Chapter 8: Q41P (page 430)

Solve by use of Fourier series. Assume in each case that the right-hand side is a periodic function whose values are stated for one period.
$y^{"} + 2 y^{'} + 2 y = | x |, - π < x < π$ .

Short Answer

Expert verified

Answer

The solution of $y^{"} + 2 y^{'} + 2 y = | x |, - π < x < π$ is $f (x) = \frac{1}{π} + \frac{1}{2} s i n x - \frac{2}{π} (\frac{c o s 2 x}{2^{2} - 1} + \frac{c o s 4 x}{4^{2} - 1} + \frac{c o s 4 x}{4^{2} - 1} + \frac{c o s 4 x}{6^{2} - 1} + . . . . . . . . . . . . . .)$ .

Step by step solution

Given information from question

The equation is $y^{"} + 2 y^{'} + 2 y = | x |, - π < x < π$ .

Quadratic formula

The quadratic formula to find out the roots is,

$\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Expand f(x) in a Fourier series

The given function is

$y^{"} + 2 y^{'} + 2 y = | x |, - π < x < π$

Here $f (x)$ is a function of period

$f (x) = \{\begin{cases} x & i f 0 \leq x < - π \\ - x & i f - π \leq x < π \end{cases}$

The auxiliary equation is

$D^{2} + 2 D + 2 = 0 m^{2} + 2 m + 2 = 0$

Apply quadratic formula

By Quadratic formula

$m = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- 2 \pm \sqrt{2^{2} - 4 \times 1 \times 2}}{2} = \frac{- 2 \pm \sqrt{4 - 8}}{2}$

Solve further

$m = \frac{- 2 \pm \sqrt{- 4}}{2} = - 1 \pm i$

The complementary function is

role="math" localid="1654080410163" $y_{c} e^{- x} (A \cos x + B \sin x) f (x) = \{\begin{cases} x & i f 0 \leq x < - π \\ - x & i f - π \leq x < π \end{cases} f (x) = \frac{1}{π} + \frac{1}{2} \sin x - \frac{2}{π} (\frac{\cos 2 x}{2^{2} - 1} + \frac{\cos 4 x}{4^{2} - 1} + \frac{\cos 6 x}{6^{2} - 1} + . . . . . . . . . . .)$

Thus, the solution of $y^{"} + 2 y^{'} + 2 y = | x |, - π < x < π$ is $f (x) = \frac{1}{π} + \frac{1}{2} \sin x - \frac{2}{π} (\frac{\cos 2 x}{2^{2} - 1} + \frac{\cos 4 x}{4^{2} - 1} + \frac{\cos 6 x}{6^{2} - 1} + . . . . . . . . . . .)$

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Solve by use of Fourier series. Assume in each case that the right-hand side is a periodic function whose values are stated for one period.
$y^{"} + 2 y^{'} + 2 y = | x |, - π < x < π$ .

Short Answer

Step by step solution

Given information from question

Quadratic formula

Expand f(x) in a Fourier series

Apply quadratic formula

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Solve by use of Fourier series. Assume in each case that the right-hand side is a periodic function whose values are stated for one period.y"+2y'+2y=|x|,-π<x<π.

Short Answer

Step by step solution

Given information from question

Quadratic formula

Expand f(x) in a Fourier series

Apply quadratic formula

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Modern Physics

Solid State Physics

Translational Dynamics

Magnetism and Electromagnetic Induction

Electrostatics

Kinematics Physics

Study anywhere. Anytime. Across all devices.

Solve by use of Fourier series. Assume in each case that the right-hand side is a periodic function whose values are stated for one period.
$y^{"} + 2 y^{'} + 2 y = | x |, - π < x < π$ .