Question

Let \(a, b, c,\) and \(\omega\) be constants, with \(a \neq 0\) and \(\omega>0\), and let $$ P(x)=p_{0}+p_{1} x+\cdots+p_{k} x^{k} \quad \text { and } \quad Q(x)=q_{0}+q_{1} x+\cdots+q_{k} x^{k} $$ where at least one of the coefficients \(p_{k}, q_{k}\) is nonzero, so \(k\) is the larger of the degrees of \(P\) and \(Q\). (a) Show that if \(\cos \omega x\) and \(\sin \omega x\) are not solutions of the complementary equation $$ a y^{\prime \prime}+b y^{\prime}+c y=0 $$ then there are polynomials $$ A(x)=A_{0}+A_{1} x+\cdots+A_{k} x^{k} \quad \text { and } \quad B(x)=B_{0}+B_{1} x+\cdots+B_{k} x^{k} $$ such that $$ \begin{aligned} \left(c-a \omega^{2}\right) A+b \omega B+2 a \omega B^{\prime}+b A^{\prime}+a A^{\prime \prime} &=P \\ -b \omega A+\left(c-a \omega^{2}\right) B-2 a \omega A^{\prime}+b B^{\prime}+a B^{\prime \prime} &=Q \end{aligned} $$ where \(\left(A_{k}, B_{k}\right),\left(A_{k-1}, B_{k-1}\right), \ldots,\left(A_{0}, B_{0}\right)\) can be computed successively by solving the systems $$ \begin{aligned} \left(c-a \omega^{2}\right) A_{k}+b \omega B_{k} &=p_{k} \\ -b \omega A_{k}+\left(c-a \omega^{2}\right) B_{k} &=q_{k} \end{aligned} $$ and, if \(1 \leq r \leq k,\) $$ \begin{aligned} \left(c-a \omega^{2}\right) A_{k-r}+b \omega B_{k-r} &=p_{k-r}+\cdots \\ -b \omega A_{k-r}+\left(c-a \omega^{2}\right) B_{k-r} &=q_{k-r}+\cdots \end{aligned} $$ where the terms indicated by "..." depend upon the previously computed coefficients with subscripts greater than \(k-r\). Conclude from this and Exercise \(36(\mathbf{b})\) that $$ y_{p}=A(x) \cos \omega x+B(x) \sin \omega x $$ is a particular solution of $$ a y^{\prime \prime}+b y^{\prime}+c y=P(x) \cos \omega x+Q(x) \sin \omega x . $$ (b) Conclude from Exercise \(36(\mathbf{c})\) that the equation $$ a\left(y^{\prime \prime}+\omega^{2} y\right)=P(x) \cos \omega x+Q(x) \sin \omega x $$ does not have a solution of the form (B) with \(A\) and \(B\) as in (A). Then show that there are polynomials $$ A(x)=A_{0} x+A_{1} x^{2}+\cdots+A_{k} x^{k+1} \quad \text { and } \quad B(x)=B_{0} x+B_{1} x^{2}+\cdots+B_{k} x^{k+1} $$ such that $$ \begin{aligned} a\left(A^{\prime \prime}+2 \omega B^{\prime}\right) &=P \\ a\left(B^{\prime \prime}-2 \omega A^{\prime}\right) &=Q \end{aligned} $$ where the pairs \(\left(A_{k}, B_{k}\right),\left(A_{k-1}, B_{k-1}\right), \ldots,\left(A_{0}, B_{0}\right)\) can be computed successively as follows: $$ \begin{aligned} A_{k} &=-\frac{q_{k}}{2 a \omega(k+1)} \\ B_{k} &=\frac{p_{k}}{2 a \omega(k+1)}, \end{aligned} $$ and, if \(k \geq 1\), $$ A_{k-j}=-\frac{1}{2 \omega}\left[\frac{q_{k-j}}{a(k-j+1)}-(k-j+2) B_{k-j+1}\right] $$ $$ B_{k-j}=\frac{1}{2 \omega}\left[\frac{p_{k-j}}{a(k-j+1)}-(k-j+2) A_{k-j+1}\right] $$ for \(1 \leq j \leq k\). Conclude that (B) with this choice of the polynomials \(A\) and \(B\) is a particular solution of \((\mathrm{C})\)

Short answer

Question: Find the polynomials A(x) and B(x) for a particular solution of the given nonhomogeneous second-order linear differential equation under the following conditions: (a) cos(ωx) and sin(ωx) are not solutions of the complementary equation; (b) cos(ωx) and sin(ωx) are solutions of the complementary equation. Solution: (a) When cos(ωx) and sin(ωx) are not solutions of the complementary equation, the particular solution is given by: $$ y_p = A(x)\cos{\omega x} + B(x)\sin{\omega x} $$ where A(x) and B(x) are found by solving the coefficient systems mentioned in the explanation. (b) When cos(ωx) and sin(ωx) are solutions of the complementary equation, the particular solution is given by: $$ y_p = \left[A_0 x + A_1 x^2 + \cdots + A_k x^{k+1} \right]\cos{\omega x} + \left[B_0 x + B_1 x^2 + \cdots + B_k x^{k+1}\right]\sin{\omega x} $$ where the coefficients A_k and B_k are computed using the formulas provided in the explanation.

