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

Electric Charge Field and Potential

Linear Momentum

Radiation

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Measurements

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