Find particular Integral of differential equation of D²+D+1 =x³

To find the particular integral of the differential equation given by \( D^2 + D + 1 = x^3 \), we first need to understand what the operator \( D \) represents. In this context, \( D \) denotes differentiation with respect to \( x \). Therefore, the equation can be interpreted as a second-order linear differential equation. Let's break down the steps to find the particular integral.

Understanding the Differential Operator

The operator \( D^2 \) means we will take the second derivative, \( D \) means the first derivative, and the constant term is simply multiplied by the function. The equation can be rewritten as:

\( (D^2 + D + 1)y = x^3 \)

Finding the Particular Integral

To find the particular integral, we need to identify a suitable form for the particular solution based on the right-hand side of the equation, which is \( x^3 \). For polynomial functions, we typically assume a polynomial solution of the same degree. Thus, we can propose:

\( y_p = Ax^3 + Bx^2 + Cx + D \)

Calculating Derivatives

Next, we need to calculate the first and second derivatives of \( y_p \):

• First derivative: \( y_p' = 3Ax^2 + 2Bx + C \)
• Second derivative: \( y_p'' = 6Ax + 2B \)

Substituting into the Differential Equation

Now, we substitute \( y_p \), \( y_p' \), and \( y_p'' \) back into the left-hand side of the differential equation:

\( (D^2 + D + 1)y_p = (6Ax + 2B) + (3Ax^2 + 2Bx + C) + (Ax^3 + Bx^2 + Cx + D) \)

Combining like terms, we get:

\( Ax^3 + (3A + B)x^2 + (6A + 2B + C)x + (2B + C + D) \)

Setting Up the Equation

We want this expression to equal \( x^3 \). Therefore, we can set up the following system of equations by equating coefficients:

• For \( x^3 \): \( A = 1 \)
• For \( x^2 \): \( 3A + B = 0 \)
• For \( x \): \( 6A + 2B + C = 0 \)
• Constant term: \( 2B + C + D = 0 \)

Solving the System of Equations

Substituting \( A = 1 \) into the second equation:

\( 3(1) + B = 0 \Rightarrow B = -3 \)

Now substituting \( A \) and \( B \) into the third equation:

\( 6(1) + 2(-3) + C = 0 \Rightarrow 6 - 6 + C = 0 \Rightarrow C = 0 \)

Finally, substituting \( B \) and \( C \) into the last equation:

\( 2(-3) + 0 + D = 0 \Rightarrow -6 + D = 0 \Rightarrow D = 6 \)

Particular Integral Result

Now we can write the particular integral:

\( y_p = x^3 - 3x^2 + 6 \)

This is the particular integral of the differential equation \( D^2 + D + 1 = x^3 \). If you have any further questions or need clarification on any of the steps, feel free to ask!