Problem 13 - A First Course in Differential Equations with Modeling Applications

Question

Verify that the given differential operator annihilates the indicated functions. $$(D-2)(D+5) ; \quad y=e^{2 x}+3 e^{-5 x}$$

Step-by-Step Solution

Answer

The operator \((D-2)(D+5)\) annihilates the function \(y=e^{2x}+3e^{-5x}\) as the result is zero.

Step 1: Understanding the Problem

We need to verify if the differential operator \((D-2)(D+5)\) annihilates the function \(y=e^{2x}+3e^{-5x}\). A differential operator annihilates a function if, after applying the operator to the function, the result is zero.

Step 2: Apply the First Operator

The first step is to apply the innermost operator, \(D+5\), to \(y = e^{2x} + 3e^{-5x}\). When we apply \(D+5\) to a function \(f(x)\), it means we differentiate \(f(x)\) and then add \(5f(x)\). Calculating this gives: \[(D+5)(e^{2x} + 3e^{-5x}) = \frac{d}{dx}(e^{2x} + 3e^{-5x}) + 5(e^{2x} + 3e^{-5x})\]This results in: \[ (2e^{2x} - 15e^{-5x}) + (5e^{2x} + 15e^{-5x}) = 7e^{2x}\] because the terms in \(e^{-5x}\) cancel each other.

Step 3: Apply the Second Operator

Now, apply the operator \(D-2\) to the result from the first step, which is \(7e^{2x}\). Applying \(D-2\), we differentiate \(7e^{2x}\) and subtract \(2\cdot 7e^{2x}\) from it:\[(D-2)(7e^{2x}) = \frac{d}{dx}(7e^{2x}) - 2(7e^{2x})\]This becomes \[14e^{2x} - 14e^{2x} = 0\].

Step 4: Conclusion

The result of applying \((D-2)(D+5)\) to \(y=e^{2x} + 3e^{-5x}\) is zero, which means the operator annihilates the function.

Key Concepts

Differential Operators

Differential operators are a crucial tool in solving differential equations. They help in transforming functions by applying operations like differentiation and multiplication by constants. In the context of our problem, we are dealing with the composite operator \((D-2)(D+5)\). Each part of the operator represents a specific operation on a function.

• \(D\) represents differentiation with respect to \(x\).
• \(D+5\) means differentiate the function and then add five times the function.
• \(D-2\) means differentiate the function and then subtract twice the function.

By applying these operations systematically, we can transform a function into another form or confirm whether it results in zero - indicating that the function is annihilated by the operator.

Annihilator Method

The annihilator method is a powerful technique in differential equations used to find solutions to non-homogeneous linear differential equations. The basic idea is to apply a differential operator that turns a given function or sum of functions into zero. This method uses the concept of annihilation to solve differential equations systematically.
When you see an expression like \( (D-2)(D+5) \), it tells us we have a sequence of operations to apply. The goal is for these operations to cancel out the function completely.

• Start with the first part of the operator, apply it, and simplify if possible.
• Move to the next part of the operator with the obtained result.
• Continue until you arrive at zero.

This technique is particularly useful when dealing with differential equations involving complex solutions that can be broken down through systematic application of operators.

Exponential Functions

Exponential functions play a significant role in differential equations due to their unique properties. They are generally of the form \(e^{ax}\), where \(a\) dictates the rate of growth or decay. These functions are easy to differentiate, an aspect that simplifies many differential operations.

• Derivatives of exponential functions retain the same form, importantly maintaining the base \(e\).
• Applying a differential operator like \(D\) to \(e^{2x}\) gives \(2e^{2x}\), reflecting its straightforward nature.

In our problem, the function \(y = e^{2x} + 3e^{-5x}\) comprises two exponential components. Each component can be individually handled by the differential operators, demonstrating the flexibility and power of these mathematical tools in working with exponential functions.