Consider the differential equation \((D-a)(D-b) y=P_{n}(x),\) where \(P_{n}(x)\) is a polynomial of degree \(n\). Show that a particular solution of this equation is given by (6.24) with \(c=0 ;\) that is, \(y_{p}\) is \(\left\\{\begin{array}{l}\text { a polynomial } Q_{n}(x) \text { of degree } n \text { if } a \text { and } b \text { are both different from zero; } \\ x Q_{n}(x) \text { if } a \neq 0, \text { but } b=0 \\ x^{2} Q_{n}(x) \quad \text { if } a=b=0\end{array}\right.\) Hint: To show that \(Q_{n}(x)=\sum a_{n} x^{n}\) is a solution of the differential equation for a given \(P_{n}=\sum b_{n} x^{n},\) you have only to show that the coefficients \(a_{n}\) can be found so that \((D-a)(D-b) Q_{n}(x) \equiv P_{n}(x) .\) Equate coefficients of \(x^{n}, x^{n-1}, \cdots,\) to see that this is always possible if \(a \neq b\). For \(b=0,\) the differential equation becomes \((D-a) D y=P_{n} ;\) what is \(D y\) if \(y=x Q_{n} ?\) Similarly, consider \(D^{2} y\) if \(y=x^{2} Q_{n}\).

Step by step solution

01

Define Polynomial Solutions

Identify the type of polynomial solution based on the values of constants a and b. The polynomial solutions are given by:1. If both a and b are non-zero: \( y_{p} = Q_{n}(x) \)2. If a is non-zero and b is zero: \( y_{p} = x Q_{n}(x) \)3. If both a and b are zero: \( y_{p} = x^{2} Q_{n}(x) \)

02

Assume Polynomial for General Case

For the case where both a and b are non-zero, assume\( Q_{n}(x) = \sum a_{n} x^{n} \). The hint suggests we need to show that \(Q_{n}(x)\) satisfies the differential equation \((D-a)(D-b) Q_{n}(x) = P_{n}(x)\), where \(P_{n}(x) = \sum b_{n} x^{n} \) is a polynomial of degree n.

03

Apply Differential Operators

Substitute \( Q_{n}(x) \) into the differential equation. First, compute \( DQ_{n}(x) \):\( DQ_{n}(x) = \frac{d}{dx}Q_{n}(x) \)Then apply the operators:\((D-a)(D-b) Q_{n}(x) = (D-a) (DQ_{(n)}(x) - bQ_{n}(x)) \)

04

Simplify the Equation

Simplify the equation by distributing and solving for the coefficients:\( = (D-a) (DQ_{n}(x) - b Q_{n}(x)) \)\( = D^2Q_{n}(x) - b DQ_{n}(x) - a DQ_{n}(x) + ab Q_{n}(x) \)\( = D^2 Q_{n}(x) - (a + b) D Q_{n}(x) + ab Q_{n}(x) \)

05

Equate Coefficients

By matching coefficients of \( x^n \) on both sides of the equation, determine that such coefficients \( a_{n} \) can always be found. This shows:\( Q_{n}(x) \) satisfies \((D-a)(D-b) Q_{n}(x) = P_{n}(x).\)

06

Special Case b=0

When \( b=0 \), the differential equation becomes: \((D-a)D y = P_{n}(x).\)Assume \( y = x Q_{n}(x) \), compute: \( Dy = D(x Q_{n}(x)) = Q_{n}(x) + x DQ_{n}(x) \) Substitute into the equation:

07

Simplify for b=0

\((D-a) [ Q_{n}(x) + xD Q_{n}(x)] = P_{n}(x) \), \( (D-a) [ Q_{n}(x) + x\frac{d}{dx} Q_{n}(x)] = P_{n}(x) \) Simplify to establish the coefficients continue to match.

08

Case a=b=0

For this differential equation becoming \( D^{2} y = P_{n}(x) \):Assume \( y = x^{2} Q_{n}(x) \), and find:\( D y = 2xQ_{n}(x) + x^{2} DQ_{n}(x) \), and\( D^2 y = 2Q_{n}(x) + 4x DQ_{n}(x) + x^2 D^{2}Q_{n}(x) \).Substitute back and simplify to match coefficients.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

particular solutions

To understand particular solutions in differential equations, it's crucial to recognize the specific form that accommodates the initial conditions or constraints given in the problem.
In the context of this exercise, we investigate polymonial solutions of the differential equation (D-a)(D-b) y = P_n(x).
Depending on the values of constants 'a' and 'b', the form of the particular solution changes:

• If both a and b are non-zero: The particular solution is a polynomial, designated as \(Q_n(x)\), of degree n.
• If a is non-zero but b is zero: The solution involves an x factor, meaning it's \(xQ_n(x)\).
• If both a and b are zero: It includes x squared, making it \(x^2Q_n(x)\).

Identifying the proper particular solution form is essential as it directly impacts the further steps in solving the differential equation.

polynomial solutions

Polynomial solutions are often used because they're easy to manipulate with differentiation.
A polynomial of degree n is expressed as \(Q_{n}(x) = \text{\textbackslash sum a_n x^n}\). In our exercise, \(Q_{n}(x)\) models the solution of \((D-a)(D-b)Q_n(x) = P_n(x)\).

• The polynomial \(Q_n(x)\) is solved by ensuring it fits within the coefficients defined by \(P_n(x)\), another polynomial of degree n, defined as \(P_{n}(x) = \text{\textbackslash sum b_n x^n}\).

Using polynomial solutions simplifies the process of matching coefficients, transforming a complex differential equation into a more manageable algebra problem.

differential operators

Differential operators simplify dealing with derivatives and their repeated applications.
The operator D is a derivative with respect to x. For instance, \(D f(x) = \frac{d}{dx} f(x)\). Differential operators can be combined, as in the problem: \((D-a)(D-b)\). This means applying these operators sequentially to a function or polynomial.

• Start by applying (D-b): This translates to \(D Q_{n}(x) - b Q_{n}(x)\).
• Then apply (D - a) to the result: This would give \(D[D Q_{n}(x) - b Q_{n}(x)] - a[D Q_{n}(x) - b Q_{n}(x)]\).

The goal is to simplify the differential equation by computing these operations on the assumed polynomial solution and matching it to the polynomial \(P_n(x)\).

coefficient matching

Coefficient matching frequently appears in solving polynomial differential equations.
It involves ensuring the coefficients of the polynomial terms on both sides of the equation are equal for each power of x.
For our exercise:

• Expand \((D-a)(D-b)Q_n(x)\) to obtain \(D^2 Q_{n}(x) - (a+b) D Q_{n}(x) + ab Q_{n}(x)\).
• Match this expansion to \(P_n(x)\) by equating coefficients of each power of x.

Let's delve into a case: Assume \(Q_n(x) = \text{\textbackslash sum a_n x^n}\).
This polynomial is substituted, and resulting coefficients of terms like \(x^n, x^{n-1}, \textbackslash cdots,\) are matched with similar terms on the right-hand side of the equation to determine the specific values for \(a_n\).
By ensuring coefficients \(a_n\) satisfy \((D-a)(D-b) Q_{n}(x) \text{\textbackslash equiv} P_{n}(x)\), we confirm that the polynomial \(Q_n(x)\) is indeed a solution.