University of Florida/Egm4313/IEA-f13-team10/R6

ODE: ${\displaystyle y''-3y'-10y=3cos7x}$

Part 1: show that cos7x and sin7x are linearly independent using the Wronskian and Gramian.

Part 2: Find 2 equations for the two unknowns M, N, and solve for M, N.

Part 3: Find the overall solution y(x) that corresponds to the initial conditions: ${\displaystyle y(0)=1,y'(0)=0}$

Plot the solution over 3 periods

Wronskian: Function is linearly independent if ${\displaystyle W(f,g)\neq 0}$
${\displaystyle W(f,g):={\begin{bmatrix}f&g\\f'&g'\end{bmatrix}}=fg'-gf'}$

${\displaystyle f=cos7x,f'=-7sin7x}$

${\displaystyle g=sin7x,g'=7cos7x}$

${\displaystyle W(cos7x,sin7x):={\begin{bmatrix}cos7x&sin7x\\-7sin7x&7cos7x\end{bmatrix}}=7cos^{2}7x+7sin^{2}7x=1}$

g(x) and f(x) are linearly independent

Gramian: Function is linearly independent if ${\displaystyle \Gamma (f,g)\neq 0}$
${\displaystyle \Gamma (f,g):={\begin{bmatrix}<f,f>&<f,g>\\<g,f>&<g,g>\end{bmatrix}}}$

${\displaystyle f(x)=cos7x,g(x)=sin7x}$
${\displaystyle <f,f>=\int _{-1}^{1}cos7x*cos7xdx=1.07}$
${\displaystyle <f,g>=\int _{-1}^{1}cos7x*sin7xdx=0}$
${\displaystyle <g,f>=\int _{-1}^{1}sin7x*cos7xdx=0}$
${\displaystyle <g,g>=\int _{-1}^{1}sin7x*sin7xdx=0.93}$

${\displaystyle \Gamma (f,g):={\begin{bmatrix}1.07&0\\0&0.93\end{bmatrix}}=0.9951}$

g(x) and f(x) are linearly independent

The particular solution for a ${\displaystyle r(x)=3cos7x}$ will be:

${\displaystyle y_{p}(x)=Mcos7x+Nsin7x}$

Differentiate to get:

${\displaystyle y_{p}'(x)=-M7sin7x+N7cos7x}$

${\displaystyle y_{p}''(x)=-M7^{2}cos7x-N7^{2}sin7x}$

Plug the derivatives into the equation:

${\displaystyle y_{p}''(x)-3y_{p}'(x)-10y_{p}(x)=3cos7x}$

${\displaystyle (-M7^{2}cos7x-N7^{2}sin7x)-3(-M7sin7x+N7cos7x)-10(Mcos7x+Nsin7x)}$

Separate the sin and cos terms to get 2 equations in order to solve for M and N

${\displaystyle -M7^{2}cos7x-3N7cos7x-10Mcos7x=3cos7x}$

${\displaystyle -N7^{2}sin7x+3M7sin7x-10Nsin7x=0}$

dividing each equation by cos7x and sin7x respectively:

${\displaystyle -49M-21N-10M=3}$

${\displaystyle -49N+21M-10N=0}$

${\displaystyle M=-0.0454}$

${\displaystyle N=-0.0154}$

So the particular solution is:

${\displaystyle y_{p}(x)=-0.0454cos7x-0.0154sin7x}$

The overall solution can be found by:

${\displaystyle y(x)=y_{h}(x)+y_{p}(x)}$

The roots given in the problem statement ${\displaystyle \lambda _{1}=-2,\ \lambda _{2}=+5}$

Lead to the homogeneous solution of:

${\displaystyle y_{h}(x)=C_{1}e^{-2x}+C_{2}e^{5x}}$

Combining the homogeneous and particular solution gives us:

${\displaystyle y(x)=C_{1}e^{-2x}+C_{2}e^{5x}-0.0454cos7x-0.0154sin7x}$

Solving for the constants by using the initial conditions ${\displaystyle y(0)=1,y'(0)=0}$

${\displaystyle y(0)=C_{1}e^{0}+C_{2}e^{0}-0.0454cos(0)-0.0154sin(0)=1}$

${\displaystyle y'(0)=-2C_{1}e^{0}+5C_{2}e^{0}+0.3178sin(0)-0.1078cos(0)=0}$

${\displaystyle C_{1}=0.73}$

${\displaystyle C_{2}=0.31}$

The overall solution is:

${\displaystyle y(x)=0.73e^{-2x}+0.31e^{5x}-0.0454cos7x-0.0154sin7x}$

Plot ${\displaystyle y(x)=0.73e^{-2x}+0.31e^{5x}-0.0454cos7x-0.0154sin7x}$

over 3 periods:

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Complete the solution to problem on p.8-6.
Find the overall solution ${\displaystyle y(x)}$
that corresponds to the initial condition ${\displaystyle y(0)=1,y'(0)=0}$
Plot solution over 3 periods.

