91Ó°ÊÓ

In the context of ordinary differential equations, the characteristic equation is a vital element, especially for solving linear homogeneous differential equations with constant coefficients. Consider our original equation: \[y^{\prime \prime} - 2 y^{\prime} + 26 y = 0\]The characteristic equation is formed by substituting each derivative term with the corresponding power of a variable, usually denoted as 'm'. For a second-order equation:

• Replace \(y''\) with \(m^2\)
• Replace \(y'\) with \(m\)
• Replace \(y\) itself with 1

This transforms our differential equation into the characteristic polynomial:\[ m^2 - 2m + 26 = 0 \]The solutions (or roots) of this quadratic equation will guide us to the form of the general solution of the differential equation. Therefore, understanding how to form and solve the characteristic equation is crucial for analyzing and solving these types of problems.

When solving the characteristic equation, the roots can be real or complex. In the step-by-step solution of our differential equation, we used the quadratic formula:\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our equation \( m^2 - 2m + 26 = 0 \), using \( a = 1 \), \( b = -2 \), and \( c = 26 \), we encounter complex roots:\[ m = 1 \pm 5i \]Here, \( i \) is the imaginary unit, indicating that our roots are complex conjugates. Complex roots appear as conjugate pairs \( a \pm bi \) where:

• \( a \) is the real part
• \( b \) is the imaginary part

These roots play a significant role in determining the behavior of the differential equation's solutions. They often indicate oscillatory solutions, which are essential in various applications, including engineering and physics.

The general solution of a second-order differential equation with constant coefficients depends on the type of roots of its characteristic equation. In the case of complex conjugate roots \( a \pm bi \), the solution takes a specific form:\[y(x) = e^{ax} (A\cos(bx) + B\sin(bx))\]Here:

• \( e^{ax} \) represents an exponential growth or decay depending on \( a \).
• \( \cos(bx) \) and \( \sin(bx) \) represent oscillations introduced by the imaginary component \( b \).
• \( A \) and \( B \) are arbitrary constants that define the specific solution according to initial conditions.

Applying our specific roots (\(1 \pm 5i\)) to this general form yields:\[y(x) = e^{x}(A \cos(5x) + B \sin(5x))\]This expressed solution captures both exponential behavior and oscillatory components, which explains a wide variety of real-world dynamical systems. Hence, knowing how to express and interpret the general solution with complex roots is fundamental in the study of differential equations.