91Ó°ÊÓ

Cubic equations are polynomial equations of degree three. They typically take the form: \[ ax^3 + bx^2 + cx + d = 0 \] where \(a\), \(b\), \(c\), and \(d\) are constants and \(a eq 0\). Solving a cubic equation involves finding the values of \(m\) (also known as roots or solutions) that satisfy the equation. In the given exercise, the cubic equation is \(m^3 - m^2 - 4m + 4 = 0\). To solve it, we might use methods such as factoring, synthetic division, or the Cubic Formula. In our case, the roots are already provided: \(m_1 = 1\), \(m_2 = 2\), and \(m_3 = -2\). This means these are the values of \(m\) for which the equation holds true, i.e., it equals zero when these roots are substituted back into the original equation.

The roots of an equation are the values that make the equation true when they replace the variable. For a cubic equation like \(m^3 - m^2 - 4m + 4 = 0\), the roots can often be found by strategic methods including factoring, the Rational Root Theorem, or direct computation. - In the given problem, the roots are given as \(m_1 = 1\), \(m_2 = 2\), and \(m_3 = -2\). - Each of these values represents a solution to the cubic equation where the left side equals zero. These roots are significant because they determine the form of the complementary solution to the related differential equation. Knowing the roots is crucial since each root corresponds to an exponential solution in the form of the differential equation's complementary solution.

The particular solution of a non-homogeneous differential equation addresses the specific part of the solution that corresponds to the non-zero right-hand side of the equation. For a differential equation of the form \[ m^3 - m^2 - 4m + 4 = 0 \], we hypothesize a particular solution based on the non-homogeneous term. In our example, the particular solution is chosen as \[ y_p = A + Bx e^{x} + Cx e^{2x} \]. This guess is informed by the structure of the complementary solution and the form of the non-homogeneous function. By substituting this form into the differential equation, one can determine the specific values of \(A\), \(B\), and \(C\) by matching coefficients from both sides. This process yields the particular solution needed to complete the general solution of the differential equation.

The complementary solution, denoted as \(y_c\), of a homogeneous differential equation is derived from the associated homogeneous equation where all terms are equated to zero. In our exercise, the complementary solution takes the form: \[ y_c = c_1 e^{x} + c_2 e^{2x} + c_3 e^{-2x} \]. This result is based on the roots of the characteristic polynomial obtained from the differential equation, which in this case are \(m_1 = 1\), \(m_2 = 2\), and \(m_3 = -2\). Each root corresponds to an exponential term in the complementary solution. - For \(m_1 = 1\), the term is \(c_1 e^{x}\). - For \(m_2 = 2\), it is \(c_2 e^{2x}\). - For \(m_3 = -2\), it is \(c_3 e^{-2x}\). By then adjusting for specific initial or boundary conditions, one can solve for the coefficients \(c_1\), \(c_2\), and \(c_3\), completing the complementary part of the general solution to the differential equation.