91Ó°ÊÓ

Differential equations are mathematical equations that relate a function to its derivatives, describing how the function changes. They play a crucial role in modeling physical phenomena such as motion, heat, and waves. A second-order differential equation, like the one given in the exercise \[ y^{\prime \prime} + 2y^{\prime} + y = e^{-x} \]involves the second derivative of the function. The equation is in the form of a non-homogeneous differential equation because of the non-zero term on the right, \( e^{-x} \).
Understanding these equations involves determining two components: the homogeneous solution, which solves the equation when the right side is zero, and the particular solution, addressing the presence of the non-zero term.

Finding the homogeneous solution is the first step in solving a non-homogeneous differential equation using the method of variation of parameters. The homogeneous equation derived from the given differential equation is:\[ y^{\prime \prime} + 2y^{\prime} + y = 0 \]This type of equation is characterized by setting the non-homogeneous part (the right-side term) to zero.
- Plug in \( y_h = e^{rx} \) as a trial solution.- This results in the characteristic equation: \( r^2 + 2r + 1 = 0 \).- Solving this equation gives: \( (r+1)^2 = 0 \), leading to a repeated root \( r = -1 \).
The solution involves exponential terms derived from the roots:\[ y_h = c_1 e^{-x} + c_2 x e^{-x} \]where \( c_1 \) and \( c_2 \) are arbitrary constants that depend on initial conditions.

The particular solution in the method of variation of parameters addresses the non-zero term on the right side of the differential equation, namely \( e^{-x} \). To find it, we assume a solution of the form\[ y_p = u_1(x) e^{-x} + u_2(x) x e^{-x} \]where \( u_1(x) \) and \( u_2(x) \) are functions to be determined. This leads to:
- For \( u_1^{\prime} \), use the formula: \( u_1^{\prime} = -\frac{y_2 g(x)}{W(y_1, y_2)} \)- For \( u_2^{\prime} \): \( u_2^{\prime} = \frac{y_1 g(x)}{W(y_1, y_2)} \)This requires calculating the Wronskian, and upon integration of these derivatives, we find:
\[ u_1(x) = -\frac{x^2}{2} \] \[ u_2(x) = x \]These values are then substituted back to construct the particular solution as\[ y_p = \frac{x^2}{2} e^{-x} \].

The Wronskian is a determinant used in the context of differential equations to determine the linear independence of solutions. For the given equation, the two solutions are \( y_1 = e^{-x} \) and \( y_2 = x e^{-x} \). Calculating the Wronskian involves:
- Computing the determinant of a matrix formed by these solutions and their derivatives.
The matrix for the functions \( y_1 \) and \( y_2 \) and their derivatives is:\[W(y_1, y_2) = \begin{vmatrix}e^{-x} & x e^{-x} \-e^{-x} & (1-x)e^{-x} \end{vmatrix}\]After solving, the result is:\[ W(y_1, y_2) = e^{-2x} \]The Wronskian not only helps in the construction of solutions but also verifies that the chosen functions are linearly independent, ensuring the validity of the variation of parameters method.