University of Florida/Egm4313/s12.team5.R5

Report 5

Find ${\displaystyle R_{c}}$ for the following series:
1. ${\displaystyle r(x)=\sum _{k=0}^{\infty }(k+1)kx^{k}}$

2. ${\displaystyle r(x)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{\gamma ^{k}}}x^{2k}}$

${\displaystyle \gamma ={\text{constant}}}$

Use (2)-(3)p.7-31 to find ${\displaystyle R_{c}}$ for the Taylor series of

3. ${\displaystyle sinx}$ at ${\displaystyle {\hat {x}}=0}$

4. ${\displaystyle log(1+x)}$ at ${\displaystyle {\hat {x}}=0}$

5. ${\displaystyle log(1+x)}$ at ${\displaystyle {\hat {x}}=1}$

1. ${\displaystyle r(x)=\sum _{k=0}^{\infty }(k+1)kx^{k}=0+2x+6x^{2}...}$

Using L'Hospital's rule and (2)p.7-31, and setting ${\displaystyle d_{k}=(k+1)k}$:

${\displaystyle R_{c}=\left[\lim _{k\to \infty }\left|{\frac {(k+2)(k+1)}{(k+1)k}}\right|\right]^{-1}=\left[\lim _{k\to \infty }\left|{\frac {k+2}{k}}\right|\right]^{-1}=\left[\lim _{k\to \infty }\left|{\frac {1}{1}}\right|\right]^{-1}=1}$

2.${\displaystyle r(x)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{\gamma ^{k}}}x^{2k}=1-{\frac {1}{\gamma }}x^{2}+{\frac {1}{\gamma ^{2}}}x^{4}...}$

Using (3)p.7-31 and setting ${\displaystyle d_{k}={\frac {(-1)^{k}}{\gamma ^{k}}}}$:

${\displaystyle R_{c}=\lim _{k\to \infty }\left[{\sqrt[{k}]{\left|{\frac {(-1)^{k}}{\gamma ^{k}}}\right|}}\right]^{-1}=\lim _{k\to \infty }{\frac {{\sqrt[{k}]{|\gamma ^{k}}}|}{\sqrt[{k}]{|(-1)^{k}|}}}=\lim _{k\to \infty }{\frac {\sqrt {|\gamma |}}{\sqrt {|-1|}}}={\sqrt {|\gamma |}}}$

3. Taylor series about ${\displaystyle {\hat {x}}=0}$:

${\displaystyle sinx=x+{\frac {1}{6}}x^{3}+{\frac {1}{120}}x^{5}...\Rightarrow \sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}x^{2k+1}}$

Using (2)p.7-31 and setting ${\displaystyle d_{k}={\frac {(-1)^{k}}{(2k+1)!}}}$:

${\displaystyle R_{c}=\left[\lim _{k\to \infty }\left|{\frac {\frac {(-1)^{k+1}}{(2(k+1)+1)!}}{\frac {(-1)^{k}}{(2k+1)!}}}\right|\right]^{-1}=\left[\lim _{k\to \infty }\left|{\frac {-1}{(2k+3)(2k+2)}}\right|\right]^{-1}=\infty }$

4. Taylor series about ${\displaystyle {\hat {x}}=0}$:

${\displaystyle log(1+x)=0+x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}...\Rightarrow \sum _{n=0}^{\infty }{\frac {(-1)^{k}(x)^{k+1}}{(k+1)}}}$

Using L'Hospital's rule and (2)p.7-31, and setting ${\displaystyle d_{k}={\frac {(-1)^{k}}{(k+1)}}}$:

${\displaystyle R_{c}=\left[\lim _{k\to \infty }\left|{\frac {\frac {(-1)^{k+1}}{(k+2)}}{\frac {(-1)^{k}}{(k+1)}}}\right|\right]^{-1}=\left[\lim _{k\to \infty }\left|{\frac {(-1)(k+1)}{(k+2)}}\right|\right]^{-1}=\left[\lim _{k\to \infty }\left|{\frac {-1}{1}}\right|\right]^{-1}=1}$

