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A characteristic equation is a key tool in solving differential equations, especially those of higher order. It is derived by assuming that the solution to a differential equation can take the form of an exponential function, typically expressed as \( y = e^{rt} \), where \( r \) is a constant. This assumption simplifies the problem considerably, as the derivatives of \( e^{rt} \) remain in exponential form:

• First derivative: \( y' = re^{rt} \)
• Second derivative: \( y'' = r^2e^{rt} \)
• Third derivative: \( y''' = r^3e^{rt} \)

By substituting these derivatives into the original differential equation, you obtain a polynomial in terms of \( r \), known as the characteristic equation. In our example, the differential equation is \( y^{\prime \prime \prime} + 3y^{\prime \prime} - 4y^{\prime} - 12y = 0 \), leading to the characteristic equation \( r^3 + 3r^2 - 4r - 12 = 0 \). Solving this polynomial is the key to finding the solution to the differential equation.

The Rational Root Theorem is a handy concept for finding potential rational roots of a polynomial equation. It states that any rational solution, expressed as a fraction \( \frac{p}{q} \), must be such that \( p \) (the numerator) is a factor of the constant term, and \( q \) (the denominator) is a factor of the leading coefficient.
To apply this theorem to our characteristic equation, \( r^3 + 3r^2 - 4r - 12 = 0 \), note that the leading coefficient is 1 and the constant term is -12. Thus, the possible rational roots are the factors of -12: \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \).
By testing these potential roots in the equation, we identify that \( r = -2 \) satisfies the equation, verifying it as a root. This result allows us to factor the cubic polynomial using synthetic division, simplifying the problem significantly.

The quadratic formula is a powerful method to find the roots of any quadratic equation of the form \( ax^2 + bx + c = 0 \). It is given by:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In our problem, after factoring out \( r + 2 \) from the characteristic polynomial, we are left with a quadratic equation \( r^2 + r - 6 = 0 \).
Here, \( a = 1 \), \( b = 1 \), and \( c = -6 \). Plugging these into the quadratic formula, we find:

• \( b^2 - 4ac = 1^2 - 4 \times 1 \times (-6) = 1 + 24 = 25 \)
• \( r = \frac{-1 \pm 5}{2} \)

This calculation results in the roots \( r = 2 \) and \( r = -3 \), which are critical for constructing the general solution to the differential equation.

The concept of a general solution in differential equations refers to the most comprehensive form of the solution, incorporating all possible specific solutions. For linear differential equations with constant coefficients, the general solution is a combination of functions derived from the roots of the characteristic equation.
In our case, the characteristic equation \( r^3 + 3r^2 - 4r - 12 = 0 \) has roots \( r = -2, 2, \) and \( -3 \). Each root corresponds to an exponential function of the form \( e^{rt} \).

• The root \( r = -2 \) results in \( e^{-2t} \)
• The root \( r = 2 \) gives \( e^{2t} \)
• The root \( r = -3 \) corresponds to \( e^{-3t} \)

Therefore, the general solution is a linear combination of these functions: \[ y(t) = C_1 e^{-2t} + C_2 e^{2t} + C_3 e^{-3t} \]Here, \( C_1, C_2, \) and \( C_3 \) are arbitrary constants, which can be determined if initial conditions are provided. This solution represents all possible outcomes for the given differential equation.