Step-by-step solution

(a) Finding A(x) and B(x) when cos(ωx) and sin(ωx) are not solutions of complementary equation #

In order to find the polynomials A(x) and B(x) as stated, we will first rewrite the given system of equation as follows: $$ \begin{cases} P(x) = (c-a\omega^2)A(x) + b\omega B(x) + 2a\omega B'(x) + b A'(x) + a A''(x) \\ Q(x) = -b\omega A(x) + (c-a\omega^2)B(x) - 2a\omega A'(x) + b B'(x) + a B''(x) \end{cases} $$ Start by solving the first coefficient system for \(A_k\) and \(B_k\): $$ \begin{cases} (c-a\omega^2)A_k + b\omega B_k = p_k \\ -b\omega A_k + (c-a\omega^2)B_k = q_k \end{cases} $$ Now, for \(1 \leq r \leq k\), solve the following coefficient system, which will depend on the previously computed coefficients with subscripts greater than \(k-r\): $$ \begin{cases} (c-a\omega^2)A_{k-r} + b\omega B_{k-r} = p_{k-r} +\cdots \\\ -b\omega A_{k-r} + (c-a\omega^2)B_{k-r} = q_{k-r} +\cdots \end{cases} $$ Now, combining the results with the particular solution computed in Exercise 36(b), we have the particular solution: $$ y_p = A(x)\cos{\omega x} + B(x)\sin{\omega x} $$

(b) Finding A(x) and B(x) when cos(ωx) and sin(ωx) are solutions of the complementary equation #

First, note that the given equation with solutions in the form of cos(ωx) and sin(ωx) can be written as: $$ a(y''+\omega^2 y) = P(x)cos{\omega x} + Q(x)sin{\omega x} $$ We are asked to find polynomials A(x) and B(x) such that: $$ \begin{cases} a(A''(x)+ 2\omega B'(x))= P(x) \\ a(B''(x)- 2\omega A'(x)) = Q(x) \end{cases} $$ Start by computing the first coefficients for \(A_k\) and \(B_k\) as follows: $$ \begin{cases} A_k = -\frac{q_k}{2a\omega(k+1)} \\ B_k = \frac{p_k}{2a\omega(k+1)} \end{cases} $$ Now, for \(1 \leq j \leq k\), compute the coefficients of A(x) and B(x) using the following formulas: $$ \begin{cases} A_{k-j} = -\frac{1}{2\omega}\left[\frac{q_{k-j}}{a(k-j+1)} - (k-j+2)B_{k-j+1}\right] \\ B_{k-j} = \frac{1}{2\omega}\left[\frac{p_{k-j}}{a(k-j+1)} - (k-j+2)A_{k-j+1}\right] \end{cases} $$ With these coefficients, the particular solution can be written as: $$ y_p = \left[A_0 x + A_1 x^2 + \cdots + A_k x^{k+1} \right]\cos{\omega x} + \left[B_0 x + B_1 x^2 + \cdots + B_k x^{k+1}\right]\sin{\omega x} $$

Key concepts

Particular Solutions

In the study of differential equations, a particular solution is a solution that satisfies not only the differential equation but also an auxiliary condition. This kind of solution is essential because while homogeneous equations provide a general solution in terms of arbitrary constants, the addition of a particular solution addresses specific conditions or non-homogeneities present in practical scenarios.
In solving the given exercise, the particular solution is introduced when dealing with a non-homogeneous differential equation of the form: \[ a y'' + b y' + c y = P(x) \cos(\omega x) + Q(x) \sin(\omega x) \]Here, the functions \( A(x) \) and \( B(x) \) are to be determined so that:

• \( y_p = A(x)\cos\omega x + B(x)\sin\omega x \)

becomes the necessary particular solution that satisfies the conditions set by the non-homogeneous term.

Polynomials in Differential Equations

Polynomials play a significant role in the analysis of differential equations since they often appear as functions or coefficients in both homogeneous and non-homogeneous terms. In the current exercise, polynomials \( P(x) \) and \( Q(x) \) are given as:\[ P(x) = p_0 + p_1 x + \cdots + p_k x^k \]\[ Q(x) = q_0 + q_1 x + \cdots + q_k x^k \]These expressions help define the non-homogeneous part of the given differential equation.

• In the process of finding particular solutions, it is necessary to express other polynomial functions, \( A(x) \) and \( B(x) \), which satisfy the modified equation with these polynomials.
• The coefficients of \( A(x) \) and \( B(x) \) can often be recursively calculated using the polynomial degree \( k \) and are crucial in systematically solving the equation step-by-step.

Complementary Equations

A complementary equation in differential equations refers to the homogeneous part of the problem excluding any non-homogeneous terms. In this exercise, the complementary equation is:\[ a y'' + b y' + c y = 0 \]

• This equation is crucial in understanding the overall behavior of solutions as it represents the general solution which is formed by the linear combination of its independent solutions.
• When solving a non-homogeneous differential equation, the complementary solution helps to define the structure upon which a particular solution can be superimposed to satisfy the entire system.

Solving for the complementary equation can reveal if specific forms like \( \cos\omega x \) and \( \sin\omega x \) are solutions, which influences the form of the particular solution. By ensuring these forms are not solutions of the complementary equation, it allows for unique solutions to be crafted as the particular solution that fit the non-homogeneous need.

Trigonometric Functions as Solutions

Trigonometric functions often surface as solutions to differential equations involving periodic coefficients. In our exercise, \( \cos\omega x \) and \( \sin\omega x \) feature prominently. They are considered when:

• Defining particular solutions, if these trigonometric functions do not appear in the complementary solutions, they are employed directly in forming the particular solution.
• In essence, \( y_p = A(x)\cos\omega x + B(x)\sin\omega x \) utilizes the periodic nature and orthogonal properties of these functions to adjust the solution to meet specific conditions dictated by the non-homogeneous part of the differential equation.
• However, if they are solutions of the complementary equation, a modified approach is adopted, such as adjusting the polynomial degrees or redefining functions \( A(x) \) and \( B(x) \) to ensure independence and completeness of the solution.

By introducing trigonometric functions into the form of the solution, the exercise leverages their properties to solve complex oscillatory system equations more effectively.