5. Taylor series about ${\displaystyle {\hat {x}}=1}$:

${\displaystyle log(1+x)=log(2)+{\frac {(x-1)}{2}}-{\frac {(x-1)^{2}}{8}}+{\frac {(x-1)^{3}}{24}}...\Rightarrow log(2)+\sum _{n=0}^{\infty }{\frac {(-1)^{k+1}(x-1)^{k}}{k2^{k}}}}$

Using L'Hospital's rule and (2)p.7-31, and setting ${\displaystyle d_{k}={\frac {(-1)^{k+1}}{k2^{k}}}}$:

${\displaystyle R_{c}=\left[\lim _{k\to \infty }\left|{\frac {\frac {(-1)^{k+2}}{(k+1)2^{k+1}}}{\frac {(-1)^{k+1}}{k2^{k}}}}\right|\right]^{-1}=\left[\lim _{k\to \infty }\left|{\frac {-1k}{2(k+1)}}\right|\right]^{-1}=\left[\lim _{k\to \infty }\left|{\frac {-1}{2}}\right|\right]^{-1}=2}$

This problem is solved and uploaded by Radina Dikova

Determine whether the following pairs of functions are linearly independent:

${\displaystyle \displaystyle \ 1.f(x)=x^{2},g(x)=x^{4}\ }$

${\displaystyle \displaystyle \ 2.f(x)=\cos(x),g(x)=\sin(3x)\ }$

First use the Wronskian method, then use the Gramian method.

The Wronskian is defined as: ${\displaystyle \displaystyle \ W(f,g):=\det {\begin{bmatrix}f&g\\f'&g'\end{bmatrix}}=fg'-gf'\ }$

If ${\displaystyle \displaystyle \ W(f,g)\neq 0\ }$ , then the functions f and g are linearly independent.

The Gramian is defined as: ${\displaystyle \displaystyle \ \Gamma (f,g):=\det {\begin{bmatrix}<f,f>&<f,g>\\<g,f>&<g,g>\end{bmatrix}}\ }$

Where ${\displaystyle \displaystyle \ <f,g>:=\int _{a}^{b}f(x)g(x)dx\ }$

If ${\displaystyle \displaystyle \ \Gamma (f,g)\neq 0\ }$ , then the functions f and g are linearly independent.

1. Using Wronskian.

${\displaystyle \displaystyle \ W(f,g)=fg'-gf'=x^{2}(4x^{3})-x^{4}(2x)=4x^{5}-2x^{5}=2x^{5}\neq 0}$

Therefore, f and g are linearly independent.

2. Using Wronskian.

${\displaystyle \displaystyle \ W(f,g)=fg'-gf'=\cos(x)(3\cos(3x))-\sin(3x)(-\sin(x))\neq 0}$

Therefore, f and g are linearly independent.

1. Using Gramian with an interval of [-1,1]

${\displaystyle \displaystyle \ <f,f>=\int _{-1}^{1}x^{2}(x^{2})dx={\frac {1}{5}}x^{5}\mid _{-1}^{1}={\frac {2}{5}}\ }$

${\displaystyle \displaystyle \ <f,g>=<g,f>=\int _{-1}^{1}x^{2}(x^{4})dx={\frac {1}{7}}x^{7}\mid _{-1}^{1}={\frac {2}{7}}\ }$

${\displaystyle \displaystyle \ <g,g>=\int _{-1}^{1}x^{4}(x^{4})dx={\frac {1}{9}}x^{9}\mid _{-1}^{1}={\frac {2}{9}}\ }$

${\displaystyle \displaystyle \ \Gamma (f,g)=<f,f><g,g>-<f,g><g,f>={\frac {2}{5}}({\frac {2}{9}})-{\frac {2}{7}}({\frac {2}{7}})={\frac {4}{45}}-{\frac {4}{49}}\neq 0\ }$

Therefore, f and g are linearly independent.

