Question

Consider the initial-value problem $$ \begin{aligned} y^{\prime} &=x+y \\ y(0) &=0 \end{aligned} $$ By the Picard method of successive approximations, show that the successive approximations \(\phi_{0}, \phi_{1}, \ldots, \phi_{n}, \ldots\) exist for all \(x .\) Compute \(\phi_{0}, \phi_{1}, \phi_{2}, \phi_{3}, \phi_{4}\) and compare with the exact solution.

Short answer

Based on the given initial-value problem \(y'(x) = x + y(x)\) with \(y(0) = 0\), we have used the Picard method of successive approximations to compute the first few approximations (\(\phi_0(x)\), \(\phi_1(x)\), \(\phi_2(x)\), \(\phi_3(x)\), and \(\phi_4(x)\)). We also found the exact solution: \(y(x) = x + x^2 + e^x - e^x\). As we increase 'n' in the Picard method, the approximations get closer to the exact solution, demonstrating that the method is working effectively for this initial-value problem.

Step-by-step solution

Define the initial condition and the formula for approximations

For our specific problem, the initial condition is given as \(y(0) = 0\). The formula for successive approximations will be: $$ \phi_{n+1}(x) = 0 + \int\limits_0^x (t +\phi_n(t)) dt $$

Compute the first few approximations

We will compute the first few approximations using the formula defined in Step 1: - \(\phi_0(x) = y(0) = 0\) - \(\phi_1(x) = \int\limits_0^x (t + \phi_0(t)) dt = \int\limits_0^x t dt = \frac{1}{2}x^2\) - \(\phi_2(x) = \int\limits_0^x (t + \phi_1(t)) dt = \int\limits_0^x (t + \frac{1}{2}t^2) dt = \frac{1}{6}x^3 + \frac{1}{4}x^2\) - \(\phi_3(x) = \int\limits_0^x (t + \phi_2(t)) dt = \int\limits_0^x (t + \frac{1}{6}t^3 + \frac{1}{4}t^2) dt = \frac{1}{8}x^4 + \frac{1}{4}x^3 + \frac{1}{2}x^2\) - \(\phi_4(x) = \int\limits_0^x (t + \phi_3(t)) dt = \int\limits_0^x (t + \frac{1}{8}t^4 + \frac{1}{4}t^3 + \frac{1}{2}t^2) dt = \frac{1}{15}x^5 + \frac{1}{8}x^4 + \frac{1}{4}x^3 + \frac{1}{2}x^2\)

Compute the exact solution

To find the exact solution, we can rewrite the given differential equation as a linear first-order differential equation: $$ y'(x) - y(x) = x $$ The integrating factor can be computed as: $$ e^{\int -1 dx} = e^{-x} $$ Now, we multiply the differential equation by the integrating factor: $$ e^{-x}y'(x) - e^{-x}y(x) = xe^{-x} $$ This is now an exact differential equation. We can then integrate both sides: $$ \int (e^{-x}y'(x) - e^{-x}y(x)) dx = \int x e^{-x} dx $$ Integrating both sides, we get: $$ -e^{-x}y(x) = \frac{-x^2}{2}e^{-x} - xe^{-x} - e^{-x} + C $$ So the exact solution is: $$ y(x) = x + x^2 + e^x + Ce^x $$ Apply the initial condition \(y(0) = 0\): $$ 0 = 0 + 0 + 1 + C $$ Thus, we have \(C = -1\), and the exact solution is: $$ y(x) = x + x^2 + e^x - e^x $$

Compare the approximations with the exact solution

We have calculated in previous steps: - \(\phi_0(x) = 0\) - \(\phi_1(x) = \frac{1}{2}x^2\) - \(\phi_2(x) = \frac{1}{6}x^3 + \frac{1}{4}x^2\) - \(\phi_3(x) = \frac{1}{8}x^4 + \frac{1}{4}x^3 + \frac{1}{2}x^2\) - \(\phi_4(x) = \frac{1}{15}x^5 + \frac{1}{8}x^4 + \frac{1}{4}x^3 + \frac{1}{2}x^2\) The exact solution is: $$ y(x) = x + x^2 + e^x - e^x $$ Comparing the approximations, we can observe that the approximations are getting closer to the exact solution as 'n' increases. It implies that the Picard method of successive approximations is working for this particular initial-value problem.

Key concepts

Initial-Value Problem

An initial-value problem is a type of problem in differential equations where we are not only interested in finding a general solution for the equation, but also a specific solution that satisfies a given initial condition—usually the value of the unknown function and sometimes its derivatives at a certain point.
For example, the problem \begin{aligned}y^{\textprime} &= x + y,\y(0) &= 0\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text{‌​}}\text{\text(Continued...)end{aligned} constitutes an initial-value problem. Solving it means both finding a general solution for the differential equation and making sure that this solution satisfies the initial condition, which is \( y(0) = 0 \) in this case.
Initial-value problems are crucial since they often represent practical situations where initial conditions are known, and we're interested in predicting future behavior of the system.

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations involving derivatives of a function with respect to one variable. In the context of the exercise we're discussing, \( y^{\textprime} = x + y \) is an ODE because it relates the derivative of \( y \) (the function we're seeking) to the variable \( x \) and to \( y \) itself.
ODEs are the backbone of many scientific and engineering disciplines because they can model physical phenomena like motion, growth, decay, and fluctuations. The method we used here, the Picard method of successive approximations, is one of several techniques to tackle ODEs. It systematically refines guesses of the solution, starting from the initial condition and gradually incorporating the behavior dictated by the ODE, until the solutions converge to a function that closely approximates the exact solution.

Exact Solution

An 'exact solution' refers to a solution of a differential equation that is found analytically and satisfies the equation over the interval of interest without requiring approximation.
In our exercise, an exact solution to the ODE \( y^{\textprime} = x + y \) is sought. To find this, an integrating factor is often used to transform a non-exact differential equation into an exact one, which can be integrated directly.
The exact solution derived in the steps considers not only the equation itself but also the initial conditions to determine the constant of integration. The resulting exact solution, \( y(x) = x + x^2 + C e^x \), becomes absolutely precise upon applying the initial condition \( y(0) = 0 \), resulting in \( C = -1 \). This solution is the standard against which approximate solutions are compared. Understanding an exact solution is essential as it provides a benchmark and deeper insight into the behavior of the system described by the differential equation.

Integrating Factor

The integrating factor is a function used to multiply both sides of a non-exact differential equation to make it exact. This technique is especially useful for first-order linear ODEs, where the integrating factor is generally of the form \( e^{\text{∫}P(x)dx} \), with \( P(x) \) being the function multiplying \( y \) in the standard form \( y^{\textprime} + P(x)y = Q(x) \).
In our example, the integrating factor \( e^{-x} \) is used to simplify the equation \( y' - y = x \). After applying the integrating factor, the left-hand side becomes an exact differential, which then makes the equation solvable through direct integration.
Knowing how to determine and apply an integrating factor is instrumental when solving ODEs because it transforms problems that are not easily solvable into a form that can be integrated directly, leading to the exact solution of the equation.