User:Egm4313.s12.team1.essenwein

Eric Essenwein

Problem R6.4

Statement

Consider the L2-ODE-CC (5) p7b-7 with the window function f(x) p.9-8 as excitation:

 $\displaystyle {y''-3y'+2y=r(x)}$ $r(x)=f(x)$ (4.0)

and the initial conditions

 $\displaystyle {y(0)=1,y'(0)=0}$ (4.1)

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

 $\displaystyle {y''_{n}+ay'_{n}+by_{n}=r_{n}(x)}$ (4.2)

with the same initial conditions (4.1).
Plot $y_{n}(x)$ for n=2,4,8, for x in [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:n=0,1

Part 1 Solution

First, we shift the excitation f(x) to the left by introducing a new independent variable, t. This allows the period to start at zero.

 $\displaystyle t=x-{\frac {1}{4}}$ (4.3)

The piecewise representation of the window function is now (in terms of t) as follows:

 $\displaystyle f(t)=A$ for $\displaystyle t=\left[0,2\right]$ $\displaystyle f(t)=0$ for $\displaystyle t=\left[2,4\right]$ (4.4)

To find the Fourier transform, the period of oscillation is determined:

 $\displaystyle p=2L=4$ (4.5)

And the frequency of oscillation:

 $\displaystyle \omega ={\frac {2\pi }{p}}={\frac {\pi }{2}}$ (4.6)

The general form of a Fourier transform is

 $\displaystyle f(t)={{a}_{0}}+\sum \limits _{1}^{n}{\left[{{a}_{n}}\cos(n\omega t)+{{b}_{n}}\sin(n\omega t)\right]}$ (4.7)

The following equations are given values of the constants in (4.7), evaluated for the function in (4.4):

 $\displaystyle {{a}_{0}}={\frac {1}{2L}}\int \limits _{0}^{2L}{f(t)dt}={\frac {1}{4}}\int \limits _{0}^{2}{Adt}={\frac {A}{4}}(2-0)={\frac {A}{2}}$ (4.8)

 $\displaystyle {{a}_{n}}={\frac {1}{L}}\int \limits _{0}^{2L}{f(t)\cos(n\omega t)dt}={\frac {1}{2}}\int \limits _{0}^{2}{A\cos({\frac {n\pi }{2}}t)dt}={\frac {A}{n\pi }}(\sin n\pi -\sin 0)$ (4.9)

 $\displaystyle {{b}_{n}}={\frac {1}{L}}\int \limits _{0}^{2L}{f(t)\sin(n\omega t)dt}={\frac {1}{2}}\int \limits _{0}^{2}{A\sin({\frac {n\pi }{2}}t)dt}=-{\frac {A}{n\pi }}(\cos n\pi -\cos 0)={\frac {A}{n\pi }}(1-\cos n\pi )$ (4.10)

In their simplest forms:

 $\displaystyle {{a}_{n}}=0$ (4.11)

 $\displaystyle {{b}_{n}}=\left\{0,{\frac {A}{\pi }},0,{\frac {A}{2\pi }},0,{\frac {A}{3\pi }},0,{\frac {A}{4\pi }},...,0,{\frac {2A}{n\pi }}\right\}$ (4.12)

Plugging in r(x)=f(t) in (4.0),

 $\displaystyle y''-3y'+2y={\frac {1}{2}}+{\frac {2}{n\pi }}\sin({\frac {n\pi }{2}}t)$ (4.13)

From the general form, a particular solution to (4.13) and its derivatives are as follows:

 $\displaystyle {{y}_{n}}={{A}_{n}}\cos({\frac {n\pi }{2}}t)+{{B}_{n}}\sin({\frac {n\pi }{2}}t)$ (4.14)

 $\displaystyle {{y}_{n}}'=-{{A}_{n}}{\frac {n\pi }{2}}\sin({\frac {n\pi }{2}}t)+{{B}_{n}}{\frac {n\pi }{2}}\cos({\frac {n\pi }{2}}t)$ (4.15)

 $\displaystyle {{y}_{n}}''=-{{A}_{n}}{\frac {{{n}^{2}}{{\pi }^{2}}}{4}}\cos({\frac {n\pi }{2}}t)-{{B}_{n}}{\frac {{{n}^{2}}{{\pi }^{2}}}{4}}\sin({\frac {n\pi }{2}}t)$ (4.16)

By plugging in (4.14), (4.15), and (4.16) into (4.13), we find An and Bn in terms of another constant, Cn:

 $\displaystyle {{A}_{n}}=6n\pi {{C}_{n}}$ (4.17)

 $\displaystyle {{B}_{n}}=(8-{{n}^{2}}{{\pi }^{2}}){{C}_{n}}$ (4.18)

 $\displaystyle {{C}_{n}}={{\left[{\tfrac {1}{8}}n\pi ({{(8-{{n}^{2}}{{\pi }^{2}})}^{2}}+{{(6n\pi )}^{2}})\right]}^{-1}}$ (4.19)

After substituting t with (4.3), the solutions are shown for n=2,4,8.

 $\displaystyle {{y}_{2}}={{y}_{1}}+{{y}_{2}}$ (4.20)

 $\displaystyle {{y}_{4}}={{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}$ (4.21)

 $\displaystyle {{y}_{8}}={{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}+{{y}_{5}}+{{y}_{6}}+{{y}_{7}}+{{y}_{8}}$ (4.22)

Part 2 Solution

The excitation for n=0 and n=1 are the same, because $a_{1}=b_{1}=0$ .

