Here, [D frac{d}{dx}] (the differential operator).

Homogeneous Part of the Differential Equation

1. Finding the Homogeneous Solution

First, we need to solve the homogeneous part of the equation:

[D^2 - 4D 4Y 0]D^2 - 4D 4Y 0

This is a second-order linear homogeneous differential equation. We can solve this by finding the roots of the characteristic equation:

[r^2 - 4r 4 0]r^2 - 4r 4 0

Factoring the quadratic equation:

[(r - 2)^2 0](r - 2)^2 0

The roots of the equation are [r 2], which are real and equal. Thus, the complementary function (the general solution to the homogeneous equation) is:

[y_h a e^{-2x} b x e^{-2x}]y_h a e^{-2x} b x e^{-2x}

Non-Homogeneous Part of the Differential Equation

2. Finding the Particular Integral

Next, we need to find the particular integral of the given non-homogeneous differential equation:

[D^2 - 4D 4Y 8x^2)D^2 - 4D 4Y 8x^2)

To find the particular integral, we propose a form for [Y_p] based on the highest power and the coefficients of the non-homogeneous term. In this case, we choose:

[Y_p ax^2 bx c]Y_p ax^2 bx c

Let us substitute this form into the differential equation:

[D^2(ax^2 bx c) 2a]D^2(ax^2 bx c) 2a

[D(2ax b) 2a)D(2ax b) 2a)

[4a(2ax b) 8ax 4b]4a(2ax b) 8ax 4b)

Equating coefficients, we get:

From [2a 8x], we obtain [a 4x].

From [4b 8x^2], we obtain [b 2x^2].

From [4c 0], we obtain [c 0].

Thus, the particular integral is:

[Y_p 2x^2 - 4x 3]Y_p 2x^2 - 4x 3

General Solution of the Differential Equation

The general solution of the differential equation is the sum of the complementary function and the particular integral:

[y y_h Y_p a e^{-2x} b x e^{-2x} 2x^2 - 4x 3)y y_h Y_p a e^{-2x} b x e^{-2x} 2x^2 - 4x 3)

Applying Initial Conditions

To determine the specific values of the constants [a] and [b], we use the initial conditions given in the problem:

[y1] at [x0]y1) at (x0

Substituting [y a e^{-2(0)} b (0) e^{-2(0)} 2(0)^2 - 4(0) 3 1], we get:

[a 3 1]a 3 1

Solving for [a], we obtain:

[a -2]a -2

Conclusion

In summary, the solution to the differential equation [D^2 - 4D 4Y 8x^2] (with the given initial condition) is:

[y -2 e^{-2x} b x e^{-2x} 2x^2 - 4x 3]y -2 e^{-2x} b x e^{-2x} 2x^2 - 4x 3

To find the value of [b], we would differentiate the solution and substitute the appropriate condition, but since the problem does not provide a second condition, we leave it in terms of [b].