We are given the differential equation:

$\frac{dy}{dx} = x - y^2$

with the initial condition $y(0) = 1$. We are tasked with finding the approximate value of $y(0.1)$ using Taylor's method.

Step 1: Taylor's series formula

Taylor's method expands $y(x)$ as a Taylor series around a point $x_0$:

$y(x) = y(x_0) + h y'(x_0) + \frac{h^2}{2!} y''(x_0) + \frac{h^3}{3!} y'''(x_0) + \cdots$

where $h$ is the step size, and $y', y'', y'''$ are the successive derivatives of $y(x)$.

Step 2: Compute successive derivatives

We need to compute $y'(x)$, $y''(x)$, and possibly higher derivatives. Let's start by calculating the necessary derivatives.

1. First derivative: $y'(x) = x - y^2$

   At $x = 0$ and $y(0) = 1$: $y'(0) = 0 - 1^2 = -1$

2. Second derivative (use the chain rule): Differentiate $y'(x) = x - y^2$ with respect to $x$: $y''(x) = 1 - 2y \cdot y'$ Substituting $y(0) = 1$ and $y'(0) = -1$: $y''(0) = 1 - 2(1)(-1) = 1 + 2 = 3$

3. Third derivative: Differentiate $y''(x) = 1 - 2y \cdot y'$ with respect to $x$: $y'''(x) = -2(y' \cdot y' + y \cdot y'')$ Substituting $y(0) = 1$, $y'(0) = -1$, and $y''(0) = 3$: $y'''(0) = -2\left((-1)^2 + 1 \cdot 3\right) = -2(1 + 3) = -8$

Step 3: Taylor expansion up to the third term

Using the Taylor expansion formula up to the second-order term:

$y(0.1) = y(0) + h y'(0) + \frac{h^2}{2!} y''(0) + \frac{h^3}{3!} y'''(0)$

where $h = 0.1$, and substituting the values we computed:

$y(0.1) = 1 + 0.1(-1) + \frac{(0.1)^2}{2}(3) + \frac{(0.1)^3}{6}(-8)$

Simplifying:

$y(0.1) = 1 - 0.1 + \frac{0.01}{2}(3) + \frac{0.001}{6}(-8)$

$y(0.1) = 1 - 0.1 + 0.015 - 0.00133$

$y(0.1) \approx 0.91367$

Thus, the approximate value of $y(0.1)$ is $\mathbf{0.91367}$.

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Do you want any details about the steps or have further questions?

Here are 5 related questions:

1. How can we extend Taylor's method for larger values of $x$?
2. What is the error bound when using Taylor's method for approximations?
3. How does Taylor's method compare to Euler's method for solving ODEs?
4. Can we solve this differential equation using an exact method, like separation of variables?
5. What are the higher-order terms in the Taylor expansion, and how do they influence the accuracy?

Tip: Higher-order terms in the Taylor series improve accuracy but require calculating more derivatives, which may get complex for non-linear differential equations.