Word Count : 1337

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To solve the initial value problem \( y' = x + \cos(y) \), \( y(0) = 0 \) using the fourth-order Taylor series method with \( h = 0.2 \), we'll iteratively approximate the solution.

First, let's write down the Taylor series expansion up to the fourth order for \( y(x+h) \) around \( x \):

\[
y(x+h) = y(x) + hy'(x) + \frac{h^2}{2!}y''(x) + \frac{h^3}{3!}y'''(x) + \frac{h^4}{4!}y''''(x) + O(h^5)
\]

Given \( y' = x + \cos(y) \), we need to find \( y'' \), \( y''' \), and \( y'''' \) to the appropriate orders of differentiation.

Differentiating \( y' \) with respect to \( x \), we get:

\[
y'' = 1 - \sin(y)\frac{dy}{dx}
\]

Differentiating \( y'' \) with respect to \( x \), we get:

\[
y''' = -\cos(y)\left(\frac{dy}{dx}\right)^2 - \sin(y)\frac{d^2y}{dx^2}
\]

Differentiating \( y''' \) with respect to \( x \), we get:

\[
y'''' = -2\cos(y)\frac{dy}{dx}\frac{d^2y}{dx^2} - \sin(y)\frac{d^3y}{dx^3}
\]

Now, plug these expressions back into the Taylor series expansion:

\[
y(x+h) = y(x) + h\left(x + \cos(y)\right) + \frac{h^2}{2!}\left(1 - \sin(y)\frac{dy}{dx}\right) + \frac{h^3}{3!}\left(-\cos(y)\left(\frac{dy}{dx}\right)^2 - \sin(y)\frac{d^2y}{dx^2}\right) + \frac{h^4}{4!}\left(-2\cos(y)\frac{dy}{dx}\frac{d^2y}{dx^2} - \sin(y)\frac{d^3y}{dx^3}\right)
\]

Now, we'll use this formula to iterate from \( x = 0 \) up to \( x = 1 \) with \( h = 0.2 \) to find the approximate values of \( y(x) \). We'll start with \( y(0) = 0 \), then use the previous \( y(x) \) and \( \frac{dy}{dx} \) to compute \( y(x+h) \) at each step.

We'll start with \( y(0) = 0 \) and compute \( y(x) \) for \( x = 0.2, 0.4, 0.6, 0.8, 1.0 \) using the given step size \( h = 0.2 \).

Given:
- \( h = 0.2 \)
- \( y(0) = 0 \)

We'll use the formula:

\[
y(x+h) = y(x) + h\left(x + \cos(y)\right) + \frac{h^2}{2!}\left(1 - \sin(y)\frac{dy}{dx}\right) + \frac{h^3}{3!}\left(-\cos(y)\left(\frac{dy}{dx}\right)^2 - \sin(y)\frac{d^2y}{dx^2}\right) + \frac{h^4}{4!}\left(-2\cos(y)\frac{dy}{dx}\frac{d^2y}{dx^2} - \sin(y)\frac{d^3y}{dx^3}\right)
\]

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