Question

We consider another way of arriving at the proper form of \(Y(t)\) for use in the method of undetermined coefficients. The procedure is based on the observation that exponential, polynomial, or sinusoidal terms (or sums and products of such terms) can be viewed as solutions of certain linear homogeneous differential equations with constant coefficients. It is convenient to use the symbol \(D\) for \(d / d t\). Then, for example, \(e^{-t}\) is a solution of \((D+1) y=0 ;\) the differential operator \(D+1\) is said to annihilate, or to be an annihilator of, \(e^{-t}\). Similarly, \(D^{2}+4\) is an annihilator of \(\sin 2 t\) or \(\cos 2 t,\) \((D-3)^{2}=D^{2}-6 D+9\) is an annihilator of \(e^{3 t}\) or \(t e^{3 t},\) and so forth. Show that linear differential operators with constant coefficients obey the commutative law, that is, $$ (D-a)(D-b) f=(D-b)(D-a) f $$ for any twice differentiable function \(f\) and any constants \(a\) and \(b .\) The result extends at once to any finite number of factors.

Short answer

Question: Prove that linear differential operators with constant coefficients obey the commutative law: \((D-a)(D-b)f = (D-b)(D-a)f\).

Step-by-step solution

Apply the first operator on both sides

Starting with the left side, apply the operator \((D-a)\) to the function \(f\). $$(D-a)f = Df - af$$ Now, on the right side, apply the operator \((D-b)\) to the function \(f\). $$(D-b)f = Df - bf$$

Apply the second operator on both sides

Now, apply the second operator to the result obtained in step 1 on both sides. Starting with the left side, apply the operator \((D-b)\) to the expression \(Df - af\). $$(D-b)(Df-af) = D(Df- af) - b(Df- af)$$ Expanding, we get: $$D^2f - aDf - bDf + abf$$ Similarly, for the right side, apply the operator \((D-a)\) to the expression \(Df - bf\). $$(D-a)(Df-bf) = D(Df - bf) - a(Df - bf)$$ Expanding, we get: $$D^2f - bDf - aDf + abf$$

Compare both sides

Now that we have simplified both sides of the equation, we can compare the results. On the left side, we have: $$D^2f - aDf - bDf + abf$$ And on the right side, we have: $$D^2f - bDf - aDf + abf$$ We can see that both sides are equal for any twice differentiable function \(f\) and any constants \(a\) and \(b\). Thus, we have proved that linear differential operators with constant coefficients obey the commutative law: $$(D-a)(D-b)f = (D-b)(D-a)f$$

Key concepts

Understanding Linear Differential Equations

A differential equation involves functions and their derivatives. A **linear** differential equation means each term is either a constant or a product of a constant and the first power of the function or its derivatives. There's no multiplication or division between variables and their derivatives. The equation typically looks like this:
\[ a_n \frac{d^n y}{dt^n} + a_{n-1} \frac{d^{n-1} y}{dt^{n-1}} + \dots + a_1 \frac{dy}{dt} + a_0 y = 0 \]

• The highest derivative involved is the order of the differential equation.
• The equation is linear because each derivative is to the first power, and there are no products of the function or its derivatives.

Linear differential equations are fundamental in modeling many physical systems, from electrical circuits to mechanical vibrations. They help predict how a system evolves over time.

Exploring Constant Coefficients

The term **constant coefficients** refers to when the coefficients of the differential equation are constants, not functions of the independent variable (e.g., time). This simplifies the solving process greatly. For example, in the equation:
\[ 2 \frac{d^2 y}{dt^2} + 3 \frac{dy}{dt} + 4y = 0 \]

• The coefficients 2, 3, and 4 are constants.

Why is this important?

• With constant coefficients, we can use techniques like the method of undetermined coefficients or characteristic equations to find solutions.
• This scenario leads to homogeneous solutions involving exponential, polynomial, and trigonometric functions.

Solutions are predictable and reliable for modeling and predicting system behaviors.

Understanding the Commutative Law in Differential Operators

The **commutative law** is quite intuitive in basic mathematics, commonly expressed as \(a + b = b + a\) for addition. In the context of differential operators, it states that the order of applying operators doesn’t affect the result.
When using operators \((D-a)\) and \((D-b)\):
\[ (D-a)(D-b) f = (D-b)(D-a) f \]

• This means that applying one operator first and the other second gives the same outcome as doing it in reverse.
• It's essential to note that this holds true when dealing with linear differential operators with constant coefficients.

The exercise confirms this property by showing that the expressions simplify to the same result, proving that these operators indeed commute. This allows flexibility and simplicity when working with differential equations.