To find the general solution of the differential equation using the variation of parameters method, we first need to find the solutions to the associated homogeneous equation.
The associated homogeneous equation is:
d²y/dx² + 4x²y = 0
We are given that two solutions to the associated homogeneous equation are e^(2x). Let's denote these solutions as y₁ = e^(2x) and y₂ = e^(2x).
Now, we can proceed with the variation of parameters method. We assume that the particular solution has the form:
y_p = u₁(x)y₁ + u₂(x)y₂
where u₁(x) and u₂(x) are unknown functions to be determined.
Next, we find the Wronskian of the solutions y₁ and y₂:
W(y₁, y₂) = |y₁ y₂| = |e^(2x) e^(2x)| = 0
The Wronskian is zero, which means we need to multiply one of the solutions by x to obtain linearly independent solutions. Let's multiply y₁ by x:
y₁ = xe^(2x)
Now, we can calculate the modified Wronskian:
W(y₁, y₂) = |xe^(2x) e^(2x)| = 2xe^(2x)
To find the particular solution, we can use the formulas:
u₁(x) = -∫(y₂f(x)) / W(y₁, y₂) dx
u₂(x) = ∫(y₁f(x)) / W(y₁, y₂) dx
where f(x) = 0 in this case.
Substituting the values into the formulas, we have:
u₁(x) = -∫(e^(2x) * 0) / (2xe^(2x)) dx = 0
u₂(x) = ∫(xe^(2x) * 0) / (2xe^(2x)) dx = 0
Therefore, the particular solution is y_p = 0.
The general solution of the differential equation is the sum of the homogeneous solution and the particular solution:
y = y_h + y_p = c₁e^(2x) + c₂e^(2x) + 0
Simplifying, we get:
y = (c₁ + c₂)e^(2x)
So, the general solution of the given differential equation is y = (c₁ + c₂)e^(2x), where c₁ and c₂ are arbitrary constants.