Given:
${\displaystyle y''+4y'+13y=2*e^{-2x}*3cos7x}$
${\displaystyle y_{h}(x)=e^{-2x}*(Acos(3x)+BNsin(3x))}$
${\displaystyle y_{P}(x)=x*e^{-2x}*(Mcos(3x)+Nsin(3x))}$


${\displaystyle y_{P}'(x)=e^{-2x}*cos(3x)(x(-2M+N)+M)+e^{-2x}*sin(3x)(x(-2N-M)+N)}$
${\displaystyle y_{P}''(x)=e^{-2x}(-3sin(3x)(x(-2M+N)+M)}$
${\displaystyle +cos(3x)(-2M+N))-2e^{-2x}cos(3x)(x(-2M+N)+M)+e^{-2x}(3cos(3x)(x(-2N-M)}$
${\displaystyle +N)+sin(3x)(-2N-M))-2e^{-2x}sin(3x)(x(-2N-M)+N)}$

${\displaystyle y_{p}''+4y_{p}'+13y_{p}=2*e^{-2x}*3cos7x}$


Solve for M and N:
${\displaystyle 4N=2,-2M+4N}$
${\displaystyle M=1,N=0.5}$

${\displaystyle y(x)=e^{-2x}*(Acos(3x)+BNsin(3x))+x*e^{-2x}*(cos(3x)+.5*sin(3x))}$
Using initial conditions given find A and B

After applying initial conditions, we get
${\displaystyle A=1,-2A+3B+1=0}$
${\displaystyle A=1,B=1/3}$

${\displaystyle y(x)=e^{-2x}*(cos(3x)+sin(3x)/3)+x*e^{-2x}*(cos(3x)+.5*sin(3x))}$

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Is the given function even or odd or neither even nor odd? Find its Fourier Series.

${\displaystyle f(x)=x^{2},(-1<x<1),p=2}$

${\displaystyle f(-x)=f(x)}$ so ${\displaystyle f(x)=x^{2}}$ is an even function.

The Fourier series is ${\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}cos(nwx)+b_{n}sin(nwx)]}$.

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

For ${\displaystyle n=1,2,...}$
${\displaystyle a_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)cos(nwx)dx}$
${\displaystyle b_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)sin(nwx)dx}$

For ${\displaystyle f(x)=x^{2}}$
${\displaystyle a_{0}={\frac {1}{2(1)}}\int _{-1}^{1}x^{2}dx={\frac {1}{3}}}$
${\displaystyle a_{n}={\frac {1}{1}}\int _{-1}^{1}x^{2}cos(n\pi x)dx}$
The above integral requires two iterations of integration by parts. Which gives
${\displaystyle a_{n}={\frac {1}{n\pi }}[sin(n\pi )-sin(-n\pi )]+{\frac {2}{n^{2}\pi ^{2}}}[cos(n\pi )+cos(-n\pi )]-{\frac {2}{n^{3}\pi ^{3}}}[sin(n\pi )-sin(-n\pi )]}$
Similarly, integration by parts needs to be used twice to solve the following integral.
${\displaystyle b_{n}={\frac {1}{1}}\int _{-1}^{1}x^{2}sin(n\pi x)dx}$
${\displaystyle b_{n}={\frac {-1}{n\pi }}[cos(n\pi )-cos(-n\pi )]+{\frac {2}{n^{2}\pi ^{2}}}[sin(n\pi )+sin(-n\pi )]+{\frac {2}{n^{3}\pi ^{3}}}[cos(n\pi )-cos(-n\pi )]}$
So the Fourier series for ${\displaystyle f(x)=x^{2}}$ is
${\displaystyle {\frac {1}{3}}+\sum _{n=1}^{\infty }cos(nwx)[{\frac {1}{n\pi }}[sin(n\pi )-sin(-n\pi )]+{\frac {2}{n^{2}\pi ^{2}}}[cos(n\pi )+cos(-n\pi )]-{\frac {2}{n^{3}\pi ^{3}}}[sin(n\pi )-sin(-n\pi )]]}$
${\displaystyle +sin(nwx)[-{\frac {1}{n\pi }}[cos(n\pi )-cos(-n\pi )]+{\frac {2}{n^{2}\pi ^{2}}}[sin(n\pi )+sin(-n\pi )]+{\frac {2}{n^{3}\pi ^{3}}}[cos(n\pi )-cos(-n\pi )]]}$

