Use the fourth-order Runge-Kutta method with 𝒉 = 𝟎. 𝟏 to obtain a four-decimal
approximation to 𝒚(𝟏. 𝟐) for the solution of 𝒚′ = 𝒙𝒚𝟐 −𝒚/𝒙, 𝒚(𝟏) = 𝟏.

To use the fourth-order Runge-Kutta method to approximate \( y(1.2) \) for the differential equation \( y' = xy^2 - \frac{y}{x} \) with the initial condition \( y(1) = 1 \) and step size \( h = 0.1 \), we follow these steps:

The fourth-order Runge-Kutta method is given by:

\[ 
y_{n+1} = y_n + \frac{h}{6} (k_1 + 2k_2 + 2k_3 + k_4) 
\]

where

\[
k_1 = f(x_n, y_n) 
\]
\[
k_2 = f\left(x_n + \frac{h}{2}, y_n + \frac{h}{2} k_1\right) 
\]
\[
k_3 = f\left(x_n + \frac{h}{2}, y_n + \frac{h}{2} k_2\right) 
\]
\[
k_4 = f(x_n + h, y_n + hk_3) 
\]

Given the differential equation:

\[
f(x, y) = xy^2 - \frac{y}{x}
\]

Let's compute the values step-by-step from \( x = 1 \) to \( x = 1.2 \) with \( h = 0.1 \).

### Step 1: From \( x = 1 \) to \( x = 1.1 \)

- \( x_0 = 1 \)
- \( y_0 = 1 \)

\[
k_1 = f(1, 1) = 1 \cdot 1^2 - \frac{1}{1} = 0 
\]

\[
k_2 = f\left(1 + \frac{0.1}{2}, 1 + \frac{0.1}{2} \cdot 0\right) = f(1.05, 1) = 1.05 \cdot 1^2 - \frac{1}{1.05} = 1.05 - 0.9524 = 0.0976 
\]

\[
k_3 = f\left(1.05, 1 + \frac{0.1}{2} \cdot 0.0976\right) = f(1.05, 1.00488) = 1.05 \cdot 1.00488^2 - \frac{1.00488}{1.05} \approx 0.1027 
\]

\[
k_4 = f(1.1, 1 + 0.1 \cdot 0.1027) = f(1.1, 1.01027) = 1.1 \cdot 1.01027^2 - \frac{1.01027}{1.1} \approx 0.2091 
\]

\[
y_1 = 1 + \frac{0.1}{6} (0 + 2 \cdot 0.0976 + 2 \cdot 0.1027 + 0.2091) \approx 1.010 
\]

### Step 2: From \( x = 1.1 \) to \( x = 1.2 \)

- \( x_1 = 1.1 \)
- \( y_1 \approx 1.010 \)

\[
k_1 = f(1.1, 1.010) = 1.1 \cdot 1.010^2 - \frac{1.010}{1.1} \approx 0.2091 
\]

\[
k_2 = f\left(1.1 + \frac{0.1}{2}, 1.010 + \frac{0.1}{2} \cdot 0.2091\right) = f(1.15, 1.02045) = 1.15 \cdot 1.02045^2 - \frac{1.02045}{1.15} \approx 0.3210 
\]

\[
k_3 = f\left(1.15, 1.010 + \frac{0.1}{2} \cdot 0.3210\right) = f(1.15, 1.02605) = 1.15 \cdot 1.02605^2 - \frac{1.02605}{1.15} \approx 0.3313 
\]

\[
k_4 = f(1.2, 1.010 + 0.1 \cdot 0.3313) = f(1.2, 1.04313) = 1.2 \cdot 1.04313^2 - \frac{1.04313}{1.2} \approx 0.4526 
\]

\[
y_2 = 1.010 + \frac{0.1}{6} (0.2091 + 2 \cdot 0.3210 + 2 \cdot 0.3313 + 0.4526) \approx 1.0427 
\]

So, the approximate value of \( y(1.2) \) is \( y_2 \approx 1.0427 \).

Would you like more details or have any questions?

Here are some questions you might ask:
1. How do you derive the Runge-Kutta method equations?
2. Can you explain each step of the Runge-Kutta method in more detail?
3. How can I implement the Runge-Kutta method in Python?
4. What are the advantages of the Runge-Kutta method over other numerical methods?
5. How does the step size \( h \) affect the accuracy of the Runge-Kutta method?

**Tip:** When using numerical methods, always consider the trade-off between step size and computational efficiency to balance accuracy and performance.

Learn how to approximate the value of y(1.2) for the differential equation y' = xy^2 - y/x using the fourth-order Runge-Kutta method with a step size of h = 0.1. This problem involves applying numerical methods to solve differential equations, specifically focusing on the Runge-Kutta method. Suitable for advanced undergraduate and graduate students.

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