Differential Equation:
The given differential equation is D²y/Dx² + 4y = 4sec²(2x).

Particular Solution:
To find the particular solution of the differential equation, we can use the method of undetermined coefficients.
Let's assume that the particular solution has the form y_p = Asec²(2x), where A is a constant to be determined.

First Derivative:
Now, let's find the first derivative of y_p:
dy_p/dx = d/dx (Asec²(2x))

Using the chain rule, we can differentiate sec²(2x) with respect to 2x and multiply by the derivative of 2x with respect to x:
= A * 2sec(2x) * (d/d(2x) sec²(2x)) * (d(2x)/dx)
= 2Asec(2x) * 2sec(2x) * (d(2x)/dx)
= 4Asec²(2x) * 2

Simplifying the expression, we have:
dy_p/dx = 8Asec²(2x)

Second Derivative:
Next, let's find the second derivative of y_p:
d²y_p/dx² = d/dx (8Asec²(2x))

Using the chain rule and the power rule for differentiation, we can simplify the expression as follows:
= 8A * (d/dx sec²(2x)) * (d(2x)/dx)

Using the chain rule and the quotient rule, we can differentiate sec²(2x) with respect to 2x:
= 8A * (2sec(2x) * (d/d(2x) sec²(2x)) * (d(2x)/dx))

Simplifying further, we have:
= 8A * 2sec(2x) * 2sec(2x) * (d(2x)/dx)
= 32Asec²(2x) * 2

Simplifying again, we obtain:
d²y_p/dx² = 64Asec²(2x)

Substitution:
Now, substitute the second derivative and the particular solution into the given differential equation:
64Asec²(2x) + 4Asec²(2x) = 4sec²(2x)

Simplifying the equation, we have:
68Asec²(2x) = 4sec²(2x)

Cancellation:
Since both sides of the equation have the same factor of sec²(2x), we can cancel it out:
68A = 4

Solution:
Solving for A, we have:
A = 4/68
A = 1/17

Therefore, the particular solution of the given differential equation is:
y_p = (1/17)sec²(2x)