2. Using Gramian with an interval of [-1,1]

${\displaystyle \displaystyle \ <f,f>=\int _{-1}^{1}\cos(x)(\cos(x))dx=\int _{-1}^{1}\cos ^{2}(x)dx={\frac {1}{2}}\int _{-1}^{1}1+\cos(2x)dx={\frac {1}{2}}(x+{\frac {1}{2}}\sin(2x))\mid _{-1}^{1}=1.4546\ }$

${\displaystyle \displaystyle <f,g>=<g,f>=\int _{-1}^{1}\sin(3x)\cos(x)dx=0\ }$

${\displaystyle \displaystyle <g,g>=<g,g>=\int _{-1}^{1}\sin ^{2}(3x)dx={\frac {1}{2}}\int _{-1}^{1}1-cos(6x)dx={\frac {1}{2}}(x-{\frac {1}{6}}\sin(6x))\mid _{-1}{1}=1.04657\ }$

${\displaystyle \displaystyle \ \Gamma (f,g)=<f,f><g,g>-<f,g><g,f>=1.4546(1.04657)-0(0)\neq 0\ }$

Therefore, f and g are linearly independent.

This problem was solved and uploaded by David Herrick.

Verify that ${\displaystyle b_{1}}$ and ${\displaystyle b_{2}}$ in (1)-(2) p.7-34 are linearly independent using the Gramian

The given Grammian:

${\displaystyle \displaystyle \Gamma (b_{1},b_{2})={\begin{bmatrix}\left\langle b_{1},b_{1}\right\rangle \left\langle b_{1},b_{2}\right\rangle \\\left\langle b_{2},b_{1}\right\rangle \left\langle b_{2},b_{2}\right\rangle \end{bmatrix}}}$

In reference to to (3) on p.8-9

${\displaystyle \displaystyle \left\langle b_{1},b_{1}\right\rangle =(b_{1}\cdot b_{2})}$

Calculating the dot products yields:

${\displaystyle \displaystyle \left\langle b_{1},b_{1}\right\rangle =4+49=53}$

${\displaystyle \displaystyle \left\langle b_{1},b_{2}\right\rangle =3+21=24}$

${\displaystyle \displaystyle \left\langle b_{2},b_{1}\right\rangle =3+21=24}$

${\displaystyle \displaystyle \left\langle b_{2},b_{2}\right\rangle =2.25+9=11.25}$

Plugging the dot products into the Grammian yields:

${\displaystyle \displaystyle \Gamma =596.25-576=20.25\neq 0}$

Therefore, b1 and b2 are linearly independent.

Solved and uploaded by Derik Bell

Show that: ${\displaystyle y_{p}(x)=\sum _{i=0}^{n}(y_{p,i}(x))}$
is the particular solution to: ${\displaystyle y''+p(x)y'+q(x)y=r(x)}$
Discuss the choice of particular solutions in the table on p8-3. In other words, for r(x) = kcos(wx), why would you need to have both cos(wx) and sin(wx) in the particular solution?

For a single excitation that satisfies ${\displaystyle r_{i}(x)}$,
${\displaystyle y_{p,i}''+p(x)y_{p,i}'+q(x)y_{p,i}=r_{i}(x)}$
for example:
${\displaystyle y_{p0}''+p(x)y_{p0}'+q(x)y_{p0}=r_{0}(x)}$
${\displaystyle y_{p1}''+p(x)y_{p1}'+q(x)y_{p1}=r_{1}(x)}$
${\displaystyle y_{p2}''+p(x)y_{p2}'+q(x)y_{p2}=r_{2}(x)}$
and so on until...
${\displaystyle y_{p,n}''+p(x)y_{p,n}'+q(x)y_{p,n}=r_{n}(x)}$ where by linearity: ${\displaystyle r(x)=\sum _{i=0}^{n}r_{i}(x)}$
Since ${\displaystyle y_{p,i}}$ is the solution to a single iteration of r(x) and ${\displaystyle r(x)=\sum _{i=1}^{n}r_{i}(x)}$ , then by linearity, the solution to r(x) is: ${\displaystyle y_{p}(x)=\sum _{i=1}^{n}y_{p,i}}$
Part 2:
kcos(wx) is a periodic function. As shown by (3) on p8-2, any periodic function can be broken down into a fourier trigonometric series:
${\displaystyle r(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}cos(nwx)+b_{n}sin(nwx)]}$
r(x) can be further broken down as the sum of:
${\displaystyle r_{a}=a_{0}}$
${\displaystyle r_{b}=\sum _{n=1}^{\infty }a_{n}cos(nwx)}$
${\displaystyle r_{c}=\sum _{n=1}^{\infty }b_{n}sin(nwx)}$
Where ${\displaystyle r(x)=r_{a}+r_{b}+r_{c}}$
Since r(x) is expressed in terms of cos(x) and sin(x) the particular solution, which is also a sum of the individual particular solutions for each iteration of ${\displaystyle r_{a}}$, ${\displaystyle r_{b}}$, and ${\displaystyle r_{c}}$, needs to be in terms of sin(x) and cos(x) as well. That applies to all periodic functions as shown on p8-2, which sin(x) is as well. Therefore that justifies why the particular solutions for kcos(wx), ksin(wx), ${\displaystyle ke^{\alpha x}\cos {wx}}$, and ${\displaystyle ke^{\alpha x}\sin {wx}}$ must all include both cos(x) and sin(x).

