Chapter 6: Q27E (page 337)

use the annihilator method to determinethe form of a particular solution for the given equation.
$y'' + 2 y' + 2 y = e^{- x} c o s x + x^{2}$

Short Answer

Expert verified

$y_{p} (x) = c_{1} + c_{2} x + c_{3} x^{2} + c_{6} x e^{- x} \cos x + c_{7} x e^{- x} \sin x$

Step by step solution

Solve the homogeneous of the given equation

The homogeneous of the given equation is

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

The solution of the homogeneous is

$y_{h} (x) = c_{1} e^{- x} \cos x + c_{2} e^{- x} \sin x$ (1)

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

Then, every solution to the given nonhomogeneous equation also satisfies

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

Then

$y (x) = c_{1} + c_{2} x + c_{3} x^{2} + c_{4} e^{- x} \cos x + c_{5} e^{- x} \sin x +$ $c_{3} x^{2} + c_{6} x e^{- x} \cos x + c_{7} x e^{- x} \sin x$ (2)

is the general solution to this homogeneous equation

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

Comparing (1) & (2)

$y_{p} (x) = c_{1} + c_{2} x + c_{3} x^{2} + c_{6} x e^{- x} \cos x + c_{7} x e^{- x} \sin x$

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use the annihilator method to determinethe form of a particular solution for the given equation.
$y'' + 2 y' + 2 y = e^{- x} c o s x + x^{2}$

Short Answer

Step by step solution

Solve the homogeneous of the given equation

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use the annihilator method to determinethe form of a particular solution for the given equation.y''+2y'+2y=e-xcosx+x2

Short Answer

Step by step solution

Solve the homogeneous of the given equation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Probability and Statistics

Decision Maths

Calculus

Theoretical and Mathematical Physics

Applied Mathematics

Study anywhere. Anytime. Across all devices.

use the annihilator method to determinethe form of a particular solution for the given equation.
$y'' + 2 y' + 2 y = e^{- x} c o s x + x^{2}$