We will use Picard’s method to approximate the solution to the differential equation:

$\frac{dy}{dx} = x - y$

with the initial condition:

$y(0) = 1$

and find an approximation for $y(0.1)$.

Step 1: Picard’s Iteration Formula

Picard's method constructs successive approximations by integrating iteratively.

$y_{n+1} = 1 + \int_0^x (t - y_n) dt$

First Approximation: $y_0(x)$

We start with the initial function from the given condition:

$y_0(x) = 1$

Second Approximation: $y_1(x)$

$y_1(x) = 1 + \int_0^x (t - 1) dt$

$= 1 + \left[ \frac{t^2}{2} - t \right]_0^x$

$= 1 + \left( \frac{x^2}{2} - x \right)$

$= 1 - x + \frac{x^2}{2}$

Third Approximation: $y_2(x)$

$y_2(x) = 1 + \int_0^x \left( t - \left(1 - t + \frac{t^2}{2} \right) \right) dt$

$= 1 + \int_0^x \left( t - 1 + t - \frac{t^2}{2} \right) dt$

$= 1 + \int_0^x \left( 2t - 1 - \frac{t^2}{2} \right) dt$

Integrating term by term:

$1 + \left[ t^2 - t - \frac{t^3}{6} \right]_0^x$

$= 1 + \left( x^2 - x - \frac{x^3}{6} \right)$

$= 1 - x + x^2 - \frac{x^3}{6}$

Step 2: Compute $y(0.1)$ Using $y_2(x)$

$y_2(0.1) = 1 - 0.1 + (0.1)^2 - \frac{(0.1)^3}{6}$

$= 1 - 0.1 + 0.01 - \frac{0.001}{6}$

$= 1 - 0.1 + 0.01 - 0.0001667$

$\approx 0.909833$

Thus, the approximate solution using Picard’s second iteration at $x = 0.1$ is:

$y(0.1) \approx 0.9098$

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Would you like a higher-order approximation or a more detailed explanation?

Related Questions

1. How does Picard’s iteration method differ from Euler’s method?
2. Can Picard’s method be used to solve nonlinear differential equations?
3. What are the advantages and limitations of Picard’s method?
4. How does the number of iterations affect the accuracy of the solution?
5. Can Picard’s method be implemented computationally for complex equations?

Tip

Picard’s method is useful for proving the existence and uniqueness of solutions to differential equations, as it builds successively better approximations to the true solution.