This problem was solved and uploaded by John North.

1. Show that cos(7x) and sin(7x) are linearly independent using the Wronskian and the Gramian (integrate over 1 period)

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

3. Find the overall solution y(x) that corresponds to the initial condition y(0)=1, y'(0)=0. Plot the solution over 3 periods.

(1)

First, using Wronskian:

For 2 functions, f and g, the Wrosnkian is defined as ${\displaystyle \displaystyle W(f,g):=det{\begin{bmatrix}f&g\\f'&g'\end{bmatrix}}=fg'-gf'}$

Where f and g are linearly independent if ${\displaystyle \displaystyle W(f,g)\neq 0}$

For ${\displaystyle \displaystyle f=cos(7x),g=sin(7x),f'=-7sin(7x),g'=7cos(7x)}$

Then, ${\displaystyle \displaystyle W(f,g):=det{\begin{bmatrix}cos(7x)&sin(7x)\\-7sin(7x)&7cos(7x)\end{bmatrix}}=7cos^{2}(7x)+7sin^{2}(7x)\neq 0}$

Therefore, f and g are linearly independent

Second, using Gramian:

Consider two functions, f and g, where the scalar product is defined as

${\displaystyle \displaystyle <f,g>:=\int _{a}^{b}f(x)g(x)dx}$

And the Gramian defined as

${\displaystyle \displaystyle \Gamma (f,g):=det{\begin{bmatrix}<f,f>&<f,g>\\<g,f>&<g,g>\end{bmatrix}}}$

Then f and g are linearly indepdent if ${\displaystyle \displaystyle \Gamma (f,g)\neq 0}$

For f=cos(7x) and g=sin(7x) and integrating over one period (${\displaystyle \displaystyle {\frac {2\pi }{7}}}$)

${\displaystyle \displaystyle <f,f>=\int _{0}^{{\frac {2\pi }{7}}{}}cos^{2}(7x)dx}$

Letting ${\displaystyle \displaystyle u=7x,du=7dx}$ and changing limits of integration by plugging in old limits into "u" equation

${\displaystyle \displaystyle <f,f>={\frac {1}{7}}\int _{0}^{2\pi }cos^{2}(u)du={\frac {1}{7}}[{\frac {u}{2}}+{\frac {1}{4}}sin2u\mathbf {\mid } _{0}^{2\pi }]={\frac {\pi }{7}}}$

${\displaystyle \displaystyle <g,g>=\int _{0}^{\frac {2\pi }{7}}sin^{2}(7x)dx}$

Letting ${\displaystyle \displaystyle u=7x,du=7dx}$ and changing the limits of integration by plugging in old limits into "u" equations

${\displaystyle \displaystyle <g,g>={\frac {1}{7}}\int _{0}^{2\pi }sin^{2}(u)du={\frac {1}{7}}[{\frac {u}{2}}-{\frac {1}{4}}sin2u\mid _{0}^{2\pi }]={\frac {\pi }{7}}}$

${\displaystyle \displaystyle <f,g>=<g,f>=\int _{0}^{\frac {2\pi }{7}}cos(7x)sin(7x)dx}$

From Kreyszig p.479, it is apparent that sin and cos are orthogonal to each other, so the above integration will equal zero

${\displaystyle \displaystyle <f,g>=<g,f>=\int _{0}^{\frac {2\pi }{7}}cos(7x)sin(7x)dx=0}$

Plugging in the results of each integral into the Gramian

${\displaystyle \displaystyle {\boldsymbol {\Gamma }}(f,g)=det{\begin{bmatrix}<f,f>&<f,g>\\<g,f>&<g,g>\end{bmatrix}}=det{\begin{bmatrix}{\frac {\pi }{7}}&0\\0&{\frac {\pi }{7}}\end{bmatrix}}={\frac {\pi ^{2}}{49}}\neq 0}$

