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Second-order linear differential equations are an important class of equations in mathematics and many physical sciences. The term "second-order" refers to the highest degree of derivative in the equation, which is the second derivative for these cases. The general form of a second-order linear differential equation is:

• \[ a y'' + b y' + c y = g(t) \]

Here, \( y'' \) is the second derivative, \( y' \) is the first derivative, and \( y \) is the function itself. The coefficients \( a \), \( b \), and \( c \) are constants, and \( g(t) \) is a function of \( t \).
In our case, the equation has the form \( 16 y'' - 24 y' + 9 y = 0 \), which is a homogeneous equation because the right side equals zero, implying \( g(t) = 0 \).
This specific structure ensures that the methods of solving are systematic and allow us to find the general solution for \( y(t) \).

The characteristic equation is a key step in solving second-order linear differential equations with constant coefficients. It helps transform the differential equation into an algebraic equation that is often easier to solve.
For a homogeneous differential equation:

• \[ a y'' + b y' + c y = 0 \]

The characteristic equation is given by:

• \[ a r^2 + b r + c = 0 \]

This is a quadratic equation in terms of \( r \). By solving for \( r \), we determine the form of the solutions to the differential equation. For our exercise, the characteristic equation was derived from \( 16 y'' - 24 y' + 9 y = 0 \) leading to:

• \[ 16 r^2 - 24 r + 9 = 0 \]

From there, the quadratic formula is often used to find the values of \( r \).

Homogeneous differential equations refer to those where every term is a function of the unknown variable and its derivatives, often equating to zero. This gives these equations a special property in terms of their solutions.
In homogeneous equations, since the differential equation is set to zero (\( g(t) = 0 \)), any solution added to itself (scaled by any constant) results in another valid solution. This linearity leads to the general solution having a particular form for different roots of the characteristic equation:

• When roots are real and distinct, solutions are linear combinations of exponential terms.
• When you have repeated or double roots, like in our exercise (where \( r = \frac{3}{4} \) appears twice), the solution includes a polynomial term that introduces an extra factor of \( t \).

Constant coefficients mean that the values \( a \), \( b \), and \( c \) in the differential equation do not depend on the variable \( t \). This characteristic simplifies the process of solving differential equations because it ensures that the characteristic equation is a classic quadratic equation, and its roots can be found using the quadratic formula.
With constant coefficients, the structure of solutions is especially predictable:

• They transform into a characteristic equation that is easier to handle.
• The solutions, once the roots are derived from the characteristic equation, take specific forms involving exponential functions.

In the case of our example, the coefficients 16, -24, and 9 form the quadratic equation that makes solving for \( r \) straight forward. Once \( r \) is determined, as in this case with a double root, the solution to the differential equation becomes straightforward to write as \( y(t) = (C_1 + C_2 t) e^{\frac{3}{4} t} \).
This shows the power of dealing with constant coefficients: they allow for formulas and methods that can be applied consistently across similar problems.