Questions: Find a general solution to each of the following, using the method of educated guess to find a particular solution. a. y'' - 3y' - 10y = [72x^2 - 1]e^(2x) b. y'' - 3y' - 10y = 4xe^(6x)

Solution Steps

To solve the differential equation \( y'' - 3y' - 10y = 4xe^{6x} \), we use the method of undetermined coefficients. First, find the complementary solution by solving the homogeneous equation \( y'' - 3y' - 10y = 0 \). Then, make an educated guess for the particular solution of the non-homogeneous equation. Since the right-hand side is \( 4xe^{6x} \), a suitable guess for the particular solution would be of the form \( (Ax + B)e^{6x} \). Substitute this guess into the differential equation to determine the coefficients \( A \) and \( B \).

To find the complementary solution, we solve the homogeneous equation:

\[ y'' - 3y' - 10y = 0 \]

The characteristic equation is:

\[ r^2 - 3r - 10 = 0 \]

Factoring gives:

\[ (r - 5)(r + 2) = 0 \]

Thus, the roots are \( r_1 = 5 \) and \( r_2 = -2 \). Therefore, the complementary solution is:

\[ y_c = C_1 e^{5x} + C_2 e^{-2x} \]

Next, we find a particular solution for the non-homogeneous equation:

\[ y'' - 3y' - 10y = 4xe^{6x} \]

We make an educated guess for the particular solution of the form:

\[ y_p = (Ax + B)e^{6x} \]

Calculating the derivatives:

\[ y_p' = (A + 6Ax + 6B)e^{6x} \] \[ y_p'' = (12A + 36Ax + 36B)e^{6x} \]

Substituting \( y_p \), \( y_p' \), and \( y_p'' \) into the left-hand side of the differential equation and equating coefficients with \( 4xe^{6x} \) allows us to solve for \( A \) and \( B \). After simplification, we find:

\[ y_p = \frac{8x - 9}{16} e^{6x} \]

The general solution \( y \) is the sum of the complementary and particular solutions:

\[ y = y_c + y_p = C_1 e^{5x} + C_2 e^{-2x} + \frac{8x - 9}{16} e^{6x} \]

Final Answer