Therefore, f and g are linearly independent

(2)

Given ${\displaystyle \displaystyle y''-3y'-10y=3cos(7x)}$

And particular solutions of the form ${\displaystyle \displaystyle y_{p}(x)=Mcos(7x)+Nsin(7x),y'_{p}(x)=-M7sin(7x)+N7cos(7x),y''_{p}(x)=-M49cos(7x)-N49sin(7x)}$

Plug particular solutions back into original ODE and collect like terms

${\displaystyle \displaystyle -M49cos(7x)-N49sin(7x)+M21sin(7x)-N21cos(7x)-M10cos(7x)-N10sin(7x)=3cos(7x)}$

${\displaystyle \displaystyle -59Mcos(7x)-59Nsin(7x)+21Msin(7x)-21Ncos(7x)=3cos(7x)}$

Equating coefficients

${\displaystyle \displaystyle -59M=3\rightarrow M=-{\frac {3}{59}}}$

${\displaystyle \displaystyle -21N=3\rightarrow N=-{\frac {1}{7}}}$

${\displaystyle \displaystyle M=-{\frac {3}{59}},N=-{\frac {1}{7}}}$

(3)

The overall solution ${\displaystyle \displaystyle y(x)=y_{p}(x)+y_{h}(x)}$ consists of the particular solution and homogeneous soloution

Homogeneous solotuion

${\displaystyle \displaystyle y''-3y'-10y=0}$

${\displaystyle \displaystyle a^{2}-4b=3^{2}-4(10)=49>0}$ so we have distinct real roots

${\displaystyle \displaystyle \lambda _{1,2}={\frac {1}{2}}[-a\pm {\sqrt {a^{2}-4b}}]={\frac {1}{2}}[3\pm 7]=5,-2}$

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

Using initial conditions ${\displaystyle \displaystyle y(0)=1,y'(0)=1}$ and ${\displaystyle \displaystyle y'_{h}(x)=5c_{1}e^{5x}-2c_{2}e^{-2x}}$

${\displaystyle \displaystyle 5c_{1}-2c_{2}=0}$

${\displaystyle \displaystyle c_{1}+c_{2}=1}$

Solving the two equations for the two unknowns yields ${\displaystyle \displaystyle c_{1}={\frac {2}{7}},c_{2}={\frac {5}{7}}}$

${\displaystyle \displaystyle y_{h}(x)={\frac {2}{7}}e^{5x}+{\frac {5}{7}}e^{-2x}}$

Particular solution

${\displaystyle \displaystyle y_{p}(x)=Kcos(\omega x)+Msin(\omega x)}$ where ${\displaystyle \displaystyle \omega =7,K=M=-{\frac {3}{59}},M=N=-{\frac {1}{7}}}$

${\displaystyle \displaystyle y_{p}(x)=-{\frac {3}{59}}cos(7x)-{\frac {1}{7}}sin(7x)}$

Giving us an overall solution of

${\displaystyle \displaystyle y(x)=y_{h}(x)+y_{p}(x)={\frac {2}{7}}e^{5x}+{\frac {5}{7}}e^{-2x}-{\frac {3}{59}}cos(7x)-{\frac {1}{7}}sin(7x)}$

Plotting the solution over three periods ${\displaystyle \displaystyle P={\frac {2\pi }{7}}\Rightarrow 3P={\frac {6\pi }{7}}}$

Matlabcode
EDU>> x=0:0.001:(6*pi)/7;
EDU>> y=(2/7).*exp(5.*x)+(5/7).*exp(-2.*x)-(3/59).*cos(7.*x)-(1/7).*sin(7.*x);
EDU>> plot(x,y)

Solved and uploaded by Joshua House

Find the solution to the following initial condition problem, and plot it over 3 periods.

