To find the particular integral (PI) for the ordinary differential equation (ODE) $y'' + y' = x^2+2x+4$, we use the method of undetermined coefficients.

Analyzing the Differential Equation

The given ODE is a second-order linear non-homogeneous differential equation with constant coefficients:

The right-hand side (RHS) is a polynomial of degree 2.

Finding the Correct Form of the Particular Integral

First, we find the roots of the auxiliary equation associated with the complementary function (CF):

Auxiliary equation: $m^2 + m = 0$

Factoring, we get $m(m+1) = 0$.

The roots are $m_1 = 0$ and $m_2 = -1$.

Since the RHS is a polynomial $P(x) = x^2+2x+4$, we would normally guess the PI as a polynomial of the same degree, $y_p = Ax^2 + Bx + C$. However, one of the roots of the auxiliary equation is $m=0$, which corresponds to the $x^0$ (constant) term in the polynomial on the RHS. Because $m=0$ is a root of the auxiliary equation (with multiplicity 1), we must multiply our initial guess by $x^1$.

Therefore, the correct form for the particular integral is:

Calculating Derivatives of the Particular Integral

We need to calculate the first and second derivatives of $y_p$:

• First derivative: $y_p' = \frac{d}{dx}(Ax^3 + Bx^2 + Cx) = 3Ax^2 + 2Bx + C$
• Second derivative: $y_p'' = \frac{d}{dx}(3Ax^2 + 2Bx + C) = 6Ax + 2B$

Substituting into the ODE

Now, substitute $y_p''$ and $y_p'$ into the original differential equation $y'' + y' = x^2+2x+4$:

$(6Ax + 2B) + (3Ax^2 + 2Bx + C) = x^2+2x+4$

Group terms by powers of $x$:

Equating Coefficients

To find the values of A, B, and C, we equate the coefficients of the corresponding powers of $x$ on both sides:

1. Coefficient of $x^2$: $3A = 1 \implies A = \frac{1}{3}$
2. Coefficient of $x$: $6A + 2B = 2$ Substitute $A = \frac{1}{3}$: $6\left(\frac{1}{3}\right) + 2B = 2$ $2 + 2B = 2$ $2B = 0 \implies B = 0$
3. Constant term: $2B + C = 4$ Substitute $B = 0$: $2(0) + C = 4$ $C = 4$

Determining the Particular Integral

Substitute the calculated values of $A$, $B$, and $C$ back into the form of $y_p$:

$y_p = Ax^3 + Bx^2 + Cx$

$y_p = \frac{1}{3}x^3 + (0)x^2 + 4x$

$y_p = \frac{x^3}{3} + 4x$

This matches the second option.