs are widely used to model real-world phenomena such as motion, heat transfer, and population dynamics. The complexity of an ODE can vary significantly depending on the presence of nonlinear terms or the order of the derivative involved. The given exercise involves an ODE with nonlinear terms, creating added complexity and requiring specific methods such as successive approximations to solve.
Integration Techniques
Integration techniques are mathematical procedures for finding antiderivatives of functions, effectively reversing the process of differentiation. These techniques are essential in solving differential equations, as they allow us to integrate the derivative function back to its original form. In our problem, integration is carried out by calculating the antiderivative of each function obtained during the successive approximations. This involves basic integration rules such as power rule, where the integral of \(x^n\) is \(\frac{x^{n+1}}{n+1}\). Correctly applying these rules is crucial to find each successive approximation accurately. Furthermore, matching the result with the initial condition to solve for any constant of integration ensures that each function in the sequence adheres to the specified starting point of the problem.