${\displaystyle \displaystyle y''+4y'+13y=2e^{-2x}cos(3x)}$
${\displaystyle \displaystyle y_{h}(x)=e^{-2x}(Acos(3x)+Bsin(3x))}$
${\displaystyle \displaystyle y_{p}(x)=xe^{-2x}(Mcos(3x)+Nsin(3x))}$
${\displaystyle \displaystyle y(0)=1,y'(0)=0}$

First, we take the first and second derivative of the particular solution:

${\displaystyle \displaystyle y'_{p}(x)=e^{-2x}(-3Msin(3x)-2Mcos(3x)+3Ncos(3x)-2Nsin(3x))=e^{-2x}[(-3M-2N)sin(3x)+(-2M+3N)cos(3x)]}$
${\displaystyle \displaystyle y''_{p}(x)=e^{-2x}(-9Mcos(3x)+6Msin(3x)-9Nsin(3x)-6Ncos(3x))-2e^{-2x}(-3Msin(3x)-2Mcos(3x)+3Ncos(3x)-2Nsin(3x))}$
${\displaystyle \displaystyle =2e^{-2x}[(12M-5N)sin(3x)+(13M-12N)cos(3x)]}$

Now, we plug the particular solution derivatives into the initial equation:

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

${\displaystyle \displaystyle 2e^{-2x}[(12M-5N)sin(3x)+(13M-12N)cos(3x)]+4[e^{-2x}[(-3M-2Nsin(3x)+(-2M+3N)cos(3x)]]}$
${\displaystyle \displaystyle +13[e^{-2x}[Mcos(3x)+Nsin(3x)]]=2e^{-2x}cos(3x)}$

${\displaystyle \displaystyle e^{-2x}[(24M-10N)sin(3x)+(26M-24N)cos(3x)+(-12M-8N)sin(3x)+(-8M+12N)cos(3x)+13Mcos(3x)+13Nsin(3x)]}$
${\displaystyle \displaystyle =2e^{-2x}cos(3x)}$

${\displaystyle \displaystyle (12M-5N)sin(3x)+(31M-12N)cos(3x)=2cos(3x)}$

Now, we equate the coefficients of sin(3x) and cos(3x) to determine the unknown coefficients M and N:

${\displaystyle \displaystyle 12M-5N=0}$
${\displaystyle \displaystyle 31M-12N=2}$
${\displaystyle \displaystyle N={\frac {24}{11}},M={\frac {10}{11}}}$

Therefore, the particular solution is:

${\displaystyle \displaystyle y_{p}(x)=e^{-2x}[{\frac {10}{11}}cos(3x)+{\frac {24}{11}}sin(3x)]}$

Now, we focus on the homogeneous part of the solution. It is given to us as:

${\displaystyle \displaystyle y_{h}(x)=e^{-2x}[Acos(3x)+Bsin(3x)]}$

As you can see, this is identical to the particular solution, except that M is now A and N is now B. Therefore, the first derivative of the homogenous solution will be in the same form as the first derivative of the particular solution:

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

Remembering the two initial conditions, y(0) = 1 and y'(0) = 0, we apply these to the homogenous equations:

${\displaystyle \displaystyle y_{h}(0)=e^{-2x}[Acos(3x)+Bsin(3x)]=1}$
${\displaystyle \displaystyle y'_{h}(x)=e^{-2x}[(-3A-2B)sin(3x)+(-2A+3B)cos(3x)]=0}$

This yields two equations that we can use to solve for the coefficients A and B:

${\displaystyle \displaystyle A=1}$
${\displaystyle \displaystyle -2A+3B=0\Rightarrow 3B=2\Rightarrow B={\frac {3}{2}}}$

Therefore, the homoegenous solution is:

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

We find the final solution, y(x), by adding the homogenous and particular solutions as seen below:

${\displaystyle \displaystyle y(x)=y_{h}(x)+y_{p}(x)=e^{-2x}[cos(3x)+{\frac {3}{2}}sin(3x)]+e^{-2x}[{\frac {10}{11}}cos(3x)+{\frac {24}{11}}sin(3x)]}$
${\displaystyle \displaystyle y(x)=e^{-2x}[{\frac {21}{11}}cos(3x)+{\frac {81}{22}}sin(3x)]}$

Matlab Code: x= 0:0.001:(2*pi/3);

y = exp(-2.*x)*(21/11).*cos(3.*x) + exp(-2.*x)*(81/22).*sin(3.*x);

plot(x,y)

