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Fundamentals Of Differential Equations And Boundary Value Problems

R. Kent Nagle, Edward B. Saff, Arthur David Snider

$Math Studyset Vaia Explanations$ Math

9th Edition · 616 pages

ISBN: 9780321977069

Chapter 6: Q16E (page 337)

find a differential operator that annihilates the given function.
$x^{2} - e^{x}$

Short Answer

Expert verified

$D^{4} - D^{3}$ is the differential operator that annihilates the given function.

Step by step solution

01

Differentiate the function continuously

Let the function be $f (x) = x^{2} - e^{x}$

Let $g (x) = x^{2}$

Then

$g' (x) = 2 x g'' (x) = 2 g''' (x) = 0 D^{3} [g] = 0$

Let $h (x) = - e^{x}$

Then

$h' (x) = - e^{x} h' (x) - h (x) = 0 (D - 1) [h] = 0$

Hence $D^{3} (D - 1) [f] = 0$

Then $D^{4} - D^{3}$ is the differential operator that annihilates the given function.

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Most popular questions from this chapter

Determine whether the given functions are linearly dependent or linearly independent on the specified interval. Justify your decisions.

${\sin^{2} x, \cos^{2} x, 1}$ on $(- \infty, \infty)$

Higher-Order Cauchy–Euler Equations. A differential equation that can be expressed in the form

$a_{n} x^{n} y^{(n)} (x) + a_{n - 1} x^{n - 1} y^{(n - 1)} (x) + . .. + a_{0} y (x) = 0$

where $a_{n}, a_{n - 1}, . ..., a_{0}$ are constants, is called a homogeneous Cauchy–Euler equation. (The second-order case is discussed in Section 4.7.) Use the substitution $y = x^{y}$ to help determine a fundamental solution set for the following Cauchy–Euler equations:

(a) $x^{3} y''' + x^{2} y'' - 2 x y' + 2 y = 0, x > 0$

(b) $x^{4} y^{(4)} + 6 x^{3} y''' + 2 x^{2} y'' - 4 x y + 4 y = 0, x > 0$

(c) $x^{3} y''' - 2 x^{2} y'' + 13 x y' - 13 y = 0, x > 0$

[Hint: $x^{α + β i} = e^{(α + β i) l n x} = x^{a} {c o s (β l n x) + i s i n (β l n x)}$ ]

Find a general solution to the Cauchy-Euler equation $x^{3} y^{'''} - 2 x^{2} y^{''} + 3 x y^{'} - 3 y = x^{2}, x > 0,$

given that $\{x, xlnx, x^{3}\}$ is a fundamental solution set for the corresponding homogeneous equation

Find a general solution for the differential equation with x as the independent variable:

$2 y''' + 5 y'' - 13 y' + 7 y = 0$

Determine the largest interval (a, b) for which Theorem 1 guarantees the existence of a unique solution on (a, b) to the given initial value problem.

$x (x + 1) y''' - 3 xy' + y = 0 y (\frac{- 1}{2}) = 1, y' (\frac{- 1}{2}) = y'' (\frac{- 1}{2}) = 0$

See all solutions

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