91Ó°ÊÓ

The complementary solution plays a crucial role when solving linear differential equations. It represents the solution to the associated homogeneous equation, which means the equation with the right-hand side set to zero. For our differential equation, \
\[y^{\prime \prime \prime}+3 y^{\prime \prime}+3 y^{\prime}+y=0\]
, the complementary solution is found by determining the roots of the characteristic equation.

This characteristic equation is derived from substituting a trial solution into the homogeneous equation. The trial is typically of the form \
\[y = e^{rx}\]
, which when substituted, leads to the algebraic equation
\
\[r^3+3r^2+3r+1=0\]
. By solving this characteristic equation, we get a triple root at \
\[r=-1\]
.

Each root corresponds to a part of the solution: a simple root gives a term like \
\[Ce^{rx}\]
, while repeated roots add polynomial factors to account for the multiplicity. Therefore, the complementary solution considering our triple root is
\
\[y_c(x)=C_1 e^{-x}+C_2 xe^{-x}+C_3 x^2 e^{-x}\]
, where \
\[C_1\], \
\[C_2\], and \
\[C_3\] are constants to be determined from initial conditions or boundary values.

The characteristic equation is a pivotal concept when solving linear differential equations with constant coefficients. It provides a direct way to find the complementary solution. The roots of the characteristic equation correspond to the exponents in the solution's exponential terms.

To illustrate, our differential equation \
\[y^{\prime \prime \prime}+3 y^{\prime \prime}+3 y^{\prime}+y\]
, leads to the characteristic equation
\
\[r^3+3r^2+3r+1=0\]
. The roots of this equation, which are the values of \
\[r\] that satisfy the equation, tell us how the solution to the homogeneous part of our differential equation behaves.

In our case, factoring the characteristic equation gave us a repeated root, which suggests that the complementary solution will include exponential functions multiplied by increasing powers of \
\[x\] to account for the multiplicity of the root. This is why the complementary solution includes terms like \
\[C_2 x e^{-x}\] and \
\[C_3 x^2 e^{-x}\]. This combination guarantees that all linearly independent solutions are included in the complementary solution.

The particular solution targets the non-homogeneous part of the differential equation—the part that doesn’t equal zero. We use the annihilator technique to 'annihilate' or get rid of specific terms on the right-hand side of the differential equation to find this solution.

In our textbook problem, we have two terms: \
\[2e^{-x}\] and \
\[3e^{2x}\]. The annihilator technique involves applying a differential operator that will turn these terms into the form of the complementary equation, effectively cancelling them out. For the first term, \
\[2e^{-x}\], because \
\[e^{-x}\] already forms part of the complementary solution, we must use \
\[(D+1)^4\], the fourth power of the operator \
\[(D+1)\], to ensure the annihilator effect considering the equation's roots and multiplicities.

Meanwhile, for the term \
\[3e^{2x}\], there is no corresponding term in the complementary solution, so the operator \
\[(D-2)\] is sufficient to find the particular solution. Ultimately, we find two specific values for \
\[A\] and \
\[B\] that give us the particular solutions \
\[y_{p1}(x)=\frac{1}{12}e^{-x}\] and \
\[y_{p2}(x)=\frac{3}{2}e^{2x}\]. These particular solutions, when combined with the complementary solution, form the general solution to our original differential equation.