This problem was solved and uploaded by Will Knapper

${\displaystyle v=4e_{1}+2e_{2}=c_{1}b_{1}+c_{2}b_{2}}$
The oblique basis vectors are:
${\displaystyle b_{1}=2e_{1}+7e_{2}}$
${\displaystyle b_{2}=1.5e_{1}+3e_{2}}$

1. Find the components ${\displaystyle c_{1},c_{2}}$ using the Gram matrix as in (1)p.8-11.
2. Verify the results by using (1)-(2)p.7c-34 in (2)p8-11, and rely on the non-zero determinant of the matrix of components of ${\displaystyle b_{1},b_{2}}$ relative to the basis ${\displaystyle e_{1},e_{2}}$, as discussed on p.7c-34.

1. Using Gram matrix as in (1) p.8-10:

${\displaystyle {\begin{bmatrix}<b_{1},b_{1}>&<b_{1},b_{2}>\\<b_{2},b_{1}>&<b_{2},b_{2}>\end{bmatrix}}}$ ${\displaystyle {\begin{Bmatrix}c_{1}\\c_{2}\end{Bmatrix}}}$ = ${\displaystyle {\begin{Bmatrix}<b_{1},v>\\<c_{2},v>\end{Bmatrix}}}$

From (3)p.8-9 we know that ${\displaystyle <b_{i},b_{j}>=b_{i}\cdot b_{j}}$. Solving the various components we get:

${\displaystyle {\begin{bmatrix}53&24\\24&11.25\end{bmatrix}}}$ ${\displaystyle {\begin{Bmatrix}c_{1}\\c_{2}\end{Bmatrix}}}$ = ${\displaystyle {\begin{Bmatrix}22\\12\end{Bmatrix}}}$

In order to solve for ${\displaystyle c_{1},c_{2}}$ we need to calculate the inverse of ${\displaystyle {\begin{bmatrix}53&24\\24&11.25\end{bmatrix}}}$.

This gives is the Gram matrix as used in (1)p.8-11:
${\displaystyle c=\Gamma ^{-1}d}$

${\displaystyle {\begin{bmatrix}0.55556&-1.18519\\-1.18519&2.61728\end{bmatrix}}}$ ${\displaystyle {\begin{Bmatrix}22\\12\end{Bmatrix}}}$ = ${\displaystyle {\begin{Bmatrix}c_{1}\\c_{2}\end{Bmatrix}}}$

Thus:
${\displaystyle c_{1}=-2;c_{2}=5.333={\frac {16}{3}}}$

2. Plugging in ${\displaystyle b_{1},b_{2}}$ into ${\displaystyle v}$ we get:

${\displaystyle c_{1}(2e_{1}+7e_{2})+c_{2}(1.5e_{1}+3e_{2})=4e_{1}+2e_{2}}$

${\displaystyle 2c_{1}e_{1}+7c_{1}e_{2}+1.5c_{2}e_{1}+3c_{2}e_{2}=4e_{1}+2e_{2}}$

${\displaystyle e_{1}(2c_{1}+1.5c_{2})+e_{2}(7c_{1}+3c_{2})=4e_{1}+2e_{2}}$

Separating into components:

${\displaystyle 2c_{1}+1.5c_{2}=4}$ and ${\displaystyle 7c_{1}+3c_{2}=2}$

Solving the two linearly independent equations we get:

${\displaystyle \ }$
${\displaystyle c_{1}=-2;c_{2}={\frac {16}{3}}}$

This problem was solved and uploaded by Radina Dikova

Find the integral (see R5.9)

${\displaystyle \displaystyle \ \int x^{n}log(1+x)dx\ }$

using integration by parts and then with the help of General binomial theorem.

