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Fundamentals Of Differential Equations And Boundary Value Problems

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

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 616 pages

ISBN: 9780321977069

Chapter 6: Q28E (page 337)

use the annihilator method to determinethe form of a particular solution for the given equation. $y'' - 6 y' + 10 y = e^{3 x} - x$

Short Answer

Expert verified

$y_{p} (x) = c_{3} + c_{4} x + c_{5} e^{3 x}$

Step by step solution

01

Solve the homogeneous of the given equation

The homogeneous of the given equation is

$(D^{2} - 6 D + 10) [y] = 0$

The solution of the homogeneous is

$y_{h} (x) = c_{1} e^{3 x} c o s x + c_{2} e^{3 x} s i n x$ (1)

Now $e^{3 x} - x$ is annihilated by $(D^{3} - 3 D^{2})$

Then, every solution to the given nonhomogeneous equation also satisfies

. $(D^{3} - 3 D^{2}) (D^{2} - 6 D + 10) [y] = 0$

Then

$y (x) = c_{1} e^{3 x} \cos x + c_{2} e^{3 x} \sin x + c_{3} + c_{4} x + c_{5} e^{3 x}$ (2)

is the general solution to this homogeneous equation

We know $u (x) = u_{h} + u_{p}$

Comparing (1) & (2)

$y_{p} (x) = c_{3} + c_{4} x + c_{5} e^{3 x}$

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Most popular questions from this chapter

(a) Derive the form $y (x) = A_{1} e^{x} + A_{2} e^{鈭� x} + A_{3} c o s x + A_{4} s i n x$ for the general solution to the equation $y^{(4)} = y$ , from the observation that the fourth roots of unity are 1, -1, i, and -i.

(b) Derive the form

$y (x) = A_{1} e^{x} + A_{2} e^{鈭� x / 2} c o s (\frac{\sqrt{3 x}}{2}) + A_{3} e^{鈭� x / 2} s i n (\frac{\sqrt{3 x}}{2})$

for the general solution to the equation $y^{(3)} = y$ from the observation that the cube roots of unity are 1, $e^{i \frac{2 蟺}{3}}$ , and $e^{鈭� i \frac{2 蟺}{3}}$ .

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

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

where $a_{n}, a_{n 鈭� 1}, . ..., a_{0}$ are constants, is called a homogeneous Cauchy鈥揈uler 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鈥揈uler equations:

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

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

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

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

find a differential operator that annihilates the given function.

$x^{2} - e^{x}$

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

$y^{(4)} + 4 y'' + 4 y = 0$

In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation.

$y^{'''} - 2 y^{''} + y^{'} = x$

See all solutions

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