We are tasked with solving the linear differential equation:

$(D^2 - D - 2)y = 2x + 1x + 1x^2$

Where $D$ is the differential operator, $D = \frac{d}{dx}$. To solve this equation, we will use the method of complementary function (YCF) and particular integral (YPI).

Step 1: Solve the Homogeneous Equation (YCF)

The homogeneous equation is:

$(D^2 - D - 2)y = 0$

To solve this, we find the roots of the characteristic equation:

$D^2 - D - 2 = 0$

This is a quadratic equation, and we solve it using the quadratic formula:

$D = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)}$

$D = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm \sqrt{9}}{2} = \frac{1 \pm 3}{2}$

Thus, the roots are:

$D = 2 \quad \text{and} \quad D = -1$

So the general solution to the homogeneous equation is:

$y_{CF} = C_1 e^{2x} + C_2 e^{-x}$

Step 2: Solve the Non-Homogeneous Equation (YPI)

Now, we solve the non-homogeneous part of the equation:

$(D^2 - D - 2)y = 2x + x + x^2$

Since the right-hand side is a polynomial $2x + x + x^2 = x^2 + 3x$, we assume a particular solution of the form:

$y_{PI} = Ax^2 + Bx + C$

Now, apply the differential operator $(D^2 - D - 2)$ to $y_{PI}$. We need to compute the first and second derivatives of $y_{PI}$:

$y_{PI}' = 2Ax + B$ $y_{PI}'' = 2A$

Now apply $D^2 - D - 2$ to $y_{PI}$:

$(D^2 - D - 2)y_{PI} = (2A) - (2Ax + B) - 2(Ax^2 + Bx + C)$ $= 2A - 2Ax - B - 2Ax^2 - 2Bx - 2C$ $= -2Ax^2 + (-2A - 2B)x + (2A - B - 2C)$

We now equate this expression to the right-hand side of the original equation, $x^2 + 3x$:

$-2Ax^2 + (-2A - 2B)x + (2A - B - 2C) = x^2 + 3x$

By comparing the coefficients of $x^2$, $x$, and the constant terms, we get the following system of equations:

1. $-2A = 1$
2. $-2A - 2B = 3$
3. $2A - B - 2C = 0$

Solving for $A$, $B$, and $C$:

1. From $-2A = 1$, we get $A = -\frac{1}{2}$.
2. Substitute $A = -\frac{1}{2}$ into $-2A - 2B = 3$: $-2(-\frac{1}{2}) - 2B = 3 \implies 1 - 2B = 3 \implies B = -1$
3. Substitute $A = -\frac{1}{2}$ and $B = -1$ into $2A - B - 2C = 0$: $2(-\frac{1}{2}) - (-1) - 2C = 0 \implies -1 + 1 - 2C = 0 \implies C = 0$

Thus, the particular solution is:

$y_{PI} = -\frac{1}{2}x^2 - x$

Step 3: General Solution

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

$y = y_{CF} + y_{PI}$ $y = C_1 e^{2x} + C_2 e^{-x} - \frac{1}{2}x^2 - x$

Final Answer:

$y = C_1 e^{2x} + C_2 e^{-x} - \frac{1}{2}x^2 - x$

Do you need more details on any part? Here are five related questions:

1. What are the steps for solving homogeneous differential equations?
2. How do you determine the roots of a characteristic equation?
3. What is the method to solve non-homogeneous differential equations?
4. Why do we assume a polynomial for the particular integral in this case?
5. How does comparing coefficients help in finding the particular solution?

Tip: Always check the form of the non-homogeneous term to correctly assume the form of the particular solution!