The question in the image asks to solve the following differential equation using Picard's method:

$\frac{dy}{dx} = x - y \quad \text{for} \quad x = 0.2 \quad \text{given that} \quad y = 1 \quad \text{when} \quad x = 0.$

Picard's Iteration Method:

Picard's method involves constructing successive approximations of the solution starting from an initial guess. The general solution to an initial value problem of the form:

$\frac{dy}{dx} = f(x, y), \quad y(0) = y_0$

can be written as:

$y(x) = y_0 + \int_{x_0}^{x} f(t, y(t)) \, dt$

In this case:

• $f(x, y) = x - y$
• The initial condition is $y(0) = 1$.

Step 1: First Approximation $y_0(x)$

Using the initial condition $y(0) = 1$, the first approximation is the constant function:

$y_0(x) = 1$

Step 2: Second Approximation $y_1(x)$

We substitute $y_0(x) = 1$ into the integral:

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

Solving this integral:

$y_1(x) = 1 + \left[ \frac{t^2}{2} - t \right]_0^x = 1 + \left( \frac{x^2}{2} - x \right)$

Thus, the second approximation is:

$y_1(x) = 1 + \frac{x^2}{2} - x$

Step 3: Third Approximation $y_2(x)$

We now use $y_1(x) = 1 + \frac{x^2}{2} - x$ in the next iteration:

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

Simplifying the expression inside the integral:

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

Now, we solve the integral:

$y_2(x) = 1 + \left[ t^2 - t - \frac{t^3}{6} \right]_0^x = 1 + \left( x^2 - x - \frac{x^3}{6} \right)$

Thus, the third approximation is:

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

Step 4: Further Iterations (Optional)

If desired, we can continue to improve the approximation by repeating the process using $y_2(x)$ in the next step.

Answer for $x = 0.2$

Now, to find the approximate value of $y$ at $x = 0.2$ using $y_2(x)$:

$y_2(0.2) = 1 + (0.2)^2 - 0.2 - \frac{(0.2)^3}{6}$

Let me compute this value.The approximate value of $y$ at $x = 0.2$ using Picard's method and the third approximation $y_2(x)$ is:

$y(0.2) \approx 0.839$

Would you like more details on any of the steps or further clarification?

Here are 5 related questions for further exploration:

1. How does Picard's method compare to Euler's method for solving differential equations?
2. Can Picard's method be applied to non-linear differential equations?
3. What are the limitations of Picard's iteration method?
4. How does the accuracy of Picard's method improve with more iterations?
5. What are some alternative methods for solving first-order differential equations numerically?

Tip: For improved accuracy, consider carrying out more iterations of Picard's method or using higher-order numerical methods.