University of Florida/Egm4313/s12.team11.R6

• Find the fundamental period of ${\displaystyle \cos(n\omega x)\!}$ and ${\displaystyle \sin(n\omega x)\!}$
  
• Show that these functions also have period ${\displaystyle p\!}$.
  
• Show that the constants ${\displaystyle a_{0}\!}$ is also a periodic function with period ${\displaystyle p\!}$.

• Find the fundamental period, and show that these functions have a period of ${\displaystyle p\!}$.

${\displaystyle p=2L=2{\frac {\pi }{\omega }}\!}$

${\displaystyle L={\frac {\pi }{\omega }}\!}$

${\displaystyle f(x+np)=f(x)\!}$

${\displaystyle f(x)=cos(n\omega x)\!}$

${\displaystyle f(x+n{\frac {2\pi }{n\omega }})=cos(n\omega (x+{\frac {2\pi }{n\omega }}))\!}$

${\displaystyle f(x+{\frac {2\pi }{\omega }})=cos(n\omega x+2\pi )\!}$

${\displaystyle f(x+2L)=cos(n\omega x+2\pi )\!}$

              ${\displaystyle p=2L\!}$ is the fundamental period for ${\displaystyle cos(n\omega x)\!}$.

${\displaystyle f(x+np)=f(x)\!}$

${\displaystyle f(x)=sin(n\omega x)\!}$

${\displaystyle f(x+n{\frac {2\pi }{n\omega }})=sin(n\omega (x+{\frac {2\pi }{n\omega }}))\!}$

${\displaystyle f(x+{\frac {2\pi }{\omega }})=sin(n\omega x+2\pi )\!}$

${\displaystyle f(x+2L)=sin(n\omega x+2\pi )\!}$

              ${\displaystyle p=2L\!}$ is the fundamental period for ${\displaystyle sin(n\omega x)\!}$.

• Show that the constants ${\displaystyle a_{0}\!}$ is also a periodic function with period ${\displaystyle p\!}$.

From Fourier Series,
${\displaystyle a_{0}={\frac {1}{2L}}\int _{-L}^{L}f(x)dx\!}$

${\displaystyle f(x)=\cos(n\omega x)\!}$

${\displaystyle a_{0}={\frac {\omega }{2\pi }}\int _{-{\frac {\pi }{\omega }}}^{\frac {\pi }{\omega }}cos(n\omega x)dx\!}$

${\displaystyle a_{0}={\frac {\omega }{2\pi }}[{\frac {sin(n\omega x)}{n\omega }}|_{-{\frac {\pi }{\omega }}}^{\frac {\pi }{\omega }}]\!}$

${\displaystyle a_{0}={\frac {\omega }{2\pi }}[{\frac {2sin(n\pi )}{n\omega }}]\!}$

          ${\displaystyle a_{0}=0\!}$

${\displaystyle f(x)=\sin(n\omega x)\!}$

${\displaystyle a_{0}={\frac {\omega }{2\pi }}\int _{-{\frac {\pi }{\omega }}}^{\frac {\pi }{\omega }}sin(n\omega x)dx\!}$

${\displaystyle a_{0}={\frac {\omega }{2\pi }}[{\frac {cos(n\omega x)}{n\omega }}|_{-{\frac {\pi }{\omega }}}^{\frac {\pi }{\omega }}]\!}$

${\displaystyle a_{0}={\frac {\omega }{2\pi }}[{\frac {-cos{n\pi }+cos{n\pi }}{n\omega }}]\!}$

          ${\displaystyle a_{0}=0\!}$

Solve the problems 11 and 12 from Kreyszig p.491

First, for problem 11:
${\displaystyle f(x)=x^{2}\!}$ from ${\displaystyle <-1<x<1>\!}$ where ${\displaystyle p=2\!}$ We know that ${\displaystyle p=2=2L\!}$ so that ${\displaystyle L=1\!}$

We check is the function is even, odd or neither.
Since ${\displaystyle f(x)\!}$ satisfies the condition ${\displaystyle f(x)=f(-x)\!}$ we know that the function is even.
Given that the function is even and that ${\displaystyle b_{n}=0\!}$ for even functions, we exclude the ${\displaystyle b_{n}\!}$ term from the Fourier Series.

