For the IVP

                    y′=x−y2,    y(0)=−1,


the value of y(1)
 using Picard’s method after 2
 approximations is

Picard’s method of successive approximations for solving an Initial Value Problem (IVP) involves creating a sequence of functions that converge to the solution of the differential equation. For the given IVP:

\[
y' = x - y^2, \quad y(0) = -1,
\]

the method starts with an initial approximation and improves iteratively.

### Step-by-step process:

1. **First Approximation:**
   Start by using the initial condition directly as the first approximation:
   \[
   y_0(x) = y(0) = -1.
   \]
   
2. **Second Approximation:**
   Now, integrate the equation:
   \[
   y_1(x) = -1 + \int_0^x (t - (-1)^2) \, dt.
   \]
   Simplifying the integrand:
   \[
   y_1(x) = -1 + \int_0^x (t - 1) \, dt = -1 + \left[ \frac{t^2}{2} - t \right]_0^x.
   \]
   So,
   \[
   y_1(x) = -1 + \left( \frac{x^2}{2} - x \right) = \frac{x^2}{2} - x - 1.
   \]

3. **Third Approximation:**
   Next, use the second approximation to improve the result:
   \[
   y_2(x) = -1 + \int_0^x \left( t - \left( \frac{t^2}{2} - t - 1 \right)^2 \right) \, dt.
   \]
   This integral becomes more complex, but for the second approximation, we only need \( y_1(x) \) as the approximation. Therefore, using \( y_1(x) = \frac{x^2}{2} - x - 1 \), let's evaluate \( y_1(1) \):

\[
y_1(1) = \frac{1^2}{2} - 1 - 1 = \frac{1}{2} - 2 = -\frac{3}{2}.
\]

Thus, the value of \( y(1) \) after 2 approximations is \( \boxed{-\frac{3}{2}} \).

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Would you like further details on Picard's method, or do you have any questions?

Here are 5 relative questions to expand on the topic:
1. What are the advantages of using Picard’s method for solving differential equations?
2. How does Picard’s method compare to other numerical methods like Euler’s method?
3. What is the convergence behavior of Picard's method?
4. How does one handle more complex initial value problems using Picard’s method?
5. What role do initial conditions play in iterative methods like Picard's?

**Tip:** Successive approximations may converge slowly for nonlinear equations, so it's crucial to check if an alternative numerical method might be more efficient in certain cases.

For the IVP y′=x−y^2, y(0)=−1, the value of y(1) using Picard’s method after 2 approximations is

Learn how to solve the initial value problem y′=x−y^2 with y(0)=-1 using Picard’s method after 2 approximations. This step-by-step guide demonstrates key differential equation concepts, numerical methods, and how to compute the value of y(1) with successive approximations.

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