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

1) Develop the Fourier series of${\displaystyle f({\bar {x}})}$. Plot${\displaystyle f({\bar {x}})}$and develop the truncated Fourier series${\displaystyle f_{n}({\bar {x}})}$.
${\displaystyle f_{n}({\bar {x}}):={\bar {a}}_{0}+\sum _{k=1}^{n}[{\bar {a}}_{k}\cos k\omega {\bar {x}}+{\bar {b}}_{k}\sin k\omega {\bar {x}}]}$
for n = 0,1,2,4,8. Observe the values of at the points of discontinuities, and the Gibbs phenomenon. Transform the variable so to obtain the Fourier series expansion of . Level 1: n=0,1.

2)Do the same as above, but using${\displaystyle f({\tilde {x}})}$to obtain the Fourier series expansion of${\displaystyle f(x)}$; compare to the result obtained above. Level 1: n=0,1.

To begin, the function ${\displaystyle f({\bar {x}})}$ was determined to be even. Even functions reduce to a cosine Fourier series. Because ${\displaystyle f({\bar {x}})}$, has a period of 4, the length is 2.

${\displaystyle a_{0}={\frac {1}{L}}\int _{0}^{2}f({\bar {x}})d{\bar {x}}={\frac {A}{2}}}$

${\displaystyle a_{k}={\frac {2}{L}}\int _{0}^{2}f({\bar {x}})\cos({\frac {k\pi {\bar {x}}}{2}})d{\bar {x}}}$

${\displaystyle a_{k}={\frac {2A}{k\pi }}\sin({\frac {k\pi }{2}})}$

${\displaystyle f_{k}({\bar {x}})=a_{0}+\sum _{k=1}^{n}[a_{n}\cos {\frac {n\pi {\bar {x}}}{2}}]}$

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

For n=0,
${\displaystyle f_{0}({\bar {x}})={\frac {A}{2}}}$

For n=1,
${\displaystyle f_{1}({\bar {x}})={\frac {A}{2}}+{\frac {2A}{\pi }}\cos({\frac {\pi {\bar {x}}}{2}})}$

${\displaystyle {\bar {x}}=x-1.25}$

${\displaystyle f_{k}(x)={\frac {A}{2}}+{\frac {A}{\pi }}\cos({\frac {\pi (x-1.25)}{2}})}$

Plot (A=1)

To begin, the function ${\displaystyle f({\bar {x}})}$ was determined to be odd. Even functions reduce to a sine Fourier series. Because ${\displaystyle f({\bar {x}})}$, has a period of 4, the length is 2.

${\displaystyle a_{0}={\frac {1}{L}}\int _{0}^{2}f({\bar {x}})d{\bar {x}}={\frac {A}{2}}}$

${\displaystyle a_{0}={\frac {1}{2L}}\int _{0}^{4}f({\tilde {x}})d{\tilde {x}}={\frac {A}{2}}}$

${\displaystyle a_{n}={\frac {1}{2}}\int _{0}^{4}f{\tilde {x}}\cos({\frac {n\pi {\tilde {x}}}{2}}d{\tilde {x}}}$

${\displaystyle a_{n}={\frac {1}{2}}f({\tilde {x}})\cos({\frac {n\pi {\tilde {x}}}{2}}}$ from 0 to 4

${\displaystyle a_{n}={\frac {A}{n\pi }}\sin(\pi k)}$

${\displaystyle b_{n}={\frac {1}{2}}\int _{0}^{4}f{\tilde {x}}\sin({\frac {n\pi {\tilde {x}}}{2}}d{\tilde {x}}}$

${\displaystyle b_{n}={\frac {1}{2}}f({\tilde {x}})\sin({\frac {n\pi {\tilde {x}}}{2}})}$ from 0 to 4

${\displaystyle b_{n}={\frac {A}{n\pi }}(1-\cos(\pi k))}$

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

For n=0,
${\displaystyle f_{0}({\tilde {x}})={\frac {A}{2}}}$

For n=1,
${\displaystyle f_{1}({\tilde {x}})={\frac {A}{2}}+{\frac {A}{\pi }}\sin({\frac {\pi {\tilde {x}}}{2}})}$