We find ${\displaystyle a_{0}\!}$:
${\displaystyle a_{0}={\frac {1}{2L}}\int _{-L}^{L}f(x)dx\!}$
Plugging in,
${\displaystyle a_{0}={\frac {1}{2}}\int _{-1}^{1}x^{2}dx={\frac {1}{2}}\left[{\frac {1}{3}}x^{3}\right]_{-1}^{1}={\frac {1}{2}}\left[{\frac {1}{3}}+{\frac {1}{3}}\right]={\frac {1}{3}}\!}$

We find ${\displaystyle a_{n}\!}$
${\displaystyle a_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)cos({\frac {n\pi x}{L}})dx\!}$
${\displaystyle a_{n}=1\int _{-1}^{1}x^{2}cos(n\pi x)dx\!}$
We can use the equivalent expression:
${\displaystyle a_{n}=2\int _{0}^{1}x^{2}cos(n\pi x)dx\!}$

Now we evaluate the integral by using integration by parts. Below are the steps:
Integration by parts uses ${\displaystyle \int fdg=fg-\int gdf\!}$
${\displaystyle {\begin{matrix}f=x^{2}&dg=cos(\pi nx)dx\\df=2xdx&g={\frac {sin(\pi nx)}{\pi n}}\end{matrix}}\!}$
Then, we have:
${\displaystyle \left[{\frac {x^{2}sin(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {2}{\pi n}}\int _{0}^{1}xsin(\pi nx)dx\!}$
Repeating for the second term:
${\displaystyle {\begin{matrix}f=x&dg=sin(\pi nx)\\df=dx&g={\frac {-cos(\pi nx)}{\pi n}}\end{matrix}}\!}$
Then, we have:
${\displaystyle \left[{\frac {-xcos(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {1}{\pi n}}\int _{0}^{1}cos(\pi nx)dx\!}$
Bringing all the terms together:
${\displaystyle 2\left[\left[{\frac {x^{2}sin(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {2}{\pi n}}\left[{\frac {-xcos(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {1}{\pi n}}\int _{0}^{1}cos(\pi nx)dx\right]\!}$
Then,
${\displaystyle 2\left[\left[{\frac {x^{2}sin(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {2}{\pi n}}\left[{\frac {-xcos(\pi nx)}{\pi n}}\right]_{0}^{1}-{\frac {1}{\pi n}}\left[{\frac {sin(\pi nx)}{\pi n}}\right]_{0}^{1}\right]\!}$
Evaluating at the boundaries and simplifying:
${\displaystyle 2\left[\left[0-0\right]+{\frac {2}{\pi ^{2}n^{2}}}\left[cos(\pi n)\right]-{\frac {1}{\pi n}}\left[0-0\right]\right]\!}$
Further,
${\displaystyle {\frac {4}{\pi ^{2}n^{2}}}cos(\pi n)\!}$
We can simplify one step further by acknowledging that ${\displaystyle cos(\pi n)=(-1)^{n}\!}$.
${\displaystyle \therefore a_{n}={\frac {4}{\pi ^{2}n^{2}}}(-1)^{n}\!}$

Finally,
${\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }a_{n}cos({\frac {\pi n}{L}})x\!}$

The Fourier series is:

                                   ${\displaystyle f(x)={\frac {1}{3}}+\sum _{n=1}^{\infty }{\frac {4}{\pi ^{2}n^{2}}}(-1)^{n}cos(\pi n)x\!}$

Now problem 12:
${\displaystyle f(x)=1-{\frac {x^{2}}{4}}\!}$ from ${\displaystyle <-2<x<2>\!}$ where ${\displaystyle p=2\!}$ We know that ${\displaystyle p=4=2L\!}$ so that ${\displaystyle L=2\!}$

We check is the function is even, odd or neither.
Since ${\displaystyle f(x)\!}$ satisfies the condition ${\displaystyle f(x)=f(-x)\!}$ we know that the function is even.
Given that the function is even and that ${\displaystyle b_{n}=0\!}$ for even functions, we exclude the ${\displaystyle b_{n}\!}$ term from the Fourier Series.

