91Ó°ÊÓ

Second-order linear differential equations involve derivatives up to the second order. These equations have the general form: \[ y'' + p(t)y' + q(t)y = g(t) \]where:

• \( y'' \) is the second derivative of \( y \)
• \( p(t) \) and \( q(t) \) are functions of \( t \)
• \( g(t) \) is the non-homogeneous term

In this instance, our equation is \[ y'' + 9y = \frac{\text{sec}(3t) \tan(3t)}{3} \].
This type of equation necessitates solving both the homogeneous and non-homogeneous parts to find the complete solution. The homogeneous part (\( y'' + 9y = 0 \)) is solved first, followed by finding a specific solution to the non-homogeneous part. Combining both gives the final solution.

The method of undetermined coefficients helps to find particular solutions to non-homogeneous differential equations. This approach assumes a solution form based on the non-homogeneous term \( g(t) \).
For example, if \( g(t) \) was a polynomial, an exponential, or a trigonometric function, we guess a corresponding solution form with unknown coefficients.
In our exercise, since \( g(t) \) is expressed through secant and tangent functions, we test forms like \( y_p = A \text{sec}(3t) \). Differentiating suitably and correcting errors reveals a particular solution.
This step's accuracy is critical as it clinches the specificity needed beyond the general solution provided by the homogeneous part.

Solving a homogeneous differential equation involves finding the general solution where \( g(t) = 0 \).
Consider our homogeneous part: \[ y'' + 9y = 0 \].
Assuming a solution form \( y = e^{rt} \), substituting into the equation, yields: \[ r^2 + 9 = 0 \].
Solving this characteristic equation, we get complex roots: \( r = \pm 3i \).
Thus, the general solution for the homogeneous differential equation is given by: \[ y_h = c_1 \text{cos}(3t) + c_2 \text{sin}(3t) \].
where \( c_1 \) and \( c_2 \) are constants determined by initial conditions.

The characteristic equation is derived from assuming an exponential form of the solution: \( y = e^{rt} \). For our homogeneous equation, \[ y'' + 9y = 0 \], substituting \( y = e^{rt} \) gives: \[ r^2 e^{rt} + 9e^{rt} = 0 \]
Dividing by \( e^{rt} \) (non-zero), we get: \[ r^2 + 9 = 0 \].
This characteristic equation is crucial in determining the nature of the solution.
Here, solving \( r^2 + 9 = 0 \) for \( r \) gives complex roots: \( r = \pm 3i \).
The complex roots indicate our solution will include sine and cosine functions, forming our homogeneous solution framework as: \[ y_h = c_1 \text{cos}(3t) + c_2 \text{sin}(3t) \].