91Ó°ÊÓ

Differential equations are mathematical equations that relate a function with its derivatives. They are used in various fields to describe dynamic systems and change over time or space. In this context, we deal with a second-order linear differential equation of the form:

\[ \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=e^{-x} \sin 2 x \]

The goal is to find the unknown function, in this case, \( y(x) \), which satisfies the equation. This type of equation often consists of two parts:

• Homogeneous part, where the function equals zero.
• Non-homogeneous or particular part, where it equals a non-zero term.

In our problem, we aim to solve it using the method of undetermined coefficients. This approach involves finding a particular solution that, when combined with the complementary solution of the homogeneous equation, forms the general solution.

The complementary solution is derived from the homogeneous part of the differential equation. For the equation given, the homogeneous version is:

\[ y'' - 2y' + 2y = 0 \]

To solve this, we first form the characteristic equation by replacing the derivatives with powers of \( r \):

\[ r^2 - 2r + 2 = 0 \]

Solving this quadratic equation gives us the roots, which in this case are complex: \( r = 1 \pm i \). This indicates that the solution involves exponential decay and oscillation, which results in:

\[ y_c = e^x (C_1 \cos x + C_2 \sin x) \]

Here, \( C_1 \) and \( C_2 \) are constants determined by boundary or initial conditions. The complementary solution captures the behavior of the system without external forces (or non-homogeneous terms).

The particular solution addresses the non-homogeneous part of the differential equation. In our problem, we assume the form of the particular solution based on the non-homogeneous term \( e^{-x} \sin 2x \). This leads us to guess a particular solution of the form:

\[ y_p = e^{-x} (A \cos 2x + B \sin 2x) \]

To find \( A \) and \( B \), we differentiate \( y_p \) to obtain expressions for \( y_p' \) and \( y_p'' \). With these derivatives, we substitute back into the original equation and solve for the coefficients by equating terms with equivalent functions on both sides. In this example, we find that:

• \( A = \frac{1}{5} \)
• \( B = 0 \)

The particular solution complements the complementary solution to form the complete solution to the differential equation.

The characteristic equation is a key component in solving linear differential equations, especially when finding the complementary solution. In our differential equation, the form \( y'' - 2y' + 2y = 0 \) transforms into the characteristic equation:

\[ r^2 - 2r + 2 = 0 \]

This is a quadratic equation, where coefficients are derived from the differential equation itself. Solving it generally involves using the quadratic formula:

\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our case, plugging values \( a = 1, \ b = -2, \ c = 2 \) into the formula gives us the complex roots \( r = 1 \pm i \). These roots reveal the nature of the solution and result in a combination of exponential and trigonometric functions for the complementary solution. Understanding the characteristic equation allows one to evaluate how solutions to the differential equation behave.