We find ${\displaystyle a_{0}\!}$:
${\displaystyle a_{0}={\frac {1}{2L}}\int _{-L}^{L}f(x)dx\!}$
Plugging in,
${\displaystyle a_{0}={\frac {1}{4}}\int _{-2}^{2}1-{\frac {x^{2}}{4}}dx={\frac {1}{4}}\left[x-{\frac {1}{12}}x^{3}\right]_{-2}^{2}={\frac {1}{4}}\left[{\frac {24-8}{12}}-{\frac {24+8}{12}}\right]={\frac {2}{3}}\!}$

We find ${\displaystyle a_{n}\!}$
${\displaystyle a_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)cos({\frac {n\pi x}{L}})dx\!}$
${\displaystyle a_{n}={\frac {1}{2}}\int _{-2}^{2}(1-{\frac {x^{2}}{4}}cos(n\pi x)dx\!}$
Now we evaluate the integral by using integration by parts. This process is similar to the one shown in the solution for problem 11 above and therefore will be performed in Wolfram Alpha.
The following command was given in Wolfram Alpha: integral (1-(x^2/4))*cos(pi*x) from -2 to 2.
Wolfram Alpha yields the following
${\displaystyle {\frac {2}{\pi ^{2}n^{2}}}\!}$

Finally,
${\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }a_{n}cos({\frac {\pi n}{L}})x\!}$

The Fourier series is:

                              ${\displaystyle f(x)={\frac {2}{3}}+\sum _{n=1}^{\infty }{\frac {-2}{\pi ^{2}n^{2}}}cos({\frac {\pi n}{2}})x\!}$

Solved by Francisco Arrieta

Find the Fourier series expansion for ${\displaystyle f(x)\!}$ on p. 9.8 as follows:

                                              ${\displaystyle f_{k}({\bar {x}})={\frac {A}{2}}+\sum _{k=1}^{n}({\frac {2A}{k\pi }}\cos {\frac {k\pi }{2}}{\bar {x}})\!}$

                                              ${\displaystyle f_{k}({\bar {x}})={\frac {A}{2}}\!}$

                                              ${\displaystyle f_{k}({\bar {x}})={\frac {A}{2}}+({\frac {2A}{\pi }}\cos {\frac {\pi }{2}}{\bar {x}})\!}$

                                              ${\displaystyle f_{k}({\bar {x}})={\frac {A}{2}}+{\frac {2A}{\pi }}(\cos {\frac {\pi }{2}}{\bar {x}}-{\frac {1}{3}}\cos {\frac {3\pi }{2}}{\bar {x}}+{\frac {1}{5}}\cos {\frac {5\pi }{2}}{\bar {x}})\!}$

                                              ${\displaystyle f_{k}(x)={\frac {A}{2}}+\sum _{k=1}^{n}({\frac {2A}{k\pi }}\cos {\frac {k\pi }{2}}(x-1.25))\!}$

                                              ${\displaystyle f_{k}({\tilde {x}})={\frac {A}{2}}+\sum _{k=1}^{n}({\frac {2A}{k\pi }}\sin {\frac {k\pi }{2}}{\tilde {x}})\!}$

                                              ${\displaystyle f_{k}({\tilde {x}})={\frac {A}{2}}\!}$

                                              ${\displaystyle f_{k}({\tilde {x}})={\frac {A}{2}}+({\frac {2A}{\pi }}\sin {\frac {\pi }{2}}{\tilde {x}})\!}$

                                              ${\displaystyle f_{k}({\tilde {x}})={\frac {A}{2}}+{\frac {2A}{\pi }}(\sin {\frac {\pi }{2}}{\tilde {x}}+{\frac {1}{3}}\sin {\frac {3\pi }{2}}{\tilde {x}}+{\frac {1}{5}}\sin {\frac {5\pi }{2}}{\tilde {x}})\!}$

                                              ${\displaystyle f_{k}(x)={\frac {A}{2}}+\sum _{k=1}^{n}({\frac {2A}{k\pi }}\sin {\frac {k\pi }{2}}(x-.25))\!}$

Page 491, #15,17. Plot the truncated Fourier series for n=2,4,8.

15.

${\displaystyle f(x)=\left\{{\begin{matrix}-x-\pi ,-\pi <x<-\pi /2\\x,-\pi /2<x<\pi /2\\\pi -x,\pi /2<x<\pi \end{matrix}}\right.}$

Since ${\displaystyle f(-x)=-f(x)}$, the function is odd.

