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, the first three members of the sequence of functions that approaches the exact solution of the initial-value problem \(y^{\prime} = xy\), with \(y(0) = 1\) are: \[ \phi_1(x) = \frac{x^2}{2} + 1, \quad \phi_2(x) = \frac{x^4}{8} + \frac{x^2}{2} + 1, \quad \phi_3(x) = \frac{x^6}{48} + \frac{x^4}{8} + \frac{x^2}{2} + 1 \]

Step-by-step solution

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

For the method of successive approximations, we will start with the initial function \(\phi_{0}(x) = 1\), which corresponds to our initial condition \(y(0) = 1\).

Find \(\phi_1(x)\) using the method of successive approximations

Now, we will find \(\phi_1(x)\), by integrating \(x \phi_0(x)\) with respect to x. \[ \phi_1(x) = \int x\phi_0(x) dx = \int x(1) dx =\frac{x^2}{2} + C_1 \] We determine the constant \(C_1\) using the initial condition, \(y(0) = 1\): \[\phi_1(0) = \frac{0^2}{2} + C_1 = 1 \implies C_1 = 1\] So, \(\phi_1(x) = \frac{x^2}{2} + 1\).

Find \(\phi_2(x)\) using the method of successive approximations

We repeat the process by finding \(\phi_2(x)\), by integrating \(x\phi_1(x)\) with respect to x. \[ \phi_2(x) = \int x(\frac{x^2}{2} + 1) dx = \int (\frac{x^3}{2} + x) dx = \frac{x^4}{8} + \frac{x^2}{2} + C_2 \] Again, we determine the constant \(C_2\) using the initial condition, \(y(0) = 1\): \[\phi_2(0) = \frac{0^4}{8} + \frac{0^2}{2} + C_2 = 1 \implies C_2 = 1\] So, \(\phi_2(x) = \frac{x^4}{8} + \frac{x^2}{2} + 1\).

Find \(\phi_3(x)\) using the method of successive approximations

Finally, let's find \(\phi_3(x)\), by integrating \(x\phi_2(x)\) with respect to x. \[ \phi_3(x) = \int x(\frac{x^4}{8} + \frac{x^2}{2} + 1) dx = \int (\frac{x^5}{8} + \frac{x^3}{2} + x) dx = \frac{x^6}{48} + \frac{x^4}{8} + \frac{x^2}{2} + C_3 \] We determine the constant \(C_3\) using the initial condition, \(y(0) = 1\): \[\phi_3(0) = \frac{0^6}{48} + \frac{0^4}{8} + \frac{0^2}{2} + C_3 = 1 \implies C_3 = 1\] Thus, \(\phi_3(x) = \frac{x^6}{48} + \frac{x^4}{8} + \frac{x^2}{2} + 1\). In conclusion, the first three members of the sequence of functions using the method of successive approximations are: \[ \phi_1(x) = \frac{x^2}{2} + 1,\quad \phi_2(x) = \frac{x^4}{8} + \frac{x^2}{2} + 1,\quad \phi_3(x) = \frac{x^6}{48} + \frac{x^4}{8} + \frac{x^2}{2} + 1 \]

Key concepts

Initial-Value Problems

In mathematics, especially in the context of differential equations, an **initial-value problem** involves finding a function that satisfies a differential equation along with a specified value, called the initial condition. This initial condition typically gives the value of the unknown function at a certain point, often at zero, like in our example. Knowing this initial condition helps in uniquely determining a solution that fits both the differential equation and the given point's value.

Consider the differential equation given in the exercise: \(y' = xy\), where \(y(0) = 1\). Here, the differential equation describes the relationship between a function \(y\) and its derivative \(y'\). The initial condition \(y(0) = 1\) specifies that when \(x = 0\), the function \(y\) takes on the value of 1.

• This ensures we find a specific and unique solution, rather than a general solution that might include arbitrary constants.
• Initial-value problems are crucial because many physical processes start from a specific initial state.

Successive Approximations

The **method of successive approximations**, also known as Picard iteration, is a constructive technique to approach the solution of an initial-value problem. We systematically build more accurate approximations of the solution using iterations. Each step relies on integrating the expression derived from the previous approximation, often incorporating initial conditions along the way to refine results.
In our exercise, we start with an initial guess \( \phi_0(x) = 1 \), matching the initial condition \( y(0) = 1 \). This is the foundation upon which later approximations are built. Then, we proceed by plugging this initial guess into the integral derived from rearranging the differential equation:

• For \( \phi_1(x) \), integrate \( x \phi_0(x) \).
• For \( \phi_2(x) \), use \( x \phi_1(x) \).
• This pattern continues, refining approximation by using \( x \phi_{n}(x) \).

Each approximation, or "phi", becomes progressively closer to the actual solution, harnessing the initial condition to adjust the constant of integration accurately each time.

Sequence of Functions

When solving differential equations using **successive approximations**, we construct a **sequence of functions** \( \phi_0(x), \phi_1(x), \phi_2(x), ... \). Each function in this sequence is an improved approximation of the solution to the differential equation. The sequence is designed to converges towards the exact solution as more iterations are applied.

• The first function \( \phi_0(x) \) is the simplest, often chosen to match the initial condition exactly.
• Subsequent functions \( \phi_1(x), \phi_2(x), ... \) are calculated by successively integrating the expression from the previous function, while ensuring each fulfills the initial condition.
• In theory, as you continue this process, the functions in the sequence will approach the exact analytical solution, making the approximation more accurate with each step.

Understanding the sequence helps visualize how an iterative process refines an estimate, turning an approximate guess into a precise value.