91Ó°ÊÓ

In the context of differential equations, the **characteristic equation** plays a crucial role in finding solutions. A characteristic equation arises from substituting a trial solution, often in the form of a power function or exponential function, into the differential equation. From our original equation, by dividing by \(9t^2\) and substituting a trial solution \(x(t) = t^m\), we reached the characteristic polynomial equation:\[m^2 + m + 1 - \frac{4}{9} = 0\]To solve this, we use the quadratic formula:\[m_{1,2} = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \]where \(a = 1\), \(b = 1\), and \(c = 1 - \frac{4}{9}\). This formula helps us discover the roots of the equation, which are fundamental to determining the solution basis for the differential equation. Here, plugging the values we computed:\[m_{1,2} = \frac{-1 \pm \sqrt{1 - 4 \times \left(1 - \frac{4}{9}\right)}}{2}\]Solving gives us complex roots, a situation signaling that the general solution will involve trigonometric functions.

The **general solution** of a differential equation is the family of all possible solutions. Once we deduce the nature of the roots of the characteristic equation, we can formulate the general solution accordingly. In this problem, the roots are complex conjugates, indicated as \(m = -\frac{1}{2} \pm \frac{\sqrt{5}i}{6}\).For complex conjugate roots \(m = p \pm qi\), the general solution to a second-order linear differential equation is given by:\[x(t) = t^p[C_1 \cdot \cos(q \cdot \ln(t)) + C_2 \cdot \sin(q \cdot \ln(t))]\]Applying this formula to our specific roots:

• \(p = -\frac{1}{2}\)
• \(q = \frac{\sqrt{5}}{6}\)

Thus, our solution becomes:\[x(t) = t^{-\frac{1}{2}}[C_1 \cdot \cos(\frac{\sqrt{5}}{6} \cdot \ln(t)) + C_2 \cdot \sin(\frac{\sqrt{5}}{6} \cdot \ln(t))]\]This expression captures the complete set of solutions to the differential equation, parametrized by the constants \(C_1\) and \(C_2\). These constants can be adjusted based on specific initial conditions or boundary conditions.

**Complex conjugate roots** appear when solving characteristic equations that yield a negative discriminant in the quadratic formula. When the roots of the characteristic polynomial are complex, they occur as conjugate pairs, which means that if \(p + qi\) is a root, \(p - qi\) is also a root.This naturally leads to solutions involving sine and cosine functions because:

• Complex numbers express oscillations well.
• Euler's formula, \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\), relates exponential functions and trigonometric functions directly.

In our differential equation, these complex conjugate roots, specifically \(-\frac{1}{2} \pm \frac{\sqrt{5}i}{6}\), determine solutions involving logarithmic arguments of cosine and sine, reflecting these oscillatory dynamicsUnderstanding these roots is key to mastering the behavior of dynamic systems, where such oscillations frequently appear, such as in electrical circuits or mechanical systems with damping effects.