${\displaystyle a_{0}=(1/\pi )*[-1\int _{-\pi }^{-\pi /2}(x+\pi )dx+\int _{-\pi /2}^{\pi /2}(x)dx+\int _{\pi /2}^{\pi }(\pi -x)dx]}$
After simplification, ${\displaystyle a_{0}=0}$
. Using Equation (5) on page 490, the exapansion becomes;
${\displaystyle f(x)=8k/\pi ^{2}[sin(\pi *x/L)-sin(3\pi *x/L)/9...)}$
With ${\displaystyle L=\pi ,k=\pi /2}$ this becomes;
${\displaystyle f(x)=4/\pi (sinx-sin(3x)/9+sin(5x)/25-sin(6x)/36....)}$ for n is "odd".
This means when n is even, ${\displaystyle f(x)=0}$

17.
${\displaystyle f(x)=1-|x|,-1\leq x\leq 1}$

${\displaystyle f(x)=\left\{{\begin{matrix}1+x-\pi ,-1<x<0\\1-x,0<x<1\\\end{matrix}}\right.}$

${\displaystyle a_{0}=1/2[\int _{-1}^{0}(1+x)dx+\int _{0}^{1}(1-x)dx]}$
${\displaystyle a_{0}=1/2[1-1/2+1-1/2]}$

 ${\displaystyle a_{0}=1/2}$

${\displaystyle a_{n}=1/1[\int _{-1}^{0}(1+x)cos(n\pi *x/1)dx+\int _{0}^{1}(1-x)cos(n\pi *x/1)dx]}$
Simplifying results in;
${\displaystyle a_{n}=2[1/(n^{2}\pi ^{2})_{(}-1)^{n}/(n^{2}\pi ^{2})]}$
So,

${\displaystyle a_{n}=4/(n^{2}\pi ^{2})}$

when n is odd.
${\displaystyle b_{n}=1/1[\int _{-1}^{0}(1+x)sin(n\pi *x/1)dx+\int _{0}^{1}(1-x)sin(n\pi *x/1)dx]}$
Simplifying results in;

${\displaystyle b_{n}=0}$

${\displaystyle f(x)=1/2+\sum _{n=0}^{\infty }4/(n^{2}\pi ^{2})*cos(n\pi *x)}$

Since ${\displaystyle a_{n}=0}$ for even n, ${\displaystyle f(x)}$ becomes ${\displaystyle f(x)=1/2}$ for even n.

Consider the L2-ODE-CC (5) p.7b-7 with the window function f(x) p.9-8 as excitation:

${\displaystyle y''-3y'+2y=f(x)\!}$

and the initial conditions

${\displaystyle y(0)=1\ ,\ y'(0)=0\!}$

1. Find ${\displaystyle y_{n}(x)\!}$ such that:

${\displaystyle y_{n}''+ay_{n}'+by_{n}=r_{n}(x)\!}$

with the same initial conditions as above.

Plot ${\displaystyle y_{n}(x)\!}$ for ${\displaystyle n=2,4,8\!}$ for x in ${\displaystyle [0,10]\!}$

2. Use the matlab command ode45 to integrate the L2-ODE-CC, and plot the numerical solution to compare with the analytical solution.

Level 1: ${\displaystyle n=0,1\!}$

One period of the window function p9.8 is described as follows

${\displaystyle f(x)={\begin{cases}0&\ \ -1.75<x<l0.25\\A&\ \ 0.25<x<2.25\\\end{cases}}}$

From the above intervals one can see that the period, ${\displaystyle p=4\!}$ and therefore ${\displaystyle L=2\!}$ Applying the Euler formulas from ${\displaystyle -1.75}$ to ${\displaystyle 2.25\!}$ the Fourier coefficients are computed:

${\displaystyle a_{0}={\frac {1}{2L}}\int _{-1.75}^{2.25}f(x)\ dx\!}$

${\displaystyle a_{0}={\frac {1}{4}}(\int _{-1.75}^{0.25}0\ dx+\int _{0.25}^{2.25}A\ dx)\!}$

${\displaystyle a_{0}=+{\frac {1}{4}}(0+\int _{0.25}^{2.25}A\ dx)\!}$

${\displaystyle a_{0}={\frac {1}{4}}(2.25A-0.25A)\!}$

${\displaystyle a_{0}={\frac {A}{2}}\!}$

The integral from ${\displaystyle -1.75\!}$ to ${\displaystyle 0.25\!}$ can be omitted from this point on since it is always zero.

