Euler Method Steps for Differential Equation

We need to solve the differential equation $\frac{dy}{dx} = x + \ln(y)$ with the initial condition $y(0) = 2$ using Euler's method. The step size is $h = 0.2$, and we want to find the approximate value of $y$ at $x = 0.4$. Euler's method formula is:

$y_{n+1} = y_n + h \cdot f(x_n, y_n)$

Where $f(x, y) = x + \ln(y)$.

The initial point is $(x_0, y_0) = (0, 2)$. We need two steps to reach $x = 0.4$. ($x_1 = 0.2, x_2 = 0.4$).

Step 1: Calculate $y_1$ at $x_1 = 0.2$

• $x_0 = 0$, $y_0 = 2$, $h = 0.2$.
• Calculate $f(x_0, y_0) = f(0, 2) = 0 + \ln(2)$.
• Using $\ln(2) \approx 0.6931$.
• $f(0, 2) \approx 0.6931$.
• Apply the formula: $y_1 = y_0 + h \cdot f(x_0, y_0)$.
• $y_1 \approx 2 + 0.2 \cdot (0.6931)$.
• $y_1 \approx 2 + 0.13862$.
• $y_1 \approx 2.13862$.

Step 2: Calculate $y_2$ at $x_2 = 0.4$

• $x_1 = 0.2$, $y_1 \approx 2.13862$, $h = 0.2$.
• Calculate $f(x_1, y_1) = f(0.2, 2.13862) = 0.2 + \ln(2.13862)$.
• Using $\ln(2.13862) \approx 0.7600$.
• $f(0.2, 2.13862) \approx 0.2 + 0.7600 = 0.9600$.
• Apply the formula: $y_2 = y_1 + h \cdot f(x_1, y_1)$.
• $y_2 \approx 2.13862 + 0.2 \cdot (0.9600)$.
• $y_2 \approx 2.13862 + 0.19200$.
• $y_2 \approx 2.33062$.

The approximate solution at $x = 0.4$ is $y_2 \approx 2.33062$. This value lies between 2.3 and 2.4.