91Ó°ÊÓ

In differential equations, the auxiliary equation is crucial for finding solutions to homogeneous linear differential equations. It is derived from the characteristic form of the differential equation by replacing derivatives with powers of a variable, usually denoted by \(m\). For example, in our exercise, the auxiliary equation is \(m^2 - 9 = 0\). This is a straightforward quadratic equation that can be solved to obtain roots, which are essential for constructing the complementary solution of the differential equation. By setting the equation \(m^2 - 9 = 0\), the process involves either factoring or using the quadratic formula to find its roots. Here, factoring gives us \((m - 3)(m + 3) = 0\), leading to roots \(m = 3\) and \(m = -3\). These roots directly inform the structure of the complementary solution since they impact the exponential components of the solution.

The Wronskian is an important determinant used to assess whether a set of solutions to a differential equation is linearly independent. It is particularly useful in the method of variation of parameters. In our given problem, the Wronskian is calculated using the known solutions \(e^{3x}\) and \(e^{-3x}\). It's represented as a determinant of a matrix consisting of these functions and their derivatives:

• First row: \(e^{3x}\) and \(e^{-3x}\)
• Second row: the derivatives \(3e^{3x}\) and \(-3e^{-3x}\)

The Wronskian is computed as:\[W = \begin{vmatrix} e^{3x} & e^{-3x} \ 3e^{3x} & -3e^{-3x} \end{vmatrix} = (-3)(e^{-3x})(e^{3x}) - (3)(e^{3x})(e^{-3x}) = -6\]A non-zero Wronskian, as in this case, indicates that the solutions are linearly independent, which is essential for applying subsequent solution techniques like the variation of parameters.

Variation of Parameters is a powerful technique for finding a particular solution to non-homogeneous linear differential equations. It involves adjusting the parameters in the complementary solution based on a given function. In our scenario, the function \(f(x) = \frac{9x}{e^{3x}}\) is used to determine how the parameters change.
To apply the method, we solve two equations:

• \(u_1' e^{3x} + u_2' e^{-3x} = 0\)
• \(u_1' e^{3x} + u_2' e^{-3x} = \frac{9x}{e^{3x}}\)

Using the computed Wronskian \(W = -6\), the new parameters are found:

• \(u_1' = \frac{3}{2} x e^{-6x}\)
• \(u_2' = -\frac{3}{2} x\)

Integration of these expressions gives the specific parameter functions \(u_1\) and \(u_2\), which are essential for constructing the particular solution to the differential equation.

The complementary solution of a differential equation is derived from the roots of the auxiliary equation. It captures the behavior of the homogeneous part of the differential equation. For our problem, with the roots \(m = 3\) and \(m = -3\), the complementary solution is expressed as:\[y_c = c_1 e^{3x} + c_2 e^{-3x}\]Here, \(c_1\) and \(c_2\) are arbitrary constants that are determined by the initial or boundary conditions of the problem. This solution forms the basis onto which the particular solution is added to get the general solution. Identifying this form is vital, as it represents solutions that will satisfy the differential equation in the absence of any forcing function.

Finding a particular solution is essential to solve non-homogeneous differential equations fully. It accounts for the present forcing function in the equation. Using the variation of parameters method, we previously calculated the necessary parameter functions \(u_1\) and \(u_2\). With these, we assemble the particular solution:\[y_p = u_1 e^{3x} + u_2 e^{-3x}\]For our exercise, after integration, the particular functions were:

• \(u_1 = -\frac{1}{24} e^{-6x} - \frac{1}{4} x e^{-6x}\)
• \(u_2 = -\frac{3}{4} x^2\)

Substituting these into the expression for \(y_p\) gives:\[y_p = -\frac{1}{24} e^{-3x} - \frac{1}{4} x e^{-3x} - \frac{3}{4} x^2 e^{-3x}\]This solution ensures the complete resolution of the differential equation by considering the non-homogeneous component.

The roots of a quadratic equation provide critical information for solving differential equations. The quadratic equation in this context is \(m^2 - 9 = 0\). Solving it requires:

• Factoring: Rewriting it as \((m - 3)(m + 3) = 0\)
• Using the quadratic formula: \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In this exercise, factoring quickly reveals the roots to be \(m = 3\) and \(m = -3\). These roots are real and distinct, indicating that the complementary solution will be a combination of exponential functions. For differential equations, especially linear homogeneous ones, the nature of these roots (real and distinct, real and repeated, or complex) directly influences the form of the complementary solution, guiding us in how to construct the general solution.