${\displaystyle a_{n}={\frac {1}{L}}\int _{0.25}^{2.25}f(x)cos({\frac {n\pi x}{L}})\ dx\!}$

${\displaystyle a_{n}={\frac {1}{2}}\int _{0.25}^{2.25}Acos({\frac {n\pi x}{2}})\ dx\!}$

${\displaystyle a_{n}={\frac {2A}{2n\pi }}(sin({\frac {2.25n\pi }{2}})-sin({\frac {0.25n\pi }{2}})\!}$

${\displaystyle a_{n}={\frac {A}{n\pi }}(sin({\frac {9n\pi }{8}})-sin({\frac {n\pi }{8}})\!}$

and

${\displaystyle b_{n}={\frac {1}{L}}\int _{0.25}^{2.25}f(x)sin({\frac {n\pi x}{L}})\ dx\!}$

${\displaystyle b_{n}={\frac {1}{2}}\int _{0.25}^{2.25}Asin({\frac {n\pi x}{2}})\ dx\!}$

${\displaystyle b_{n}={\frac {2A}{2n\pi }}(cos({\frac {2.25n\pi }{2}})-cos({\frac {0.25n\pi }{2}})\!}$

${\displaystyle b_{n}={\frac {A}{n\pi }}(Acos({\frac {n\pi }{8}})-Acos({\frac {9n\pi }{8}})\!}$

The coefficients give the Fourier series:

${\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}cos({\frac {n\pi x}{L}})+b_{n}sin({\frac {n\pi x}{L}})]\!}$

${\displaystyle f(x)={\frac {A}{2}}+\sum _{n=1}^{\infty }[{\frac {A}{n\pi }}(sin({\frac {9n\pi }{8}})-sin({\frac {n\pi }{8}}))cos({\frac {n\pi x}{2}})\!}$

${\displaystyle +{\frac {A}{n\pi }}(cos({\frac {n\pi }{8}})-cos({\frac {9n\pi }{8}}))sin({\frac {n\pi x}{2}})]\!}$

Considering the homogeneous case of our ODE:

${\displaystyle y''-3y'+2y=0\!}$

The characteristic equation is

${\displaystyle \lambda ^{2}-3\lambda +2=0\!}$

${\displaystyle (\lambda -1)(\lambda -2)=0\!}$

${\displaystyle \lambda _{1}=1,\lambda _{1}=2\!}$

Therefore our homogeneous solution is of the form

${\displaystyle y_{h}=c_{1}e^{x}+c_{2}e^{2x}\!}$

Considering the case with f(x) as excitation

${\displaystyle y''-3y'+2y={\frac {A}{2}}+\sum _{k=1}^{n}{\frac {A}{k\pi }}[(sin({\frac {9k\pi }{8}})-sin({\frac {k\pi }{8}}))cos({\frac {k\pi x}{2}})\!}$

${\displaystyle +(cos({\frac {k\pi }{8}})-cos({\frac {9k\pi }{8}}))sin({\frac {k\pi x}{2}})]\!}$

The solution will be of the form

${\displaystyle y_{n}=A_{0}+\sum _{k=1}^{n}A_{k}cos({\frac {k\pi x}{2}})+\sum _{k=1}^{n}B_{k}sin({\frac {k\pi x}{2}})\!}$

Taking the derivatives

${\displaystyle y_{n}'=-A_{n}{\frac {\pi }{2}}\sum _{k=2}^{n}ksin({\frac {k\pi x}{2}})+B_{n}{\frac {\pi }{2}}\sum _{k=2}^{n}kcos({\frac {k\pi x}{2}})\!}$

${\displaystyle y_{n}''=-A_{n}{\frac {\pi ^{2}}{4}}\sum _{k=3}^{n}k^{2}cos({\frac {k\pi x}{2}})-B_{n}{\frac {\pi ^{2}}{4}}\sum _{k=3}^{n}k^{2}sin({\frac {k\pi x}{2}})\!}$

Plugging these back into the ODE:

${\displaystyle -{\frac {\pi ^{2}}{4}}\sum _{k=3}^{n}A_{k}k^{2}cos({\frac {k\pi x}{2}})-{\frac {\pi ^{2}}{4}}\sum _{k=3}^{n}B_{k}k^{2}sin({\frac {k\pi x}{2}})-3[-{\frac {\pi }{2}}\sum _{k=2}^{n}A_{k}ksin({\frac {k\pi x}{2}})\!}$

${\displaystyle +{\frac {\pi }{2}}\sum _{k=2}^{n}B_{k}kcos({\frac {k\pi x}{2}})]+2[A_{0}+\sum _{k=1}^{n}A_{k}cos({\frac {k\pi x}{2}})+\sum _{k=1}^{n}B_{k}sin({\frac {k\pi x}{2}})]\!}$

