Question

Evaluate the following functions in differential operator form.

1. D(x²+2x-3) Ans. (2x+2)
2. D²(xe^3x-e^4x) Ans. 9xe^3x+6e^3x-16e^4x

Answer 1

$D(x^2 + 2x + 3) \\ \text{Since}\, D = \dfrac{d}{dx} \\ \text{So, we have,}\\ \dfrac{d}{dx} (x^2 + 2x + 3) \\ = \dfrac{d}{dx} (x^2) + \dfrac{d}{dx} (2x) + \dfrac{d}{dx} (3) \\ = 2x + 2 \\ \\ D^2(xe^{3x} - e^{4x})\\ \text{Since} \, D = \dfrac{d}{dx} \\ \text{So, we have,}\\ \dfrac{d^2}{dx^2} (xe^{3x} - e^{4x})\\ \dfrac{d}{dx} (\dfrac{d}{dx} (xe^{3x} - e^{4x}))\\ = \dfrac{d}{dx} (\dfrac{d}{dx} xe^{3x} - \dfrac{d}{dx} e^{4x})\\ \text{Recall the product rule of differentiation}\\ UV = Vdu + Udv\\ \text{By comparison let} \, u = x, v = e^{3x}\\ \text{Upon differentiation. We have,}\\ \dfrac{d}{dx} (e^{3x} + 3xe^{3x} - 4e^{4x})\\ = \dfrac{d}{dx} (e^{3x}) + 3 \dfrac{d}{dx} (xe^{3x}) - 4 \dfrac{d}{dx} (e^{4x})\\ = 3e^{3x} + 3(e^{3x} + 3xe^{3x}) - 4(4e^{4x})\\ = 3e^{3x} + 3e^{3x} + 9xe^{3x} - 16e^{4x}\\ = 9xe^{3x} + 6e^{3x} - 16e^{4x}$