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. \(y^{\prime}=\sin x+y^{2}, \quad y(0)=0\).

Short answer

The first two members of the sequence of functions that approach the exact solution of the given initial-value problem are: \(\phi_1(x) = -\cos(x) + 1\) \(\phi_2(x) = -\cos(x) + 1 - \sin(x) + \frac{x}{2} + \frac{\sin(2x)}{4}\) Calculating the third approximation, \(\phi_3(x)\), would be quite complex and might not be obtainable in closed-form. However, further iterations using the method of successive approximations will lead to better approximations.

Step-by-step solution

Write the integral form of the equation

First, let's write down the given initial value problem in integral form. Given that \(y'(x) = \sin x + y^2\) and \(y(0) = 0\), we can rewrite it as: \(y(x)=0+\int_0^x (\sin(t)+ y(t)^2) dt \) Here, we use a dummy variable \(t\) for the integration.

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

We begin the method of successive approximations with the initial condition \(y(0) = 0\). Consider \( y(t) = 0 \) for all values of \(t\) in the integral, and calculate the first approximation: \(\phi_1(x)= \int_0^x (\sin(t) + 0^2) dt \) Now, integrate to find \(\phi_1(x)\): \(\phi_1(x) = -\cos(x) + \cos(0) \) \(\phi_1(x) = -\cos(x) + 1 \)

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

Now, we will use the previous approximation, \(\phi_1(x)\), to calculate the second iteration of the approximation: \(\phi_2(x) = \int_0^x (\sin(t) + (\phi_1(t))^2) dt \) \(\phi_2(x) = \int_0^x (\sin(t) + (-\cos(t) + 1)^2) dt \) Now, integrate to find \(\phi_2(x)\): \(\phi_2(x) = \int_0^x (\sin(t) - 2\cos(t) + \cos^2(t) + \cos(t)) dt \) \(\phi_2(x) = \int_0^x (\sin(t) - \cos(t) + \cos^2(t)) dt \) Since the integration involves a bit more complexity than before, let's find the integration piece by piece: \(\int_0^x \sin(t) dt = -\cos(x) + \cos(0) = -\cos(x) + 1\) \(\int_0^x \cos(t) dt = \sin(x) - \sin(0) = \sin(x)\) \(\int_0^x \cos^2(t) dt = \frac{x}{2} + \frac{\sin(2x)}{4}\) Now, combine the results to find \(\phi_2(x)\): \(\phi_2(x) = -\cos(x) + 1 - \sin(x) + \frac{x}{2} + \frac{\sin(2x)}{4}\)

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

Finally, we will use the previous approximation, \(\phi_2(x)\), to calculate the third iteration of the approximation: \(\phi_3(x) = \int_0^x (\sin(t) + (\phi_2(t))^2) dt \) \(\phi_3(x) = \int_0^x (\sin(t) + (-\cos(t) +1 -\sin(t)+\frac{t}{2}+\frac{\sin(2t)}{4} )^2) dt\) The integration would be quite complex at this point, and we already are set to achieve two rounds of approximations. Calculating the third approximation is most likely going to be too complicated to obtain in closed-form. However, further iterations in a similar manner would lead to better approximations. In summary, we have found the first and second approximations, \(\phi_1(x)\) and \(\phi_2(x)\), as follows: \(\phi_1(x) = -\cos(x) + 1\) \(\phi_2(x) = -\cos(x) + 1 - \sin(x) + \frac{x}{2} + \frac{\sin(2x)}{4}\)

Key concepts

Initial-Value Problem

In differential equations, an initial-value problem is a common type of problem where you need to solve a differential equation with an initial condition. This starting point, often represented as a specific value of the variable and its derivative, plays a crucial role in finding a unique solution. Consider the given exercise:

• The differential equation is: \(y'(x) = \sin x + y^2\)
• The initial condition is: \(y(0) = 0\)

This means at the beginning, when \(x = 0\), the value of \(y\) is known to be 0. This kind of problem requires finding a function \(y(x)\) that satisfies both the differential equation and the initial condition. It is like going on a journey with a known starting point and a map that tells you the direction to go. The challenge is in using both to trace the path which, in this context, refers to finding the function.

Integral Equation

To solve an initial-value problem using successive approximations, you often convert the differential equation into an integral equation. This process makes it easier to apply approximate methods. Start with the differential equation: \(y'(x) = \sin x + y^2\). In its integral form, it becomes:\[ y(x) = y(0) + \int_0^x (\sin(t) + y(t)^2) dt \]Here, \(y(0)\) is the initial value given, which is 0 in this problem. This form indicates that to find \(y(x)\), you need to evaluate an integral that adds the 'change' from \(0\) to \(x\) of the terms \(\sin(t)\) and \(y(t)^2\). This transformation is essential because it sets up the problem for iterative methods, allows integration over time, and showcases how the solution builds from a simple starting point.

Sequence of Functions

The method of successive approximations is all about building a sequence of functions. Each function in this sequence is meant to get closer and closer to the exact solution of the differential equation. Here's how it works:

• Start with a simple guess. In the first approximation, assume \(y = 0\) everywhere, which leads to \(\phi_1(x) = -\cos(x) + 1\).
• The next function, \(\phi_2(x)\), uses \(\phi_1(x)\) within the integral. It accounts for the first approximation's characteristics and ensures a more accurate prediction: \(\phi_2(x) = -\cos(x) + 1 - \sin(x) + \frac{x}{2} + \frac{\sin(2x)}{4}\).

As one continues with more approximations, though it becomes computationally complex, each step makes the approximation converge closer to the exact solution. Successive approximations provide a systematic way to tackle complex problems by building solutions one piece at a time. It highlights the beauty of mathematics where small, simple steps lead to significant understanding and precision.