To solve the given initial value problem, we can use the method of integrating factors.

Step 1: Write the given differential equation in standard form
The given differential equation is dy/dx = 2x - y.

Step 2: Identify the coefficients
The coefficients of x and y in the given differential equation are 2 and -1, respectively.

Step 3: Find the integrating factor
The integrating factor (IF) is given by e^(∫P(x)dx), where P(x) is the coefficient of x. In this case, P(x) = 2. Therefore, the integrating factor is IF = e^(∫2dx) = e^(2x).

Step 4: Multiply the differential equation by the integrating factor
Multiplying both sides of the differential equation dy/dx = 2x - y by the integrating factor e^(2x), we get e^(2x)dy/dx - e^(2x)y = 2xe^(2x) - ye^(2x).

Step 5: Apply the product rule of differentiation
The left side of the equation can be written as d(ye^(2x))/dx using the product rule of differentiation.

Step 6: Integrate both sides of the equation
Integrating both sides of the equation, we get ∫d(ye^(2x))/dx dx = ∫(2xe^(2x) - ye^(2x)) dx.

Simplifying the integral on the left side, we get ye^(2x) = ∫(2xe^(2x) - ye^(2x)) dx.

Integrating the terms on the right side, we get ye^(2x) = x^2e^(2x) + C, where C is the constant of integration.

Step 7: Solve for y
Solving for y, we divide both sides of the equation by e^(2x), y = x^2 + Ce^(-2x).

Step 8: Apply the initial condition
Using the initial condition y(0) = 1, we can substitute x = 0 and y = 1 into the equation y = x^2 + Ce^(-2x) to find the value of C.

1 = 0^2 + Ce^(-2*0)
1 = 0 + C

Therefore, C = 1.

Step 9: Substitute the value of C
Substituting the value of C into the equation y = x^2 + Ce^(-2x), we get y = x^2 + e^(-2x).

Step 10: Find the value of y at x = ln 2
To find the value of y at x = ln 2, we substitute x = ln 2 into the equation y = x^2 + e^(-2x).

y = (ln 2)^2 + e^(-2*ln 2)
y = ln^2 2 + e^(-ln 4)
y = ln^2 2 + 1/4
y = ln^2 2 + 0.25
y ≈ 0.693 + 0.