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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: Q23E (page 337)

use the annihilator method to determinethe form of a particular solution for the given equation $y'' - 5 y' + 6 y = e^{3 x} - x^{2}$

Short Answer

Expert verified

$y'' - 5 y' + 6 y = e^{3 x} - x^{2}$

Step by step solution

01

Solve the homogeneous of the given equation

The homogeneous of the given equation is

$(D^{2} - 5 D + 6) [y] = (D - 2) (D - 3) [y] = 0$

The solution of the homogeneous is

$y_{h} (x) = c_{1} e^{2 x} + c_{2} e^{3 x}$ (1)

Let $g (x) = e^{3 x}$

Then

$(D - 3) [g] = 0$

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

Then

$D^{3} [h] = 0$

Hence

$D^{3} (D - 3) [g - h] = 0$

Then, every solution to the given nonhomogeneous equation also satisfies

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

Then

$y (x) = c_{1} e^{2 x} + c_{2} e^{3 x} + c_{3} x e^{3 x} + c_{4} + c_{5} x + c_{6} x^{2}$ (2)

is the general solution to this homogeneous equation

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

Comparing (1) & (2)

$y_{p} (x) = c_{3} x e^{3 x} + c_{4} + c_{5} x + c_{6} x^{2}$

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Most popular questions from this chapter

use the annihilator method to determinethe form of a particular solution for the given equation.

$y'' + 6 y' + 8 y = e^{3 x} - s i n x$

As an alternative proof that the functions $e^{r_{1} x}, e^{r_{2} x}, e^{r_{3} x}, ..., e^{r_{n} x}$ are linearly independent on (鈭�,-鈭�) when $r_{1}, r_{2}, . .. r_{n}$ are distinct, assume $C_{1} e^{r_{1} x} + C_{2} e^{r_{2} x} + C_{3} e^{r_{3} x} + . .. + C_{n} e^{r_{n} x}$ holds for all x in (鈭�,-鈭�) and proceed as follows:

(a) Because the ri鈥檚 are distinct we can (if necessary)relabel them so that $r_{1} > r_{2} > r_{3} > . .. > r_{n}$ .Divide equation (33) by to obtain $C_{1} + \frac{C_{2} e^{r_{2} x}}{e^{r_{2} x}} + \frac{C_{3} e^{r_{3} x}}{e^{r_{3} x}} + . .. + \frac{C_{n} e^{r_{n} x}}{e^{r_{n} x}} = 0$ Now let x鈫掆垶 on the left-hand side to obtainC1 = 0.(b) Since C1 = 0, equation (33) becomes

$C_{2} e^{r_{2} x} + C_{3} e^{r_{3} x} + . .. + C_{n} e^{r_{n} x}$ = 0for all x in(鈭�,-鈭�). Divide this equation by $e^{r_{2} x}$

and let x鈫掆垶 to conclude that C2 = 0.

(c) Continuing in the manner of (b), argue that all thecoefficients, C1, C2, . . . ,Cn are zero and hence $e^{r_{1} x}, e^{r_{2} x}, e^{r_{3} x}, . .., e^{r_{n} x}$ are linearly independent on(鈭�,-鈭�).

find a differential operator that annihilates the given function.

$x^{2} e^{- x} s i n 2 x$

find a differential operator that annihilates the given function.

$e^{2 x} - 6 e^{x}$

Reduction of Order. If a nontrivial solution f(x) is known for the homogeneous equation

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

the substitution $y (x) = v (x) f (x)$ can be used to reduce the order of the equation for second-order equations. By completing the following steps, demonstrate the method for the third-order equation

(35) $y''' - 2 y'' - 5 y' + 6 y = 0$

given that $f (x) = e^{x}$ is a solution.

(a) Set $y (x) = v (x) e^{x}$ and compute y鈥�, y鈥�, and y鈥�.

(b) Substitute your expressions from (a) into (35) to obtain a second-order equation in. $w = v'$

(c) Solve the second-order equation in part (b) for w and integrate to find v. Determine two linearly independent choices for v, say, $v_{1}$ and $v_{2}$ .

(d) By part (c), the functions $y_{1} (x) = v_{1} (x) e^{x}$ and $y_{2} (x) = v_{2} (x) e^{x}$ are two solutions to (35). Verify that the three solutions $e^{x}, 鈥� y_{1} (x)$ , and $y_{2} (x)$ are linearly independent on $(- 鈭�, 鈥� 鈭�)$

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