11 Let \( y=g(x) \) be the solution to the differential equation \( \frac{d y}{d x}=x y-2 \) with the initial condition \( g(2)=-1 \). What is the approximation for \( g(1) \) if Euler's method is used, starting at \( x=2 \) with two steps of equal size?

To apply Euler's method to approximate \( g(1) \), starting at \( x = 2 \) with two steps of equal size, we first determine the step size. We want to find \( g(1) \) and start from \( x=2 \) to \( x=1 \). The interval from \( 2 \) to \( 1 \) has a length of \( 1 \), so for two equal steps, the step size \( h \) is: \[ h = \frac{2 - 1}{2} = 0.5 \] Now we know that we will calculate approximations at \( x=2 \) and \( x=1.5 \), and then at \( x=1 \). 1. Starting point: \( g(2) = -1 \) 2. The derivative from the differential equation gives us \( \frac{dy}{dx} = xy - 2 \). At \( x = 2 \): \[ \frac{dy}{dx} = 2(-1) - 2 = -2 - 2 = -4 \] Using Euler's method to find \( g(1.5) \): \[ g(1.5) \approx g(2) + h \cdot \frac{dy}{dx} = -1 + 0.5 \cdot (-4) = -1 - 2 = -3 \] Now we use the new value \( g(1.5) = -3 \) to find \( g(1) \): 3. At \( x = 1.5 \): \[ \frac{dy}{dx} = 1.5(-3) - 2 = -4.5 - 2 = -6.5 \] Using this to calculate \( g(1) \): \[ g(1) \approx g(1.5) + h \cdot \frac{dy}{dx} = -3 + 0.5 \cdot (-6.5) = -3 - 3.25 = -6.25 \] Thus, the approximation for \( g(1) \) using Euler's method is \[ \boxed{-6.25} \]