Question

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.

${\mathit{y}}^{\mathbf{"}}\mathbf{+}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\left|x\right|\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{\pi }\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{\pi }$.

Short answer

Answer

The solution of $y^{"}+2y^{\text{'}}+2y=\left|\mathrm{x}\right|,-\mathrm{\pi }<\mathrm{x}<\mathrm{\pi }$is $\mathit{f}\left(x\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{x}\mathbf{-}\frac{\mathbf{2}}{\mathbf{\pi }}\left(\frac{cos2x}{2^2-1}+\frac{cos4x}{4^2-1}+\frac{cos4x}{4^2-1}+\frac{cos4x}{6^2-1}+..............\right)$.

Step-by-step solution

Given information from question

The equation is$y^{"}+2y^{\text{'}}+2y=\left|\mathrm{x}\right|,-\mathrm{\pi }<\mathrm{x}<\mathrm{\pi }$.

Quadratic formula

The quadratic formula to find out the roots is,

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

Expand f(x) in a Fourier series

The given function is

$y^{"}+2y^{\text{'}}+2y=\left|\mathrm{x}\right|,-\mathrm{\pi }<\mathrm{x}<\mathrm{\pi }$

Here $f\left(x\right)$is a function of period

$f\left(x\right)=\left\{\begin{array}{ll}x& if0\le x<-\mathrm{\pi }\\ -x& if-\mathrm{\pi }\le \mathrm{x}<\mathrm{\pi }\end{array}\right.$

The auxiliary equation is

$$\begin{gathered} D^2+2D+2=0 \\ {m}^2+2m+2=0 \end{gathered}$$

Apply quadratic formula

By Quadratic formula

$$\begin{gathered} m=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ =\frac{-2\pm \sqrt{2^2-4\times 1\times 2}}{2} \\ =\frac{-2\pm \sqrt{4-8}}{2} \end{gathered}$$

Solve further

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

The complementary function is

role="math" localid="1654080410163" $$\begin{gathered} y_c{e}^{-x}\left(A\cos x+B\sin x\right) \\ f\left(x\right)=\left\{\begin{array}{ll}x& if0\le x<-\mathrm{\pi }\\ -x& if-\mathrm{\pi }\le \mathrm{x}<\mathrm{\pi }\end{array}\right. \\ f\left(x\right)=\frac{1}{\mathrm{\pi }}+\frac{1}{2}\sin x-\frac{2}{\mathrm{\pi }}\left(\frac{\cos 2x}{2^2-1}+\frac{\cos 4x}{4^2-1}+\frac{\cos 6x}{6^2-1}+...........\right) \end{gathered}$$

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