SOLVE (D²+2D+1)=X²+2X+1AND FIND COMPLEMENTARY FUNCTION AND PARTICULAR INTEGRAL OF THE DIFFERENTIAL EQUATION

To solve the equation \( D^2 + 2D + 1 = X^2 + 2X + 1 \), we first need to clarify what \( D \) represents. In this context, \( D \) is the differential operator, which means \( D^n \) indicates the \( n \)-th derivative with respect to \( x \). The left side of the equation can be rewritten as a differential operator acting on a function, while the right side is a polynomial expression. Let's break this down step by step.

Step 1: Understanding the Differential Operator

The operator \( D^2 + 2D + 1 \) can be factored as follows:

• Recognize that \( D^2 + 2D + 1 = (D + 1)^2 \).

This means we are looking for a function \( y \) such that:

(D + 1)^2 y = X^2 + 2X + 1.

Step 2: Finding the Complementary Function

The complementary function (CF) is found by solving the homogeneous part of the equation:

(D + 1)^2 y = 0.

The characteristic equation corresponding to this is:

r^2 + 2r + 1 = 0.

This factors to:

(r + 1)^2 = 0,

which has a double root at r = -1. Therefore, the complementary function is:

y_c = C_1 e^{-x} + C_2 x e^{-x},

where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.

Step 3: Finding the Particular Integral

Next, we need to find a particular integral (PI) for the non-homogeneous equation:

(D + 1)^2 y = X^2 + 2X + 1.

Since the right-hand side is a polynomial of degree 2, we can assume a particular solution of the form:

y_p = Ax^2 + Bx + C.

Now, we will apply the operator (D + 1)^2 to \( y_p \):

• First, calculate \( D y_p = 2Ax + B \).
• Then, \( D^2 y_p = 2A \).
• Now substitute into (D + 1)^2 y_p:

(D + 1)^2 y_p = D^2 y_p + 2D y_p + y_p = 2A + 2(2Ax + B) + (Ax^2 + Bx + C).

Combine like terms:

y_p = Ax^2 + (4A + B)x + (2A + 2B + C).

Set this equal to the right-hand side \( X^2 + 2X + 1 \):

Ax^2 + (4A + B)x + (2A + 2B + C) = x^2 + 2x + 1.

From here, we can equate coefficients:

• For \( x^2 \): A = 1.
• For \( x \): 4A + B = 2.
• For the constant term: 2A + 2B + C = 1.

Substituting A = 1 into the second equation gives:

4(1) + B = 2 → B = -2.

Now substitute A and B into the third equation:

2(1) + 2(-2) + C = 1 → 2 - 4 + C = 1 → C = 3.

Thus, the particular integral is:

y_p = x^2 - 2x + 3.

Final Solution

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

y = y_c + y_p = C_1 e^{-x} + C_2 x e^{-x} + x^2 - 2x + 3.

This solution combines the behavior of the system described by the homogeneous part with the specific response to the polynomial input on the right side. If you have any further questions or need clarification on any part of this process, feel free to ask!