${\displaystyle ={\frac {A}{2}}+\sum _{k=1}^{n}{\frac {A}{k\pi }}[(sin({\frac {9k\pi }{8}})-sin({\frac {k\pi }{8}}))cos({\frac {k\pi x}{2}})+(cos({\frac {k\pi }{8}})-cos({\frac {9k\pi }{8}}))sin({\frac {k\pi x}{2}})]\!}$

Setting the two constants equal

${\displaystyle 2A_{0}={\frac {A}{2}}\!}$

${\displaystyle A_{0}={\frac {A}{4}}\!}$

This is valid for all values of n. Since the coefficients of the excitation ${\displaystyle sin({\frac {9k\pi }{8}})-sin({\frac {k\pi }{8}})\!}$ and ${\displaystyle cos({\frac {k\pi }{8}})-cos({\frac {9k\pi }{8}})\!}$ are zero for all even n, then the coefficients ${\displaystyle A_{k}\!}$ and ${\displaystyle B_{k}\!}$ will also be zero, so we must only find these coefficients for odd n's. Now carrying out the sum to ${\displaystyle n=1\!}$ and comparing like terms yields the following sets of equations. Written in matrix form:

${\displaystyle {\begin{bmatrix}2&0\\\\0&2\end{bmatrix}}*{\begin{bmatrix}A_{1}\\B_{1}\end{bmatrix}}={\begin{bmatrix}\ {\frac {A}{\pi }}(sin({\frac {9\pi }{8}})-sin({\frac {\pi }{8}}))\\\ {\frac {A}{\pi }}(cos({\frac {\pi }{8}})-cos({\frac {9\pi }{8}}))\end{bmatrix}}\!}$

Assuming ${\displaystyle A=1\!}$ this matrix can be solved to obtain

${\displaystyle A_{1}=-0.1218\ ,\ B_{1}=0.2941\!}$

For the remaining coefficients to be solved all sums will be used so a more general equation may be written:

${\displaystyle {\begin{bmatrix}2-{\frac {(\pi k)^{2}}{4}}&{\frac {-3\pi k}{2}}\\{\frac {-3\pi k}{2}}&2-{\frac {(\pi k)^{2}}{4}}\end{bmatrix}}*{\begin{bmatrix}A_{k}\\B_{k}\end{bmatrix}}={\begin{bmatrix}\ {\frac {A}{\pi k}}(sin({\frac {9\pi k}{8}})-sin({\frac {\pi k}{8}}))\\\ {\frac {A}{\pi k}}(cos({\frac {\pi k}{8}})-cos({\frac {9\pi }{8}}))\end{bmatrix}}\!}$

Results of these calculations are shown below:

${\displaystyle A={\begin{bmatrix}A_{1}\\A_{2}\\.\\.\\A_{7}\\A_{8}\end{bmatrix}}={\begin{bmatrix}-0.1218\\0\\0.0084\\0\\0.0014\\0\\0.0001\\0\end{bmatrix}},\ B={\begin{bmatrix}B_{1}\\B_{2}\\.\\.\\B_{7}\\B_{8}\end{bmatrix}}={\begin{bmatrix}0.2941\\0\\0.0019\\0\\0.0014\\0\\0.0007\\0\end{bmatrix}}\!}$

The solution to the particular case can be written for all n (assuming A=1):

      ${\displaystyle y_{n}={\frac {1}{4}}+\sum _{k=1}^{n}A_{k}cos({\frac {k\pi x}{2}})+\sum _{k=1}^{n}B_{k}sin({\frac {k\pi x}{2}})\!}$

The general solution is

${\displaystyle y=y_{h}+y_{p}\!}$

where

${\displaystyle y_{h}=c_{1}e^{x}+c_{2}e^{2x}\!}$

Different coefficients ${\displaystyle c_{1}\ ,\ c_{2}\!}$ will be calculated for each n. These coefficients are easily solved for by applying the given initial conditions. Below are the calculations for n=2.

${\displaystyle y=c_{1}e^{x}+c_{2}e^{2x}+{\frac {1}{4}}+\sum _{k=1}^{2}A_{k}cos({\frac {k\pi x}{2}})+\sum _{k=1}^{2}B_{k}sin({\frac {k\pi x}{2}})\!}$

Applying the first initial condition ${\displaystyle y(0)=1\!}$

${\displaystyle y(0)=c_{1}+c_{2}+A_{1}+A_{2}=1\!}$