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.

${\mathit{y}}^{\mathbf{"}}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}\{\begin{array}{l}x,0<x<1\\ 0,-1<x<0\end{array}$

Short answer

Answer

The solution of function $y^{"}+9y=\{\begin{array}{l}\mathrm{x},0<\mathrm{x}<1\\ 0,-1<\mathrm{x}<0\end{array}$is $a_n=\frac{1}{\mathrm{\pi }}\left(1-\sin x-\cos x\right)$.

Step-by-step solution

Given information from question

The function is given as$y^{"}+9y=\{\begin{array}{l}\mathrm{x},0<\mathrm{x}<1\\ 0,-1<\mathrm{x}<0\end{array}$.

The 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

$y^{"}+9y=\{\begin{array}{l}\mathrm{x},0<\mathrm{x}<1\\ 0,-1<\mathrm{x}<0\end{array}$

Hereis a function of period

$f\left(x\right)=\{\begin{array}{l}\mathrm{x},if0<\mathrm{x}<1\\ 0,if-1<\mathrm{x}<0\end{array}$

The auxiliary equation is

$$\begin{gathered} D^2+9=0 \\ {D}^2=-9 \\ D=\pm 3i \end{gathered}$$

Calculate the complementary function

The complementary function is

$$\begin{gathered} y_c=e^{3x}\left(A\cos 3x+B\sin 3x\right) \\ f\left(x\right)=\left\{\begin{array}{l}xif<x<1\\ 0if-1<x<0\end{array}\right. \\ {a}_n=\frac{1}{\mathrm{\pi }}{\int }_1^{1}f\left(x\right)\cos nxd \\ {a}_n=\frac{1}{\mathrm{\pi }}{\int }_1^{0}0\cos nxdx+{\int }_{}^{1}x\cos nxdx \end{gathered}$$

Solve equation further

$$\begin{gathered} a_n=\frac{1}{\mathrm{\pi }}{\int }_0^{1}x\cos nxdx \\ {a}_n=\frac{1}{\mathrm{\pi }}{\left(x\cos nx+nx\right)}_0^1 \\ {a}_n=\frac{1}{\mathrm{\pi }}\left(1-\sin x-\cos x\right) \end{gathered}$$

Thus, the solution of functionis.