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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: Q3E (page 337)

use the method of undetermined coefficients to determine the form of a particular solution for the given equation.
$y''' + 3 y'' - 4 y = e^{- 2 x}$

Short Answer

Expert verified

$y_{p} (x) = - \frac{1}{6} x^{2} e^{- 2 x}$

Step by step solution

01

Find the corresponding auxiliaryequation

Theauxiliary equationof corresponding homogeneous equation

$r^{3} + 3 r^{2} - 4 = (r - 1) (r + 2)^{2} = 0$

The solutions of the auxiliary equation are

$r = - 2, r = - 2, r = 1$

02

Find particular solution

Let the particular solution be

$y_{p} (x) = a x^{2} e^{- 2 x}$

Then

$y_{p}^{'} (x) = 2 a x e^{- 2 x} - 2 a x^{2} e^{- 2 x} y_{p}^{''} (x) = 2 a e^{- 2 x} - 8 a x e^{- 2 x} + 4 a x^{2} e^{- 2 x} y_{p}^{'''} (x) = - 12 a e^{- 2 x} + 24 a x e^{- 2 x} - 8 a x^{2} e^{- 2 x}$

Then

$y_{p}^{'''} (x) + 3 y_{p}^{''} (x) - 4 y_{p} (x) = - 12 a e^{- 2 x} + 24 a x e^{- 2 x} - 8 a x^{2} e^{- 2 x} + 6 a e^{- 2 x} - 24 a x e^{- 2 x} + 12 a x^{2} e^{- 2 x} - 4 a x^{2} e^{- 2 x} = - 6 a e^{- 2 x} .$

If $- 6 a e^{- 2 x} = e^{- 2 x} t h e n - 6 a = 1$

Then $a = - \frac{1}{6}$

Hence $y_{p} (x) = - \frac{1}{6} x^{2} e^{- 2 x}$

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Most popular questions from this chapter

On a smooth horizontal surface, a mass of m₁ kg isattached to a fixed wall by a spring with spring constantk₁ N/m. Another mass of m₂ kg is attached to thefirst object by a spring with spring constant k₂ N/m. Theobjects are aligned horizontally so that the springs aretheir natural lengths. As we showed in Section 5.6, thiscoupled mass鈥搒pring system is governed by the systemof differential equations

$m_{1} \frac{d^{2} x}{d t^{2}} + (k_{1} + k_{2}) x 鈭� k_{2} y = 0$

$m_{2} \frac{d^{2} y}{d t^{2}} 鈭� k_{2} x + k_{2} y = 0$

Let鈥檚 assume that m₁ = m₂ = 1, k₁ = 3, and k₂ = 2.If both objects are displaced 1 m to the right of theirequilibrium positions (compare Figure 5.26, page 283)and then released, determine the equations of motion forthe objects as follows:

(a)Show that x(2) satisfies the equation $x^{(4)} (t) + 7 x'' (t) + 6 x (t) = 0$

(b) Find a general solution x(2) to (36).

(c) Substitute x(2) back into (34) to obtain a generalsolution for y(2)

(d) Use the initial conditions to determine the solutions,x(2) and y(2), which are the equations of motion.

In Problems 38 and 39, use the elimination method of Sectionto find a general solution to the given system.

$x - d^{2} y / d t^{2} = t + 1$

$d x / d t + d y / d t - 2 y = e^{t}$

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

$y''' + 2 y'' - 8 y' = 0$

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

$y^{(4)} + 2 y''' + 10 y'' + 18 y' + 9 y = 0$

(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 $e^{鈭� i \frac{2 蟺}{3}}$ .

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