The homogeneous solution to (4.0) is

 $\displaystyle {{y}_{h}}={{C}_{1}}{{\operatorname {e} }^{2t}}+{{C}_{2}}{{\operatorname {e} }^{t}}$ (4.23)

For $r(x)=a_{0}$ , the particular solution is

 $\displaystyle {{y}_{p}}={{C}_{3}}$ (4.24)

Evaluating, we find that

 ${{C}_{1}}=-{\tfrac {3}{4}}$ ${{C}_{2}}={\tfrac {3}{2}}$ ${{C}_{3}}={\tfrac {1}{4}}$ (4.25)

Thus, the complete solution $y_{0}$ is

 ${{y}_{0}}={\tfrac {1}{4}}-{\tfrac {3}{4}}{{\operatorname {e} }^{2(x-{\tfrac {1}{4}})}}+{\tfrac {3}{2}}{{\operatorname {e} }^{x-{\tfrac {1}{4}}}}$ (4.26)

Integrating $y_{0}$ using MATLAB's ode45 command, the following plot is obtained for y:

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Problem R5.6

Statement

Consider the following L2-ODE-CC; see p.6-6:

 $\displaystyle {y}''-4{y}'+13{y}=2e^{-2x}\cos(3x)$ (6.0)

Homogeneous solution:

 $\displaystyle {y}_{h}(x)=e^{-2x}[A\cos 3x+B\sin 3x]$ (6.1)

Particular solution:

 $\displaystyle {y}_{p}(x)=xe^{-2x}[M\cos 3x+N\sin 3x]$ (6.2)

Complete the solution for this problem.

Find the overall solution $y(x)$ that corresponds to the initial condition (3b) p.3-7

 $\displaystyle y(0)=1,y'(0)=0$ (6.3)

Solution

Start by finding $y_{p}'$ and $y_{p}''$ .

 $\displaystyle {y}_{p}'={{e}^{-2x}}[(-3Mx-2Nx+N)\sin 3x+(-2Mx+3Nx+M)\cos 3x]$ (6.4)
 $\displaystyle {y}_{p}''={{e}^{-2x}}[(12Mx-6M-5Nx-4N)\sin 3x+(-15Mx-12M-12Nx+6N)\cos 3x]$ (6.5)

Substitute $y_{p}$ and its derivatives into (6.0) to find $M$ and $N$ .

 $\displaystyle {y}_{p}''+4{y}_{p}'+13{y}_{p}={{e}^{-2x}}[(-6M)\sin 3x+(-10Mx-8M+6N)\cos 3x]$ (6.6)

Separating terms and setting equal to the excitation from (6.0):

 $\displaystyle -6M{{e}^{-2x}}\sin 3x+(-8M+6N){{e}^{-2x}}\cos 3x-10Mx{{e}^{-2x}}\cos 3x=2{{e}^{-2x}}\cos 3x$ (6.7)

From (6.7), we solve coefficients to get

 $\displaystyle -6M=0$ $\displaystyle -8M+6N=2$ $\displaystyle -10M=0$ (6.8)

Solving for $M$ and $N$ :

 $\displaystyle M=0$ $\displaystyle N={\frac {1}{3}}$ (6.9)

Which gives us the particular solution:

 $\displaystyle {y}_{p}={\frac {1}{3}}x{{e}^{-2x}}\sin 3x$ (6.10)

For the general solution,

 $\displaystyle y={{y}_{h}}+{{y}_{p}}$ (6.11)
 $\displaystyle y=e^{-2x}[A\cos 3x+B\sin 3x]+{\frac {1}{3}}x{{e}^{-2x}}\sin 3x$ (6.12)
 $\displaystyle y=e^{-2x}[A\cos 3x+(B+{\frac {1}{3}}x)\sin 3x]$ (6.13)

To solve for $A$ and $B$ , we use initial conditions from (6.3):

 $\displaystyle y(0)=e^{-2(0)}[A\cos 3(0)+(B+{\frac {1}{3}}(0))\sin 3(0)]=1$ (6.14)

Which simplifies to:

 $\displaystyle A=1$ (6.15)

For the second initial condition from (6.3):

 $\displaystyle y'={{e}^{-2x}}[(-3A-2B+{\frac {1}{3}}+{\frac {2}{3}}x)\sin 3x+(3B+x-2)\cos 3x]$ (6.16)
 $\displaystyle y'(0)={{e}^{-2(0)}}[(-3A-2B+{\frac {1}{3}}+{\frac {2}{3}}(0))\sin 3(0)+(3B+(0)-2)\cos 3(0)]=0$ (6.17)
 $\displaystyle 3B-2=0$ (6.18)
 $\displaystyle B={\frac {2}{3}}$ (6.19)

We can now write $y$ with all coefficients known.

 $\displaystyle y={{e}^{-2x}}[(1)\cos 3x+({\frac {2}{3}}+{\frac {1}{3}}x)\sin 3x]$ (6.20)

Final Equation

 $\displaystyle y={{e}^{-2x}}[\cos 3x+({\frac {2}{3}}+{\frac {1}{3}}x)\sin 3x]$ (6.21)

Author

Solved and Typed By - Egm4313.s12.team1.essenwein (talk) 23:45, 19 March 2012 (UTC)
Reviewed By -