${\displaystyle {\tilde {x}}=x-.25}$

${\displaystyle f_{1}(x)={\frac {A}{2}}+{\frac {A}{\pi }}\sin({\frac {\pi (x-.25)}{2}})}$
Plot (A=1)

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Find the separated ODE's for the Heat Equation:

${\displaystyle {\frac {\partial ^{2}u}{\partial t^{2}}}=k{\frac {\partial ^{2}u}{\partial x^{2}}}}$ (1)

${\displaystyle k=}$ heat capacity

Separation of Variables:

Assume: ${\displaystyle u(x,t)=F(x)\cdot G(t)}$

${\displaystyle {\frac {\partial u(x,t)}{\partial x}}=F'(x)\cdot G(t)}$ (2)

${\displaystyle {\frac {\partial ^{2}u(x,t)}{\partial x^{2}}}=F''(x)\cdot G(t)}$ (3)

${\displaystyle {\frac {\partial u(x,t)}{\partial t}}=F(x)\cdot {\dot {G}}(t)}$ (4)

${\displaystyle {\frac {\partial ^{2}u(x,t)}{\partial t^{2}}}=F(x)\cdot {\ddot {G}}(t)}$ (5)

Plug (2) and (3) into Heat Equation (1):

${\displaystyle F(x)\cdot {\dot {G}}(t)=kF''(x)\cdot G(t)}$ (6)

Rearrange (6) to combine like terms:

${\displaystyle {\frac {{\dot {G}}(t)}{kG(t)}}={\frac {F''(x)}{F(x)}}=c{\text{ (constant)}}}$

${\displaystyle {\frac {{\dot {G}}(t)}{kG(t)}}=c}$

${\displaystyle {\dot {G}}(t)=k\,c\,G(t)}$

${\displaystyle {\dot {G}}(t)-k\,c\,G(t)=0}$

${\displaystyle {\frac {F''(x)}{F(x)}}=c}$

${\displaystyle F''(x)=c\,F(x)}$

${\displaystyle F''(x)-c\,F(x)=0}$

Solution:

Separated ODE's for Heat Equation:

${\displaystyle {\dot {G}}(t)-k\,c\,G(t)=0}$

${\displaystyle F''(x)-c\,F(x)=0}$

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

Verify (4)-(5) p.19-9
(4)${\displaystyle <\phi _{i},\phi _{j}>=0}$ for ${\displaystyle i\neq j}$
(5)${\displaystyle <\phi _{i},\phi _{j}>=L/2}$ for ${\displaystyle i=j}$

Using the integral scalar product calculation,
${\displaystyle \int _{0}^{L}\phi _{i}(x)\phi _{j}(x)dx}$

Substituting in sin values,
${\displaystyle \int _{0}^{L}sin(\omega _{i}(x))sin(\omega _{j}(x))dx}$

Using ${\displaystyle z={\frac {\pi x}{L}}}$ and ${\displaystyle dz={\frac {\pi }{L}}dx}$

You can substitute z into the integral instead of x.
${\displaystyle \int _{0}^{\pi }sin(iz)sin(jz)dz{\frac {L}{\pi }}}$

Integrating,
${\displaystyle {\frac {1}{2}}[{\frac {sin(i-j)z}{i-j}}-{\frac {sin(i+j)z}{i+j}}]}$ from ${\displaystyle z=0}$ to ${\displaystyle z=\pi }$
Since ${\displaystyle i\neq j}$, the equation with its sin values turns into 0-0=0

You can use the same equation from the verification of (4) from this point:
${\displaystyle {\frac {1}{2}}[{\frac {sin(i-j)z}{i-j}}-{\frac {sin(i+j)z}{i+j}}]}$ from ${\displaystyle z=0}$ to ${\displaystyle z=\pi }$
Putting those values in and substituting L back in the equation, it turns into ${\displaystyle {\frac {L}{2}}}$

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

${\displaystyle u(x,t)=\sum _{j=1}^{\infty }a_{j}\cos(cw_{j}t)\sin(w_{j}x)}$
Plot the truncated series for n=5. ${\displaystyle t=\alpha p_{1}={\frac {\alpha 2L}{c}}}$

${\displaystyle \alpha =.5,1,1.5,2}$

${\displaystyle a_{j}={\frac {2[((-1)^{j})-1]}{\pi ^{3}j^{3}}}}$

${\displaystyle w_{j}={\frac {j\pi }{L}}}$

C=3 and L=2
Plot ${\displaystyle u(x,2/3)}$



Plot ${\displaystyle u(x,4/3)}$



Plot ${\displaystyle u(x,2)}$



Plot ${\displaystyle u(x,8/3)}$

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.