Dy/dx + y = 2x

Given information:
The initial value problem is given as follows:
dy/dx = 2x - y
y(0) = 1

To find:
The value of y at x = ln2 (rounded off to one decimal place)

Solution:

To solve the given initial value problem, we can use the method of solving first-order linear differential equations.

Step 1: Separate the variables
The given equation is in the form dy/dx = f(x) - g(x)y, where f(x) = 2x and g(x) = 1.
We can rewrite the equation as dy/dx + g(x)y = f(x).

Step 2: Determine the integrating factor
The integrating factor is given by the formula: IF(x) = e^(∫g(x)dx).
In this case, g(x) = 1, so the integrating factor becomes IF(x) = e^(∫1dx) = e^(x).

Step 3: Multiply both sides of the equation by the integrating factor
e^(x)dy/dx + e^(x)y = 2xe^(x).

Step 4: Apply the product rule on the left-hand side of the equation
(d/dx)(e^(x)y) = 2xe^(x).

Step 5: Integrate both sides of the equation
∫(d/dx)(e^(x)y)dx = ∫2xe^(x)dx.

Using the fundamental theorem of calculus, we can integrate the right-hand side of the equation to obtain:
e^(x)y = 2xe^(x) - 2e^(x) + C1,
where C1 is the constant of integration.

Step 6: Solve for y
Divide both sides of the equation by e^(x) to solve for y:
y = 2x - 2 + C1e^(-x).

Step 7: Apply the initial condition
Using the initial condition y(0) = 1, we can substitute x = 0 and y = 1 into the equation and solve for C1:
1 = 2(0) - 2 + C1e^(0),
1 = -2 + C1,
C1 = 3.

Step 8: Substitute the value of C1 back into the equation for y
y = 2x - 2 + 3e^(-x).

Step 9: Calculate the value of y at x = ln2
Substituting x = ln2 into the equation for y, we have:
y = 2(ln2) - 2 + 3e^(-ln2),
y = 2ln2 - 2 + 3(1/2),
y = 2ln2 - 2 + 3/2,
y ≈ 0.693 - 2 + 1.5,
y ≈ 0.193.

Therefore, the value of y at x = ln2 (rounded off to one decimal place) is approximately 0.2.