91Ó°ÊÓ

A homogeneous differential equation has all its terms involving the unknown function and its derivatives equal to zero. In our problem, the associated homogeneous differential equation is given by:
\[ y'' - y' - 2y = 0 \]
To solve this, we need to find the characteristic equation, which is a quadratic equation formed from the coefficients of the differential equation. The characteristic equation here is:
\[ r^2 - r - 2 = 0 \]
Solving this using the quadratic formula, we get roots \( r_1 = 2 \) and \( r_2 = -1 \). These roots help us construct the general solution to the homogeneous equation:
\[ y_h = C_1 e^{2x} + C_2 e^{-x} \]
Here, \( C_1 \) and \( C_2 \) are arbitrary constants that will be determined later by initial conditions.

The particular solution is a specific solution that fits the non-homogeneous differential equation. The given non-homogeneous equation is:
\[ y'' - y' - 2y = 2x - 1 \]
To find the particular solution, we make an educated guess depending on the right-hand side of our differential equation. Since we have a linear term \( 2x - 1 \), we assume a particular solution of the form:
\[ y_p = Ax + B \]
We then compute the first and second derivatives:

\( y_p' = A \)
\( y_p'' = 0 \)
Substituting these back into the original non-homogeneous differential equation gives us:
\[ -A - 2(Ax + B) = 2x - 1 \]
Applying algebra, we equate coefficients to find \( A \) and \( B \):
\( -2A = 2 \rightarrow A = -1 \)
\( -2B - A = -1 \rightarrow B = -\frac{1}{2} \).
Thus, our particular solution is:
\[ y_p = -x - \frac{1}{2} \]

Initial conditions are values given at a specific point to determine the arbitrary constants in our general solution. In this problem, we are given:
\( y(0) = 6 \)
\( y'(0) = 0 \)
Combining our homogeneous and particular solutions gives us the general solution:
\[ y = C_1 e^{2x} + C_2 e^{-x} - x - \frac{1}{2} \]
To use the initial conditions, we substitute \( x = 0 \) into this general solution:
\[ 6 = C_1 + C_2 - \frac{1}{2} \rightarrow C_1 + C_2 = 6.5 \]
We then find the first derivative:
\[ y' = 2C_1 e^{2x} - C_2 e^{-x} - 1 \]
Substituting \( x = 0 \) and using \( y'(0) = 0 \) gives us:
\[ 0 = 2C_1 - C_2 - 1 \]
Solving these two equations together, we find:
\( C_1 = 1 \)
\( C_2 = 5.5 \)

The characteristic equation helps in solving the homogeneous part of a linear differential equation. To form it, we use the coefficients of the derivatives in the homogeneous differential equation. For the equation:
\[ y'' - y' - 2y = 0 \]
The characteristic equation is:
\[ r^2 - r - 2 = 0 \]
This is a quadratic equation and can be solved using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -1 \), and \( c = -2 \). Plugging in these values, we get:
\[ r = \frac{1 \pm \sqrt{9}}{2} \rightarrow r_1 = 2, r_2 = -1 \]
These roots are used to form the general solution of the homogeneous equation:
\[ y_h = C_1 e^{2x} + C_2 e^{-x} \]
Understanding the characteristic equation helps in breaking down complex problems involving linear differential equations.