Problem 32 - Differential Equations and Linear Algebra

Question

Determine the general solution to the given differential equation. Derive your trial solution using the annihilator technique. $$y^{\prime \prime \prime}+3 y^{\prime \prime}+3 y^{\prime}+y=2 e^{-x}+3 e^{2 x}$$.

Step-by-Step Solution

Answer

The general solution to the given differential equation is: \[y(x)=C_1 e^{-x}+C_2 x e^{-x}+C_3 x^2 e^{-x}+\frac{1}{12}e^{-x}+\frac{3}{2}e^{2x}\] where \(C_1\), \(C_2\), and \(C_3\) are constants.

Step 1: Find the complementary solution

First, we solve the homogeneous equation by setting the right-hand side to zero: \[y^{\prime \prime \prime}+3 y^{\prime \prime}+3 y^{\prime}+y=0\] The characteristic equation is: \[r^3+3r^2+3r+1=0\] Factor this equation to find its roots: \[(r+1)^3=0\] Hence, r = -1 (a triple root). The complementary solution 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.

Step 2: Apply the annihilator technique

Now, we will apply the annihilator technique to find particular solutions for each term on the right-hand side. For the term \(2e^{-x}\), since we already have an \(e^{-x}\) term in the complementary solution, we multiply it by the derivative operator \((D+1)\) to annihilate it: \[(D+1)^4(y_p) = 2e^{-x}\] For the term \(3e^{2x}\), since it is not present in the complementary solution, we apply the derivative operator \((D-2)\) to annihilate it: \[(D-2)(y_p)=3e^{2x}\]

Step 3: Find the particular solutions

Now we will find the particular solutions for each term on the right-hand side. For the term \(2e^{-x}\), we have: \[(D+1)^4(y_p) = 2e^{-x}\] \(y_p = Ae^{-x}\) Differentiating four times and substituting back into the equation, we get: \[A(4!)=2\] \[A=\frac{1}{12}\] So, the particular solution is: \[y_{p1}(x)=\frac{1}{12}e^{-x}\] For the term \(3e^{2x}\), we have: \[(D-2)(y_p)=3e^{2x}\] \(y_p=Be^{2x}\) Differentiating once and substituting back into the equation, we get: \[B(2)=3\] \[B=\frac{3}{2}\] So, the particular solution is: \[y_{p2}(x)=\frac{3}{2}e^{2x}\]

Step 4: Combine the complementary and particular solutions

Now, we combine the complementary and particular solutions to find the general solution: \[y(x)=y_c(x)+y_{p1}(x)+y_{p2}(x)\] \[y(x)=C_1 e^{-x}+C_2 x e^{-x}+C_3 x^2 e^{-x}+\frac{1}{12}e^{-x}+\frac{3}{2}e^{2x}\] This is the general solution to the given differential equation.

Key Concepts

Complementary Solution

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.

Characteristic Equation

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.

Particular 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.