${\displaystyle \displaystyle \ (x+y)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{n-k}y^{k}\ }$

${\displaystyle \ {\binom {n}{k}}={\frac {n!}{k!(n-k)!}}={\frac {n(n-1)...(n-k+1)}{k!}}\ }$

The indefinite integral of ${\displaystyle \ x^{n}\ }$ is

${\displaystyle \ {\frac {x^{n+1}}{n+1}}+C\ }$

For n = 0, we get ${\displaystyle \ \int x^{0}=x\ }$

For n = 1, we get ${\displaystyle \ \int x^{1}={\frac {x^{2}}{2}}\ }$

And the indefinite integral of ${\displaystyle \ log(1+x)\ }$ is

${\displaystyle \ (x+1)log(1+x)-x+C\ }$

Integration by parts

- For n = 0 ${\displaystyle \ \int x^{0}log(1+x)dx=log(1+x)[x+1]-x\ }$

- For n = 1 ${\displaystyle \ \int x^{1}log(1+x)dx={\frac {1}{2}}*[(x-log(1+x))+x^{2}log(1+x)-{\frac {x^{2}}{2}}]\ }$

Solved and uploaded by Mike Wallace

Consider the L2-ODE-CC(5)p.7b-7 with log(1+x) as excitation:

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

${\displaystyle \ r(x)=log(1+x)\ }$

and the initial conditions

${\displaystyle \ y(-{\frac {3}{4}})=1,y'(-{\frac {3}{4}})=0\ }$

Project the excitation r(x) on the polynomial basis ${\displaystyle \ [b_{j}(x)=x^{j},j=0,1,...,n]\ }$

i.e., find ${\displaystyle \ d_{j}\ }$ such that

r(x) ≈ ${\displaystyle \ r_{n}(x)=\sum _{j=0}^{n}d_{j}x^{j}\ }$

for x in ${\displaystyle \ [-{\frac {3}{4}},3],\ }$ and for n = 3, 6, 9

Plot ${\displaystyle \ r(x),r_{n}(x)\ }$ to show uniform approximation and convergence

In a separate series of plots, compare the approximation of the function log(1+x) by 2 methods:

A. Projection on polynomial basis (1) p.8-17

B. Taylor series expansion about x = 0

Observe and discuss the pros and cons of each method

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

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

with the same initial conditions (2) p.7c-28

Plot ${\displaystyle \ y_{n}(x)\ }$ for n = 3, 6, 9, for x in ${\displaystyle \ [-{\frac {3}{4}},3].\ }$

In a series of separate plots, compare the results obtained with the projected excitation on polynomial basis to those with truncated Taylor series of the excitation. Plot also the numerical solution as a baseline for comparison.

Using ${\displaystyle \ y''-3y'+2y=log(1+x)\ }$

For n = 0

${\displaystyle \ {b_{0}}={x^{0}}\ }$

${\displaystyle \ r=log(1+x)=C_{0}x^{0}\ }$

${\displaystyle \ d_{0}=<x^{0},log(1+x)>=\int _{-{\frac {3}{4}}}^{3}log(1+x)dx=2.14175\ }$

${\displaystyle \ <x^{0},x^{0}>=\int _{-{\frac {3}{4}}}^{3}x^{0}x^{0}dx=x|_{-{\frac {3}{4}}}^{3}={\frac {15}{4}}\ }$

${\displaystyle \ d_{0}=C_{0}<x^{0},x^{0}>=C_{0}({\frac {15}{4}})=2.14175\ }$

${\displaystyle \ C_{0}=0.57113\ }$

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${\displaystyle \ Y=Y_{h}+Y_{p}=C_{1}e^{2x}+C_{2}e^{x}+0.57113\ }$

${\displaystyle \ Y'=2C_{1}e^{2x}+C_{2}e^{x}\ }$

${\displaystyle \ y({\frac {3}{4}})=C_{1}e^{\frac {6}{4}}+C_{2}e^{\frac {3}{4}}+0.57113=1\ }$

${\displaystyle \ y'(-{\frac {3}{4}})=-{\frac {6}{4}}C_{1}e^{-{\frac {6}{4}}}+C_{2}e^{-{\frac {3}{4}}}=0\ }$

Subtracting Y from Y' in order to find the coefficients

${\displaystyle \ C_{1}e^{-{\frac {6}{4}}}-0.57113=-1\ }$

${\displaystyle \ C_{1}=--1.92205\ }$

${\displaystyle \ C_{2}=1.81582\ }$

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For n = 0 the final solution will be

${\displaystyle \ Y=-1.92205e^{2x}+1.81582e^{x}+0.57113\ }$

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For n = 1

${\displaystyle \ C_{0}=-.09399,C_{1}=0.591217\ }$

We need to find the homogeneous Y

${\displaystyle \ \lambda ^{2}-3\lambda +2=0=>(\lambda -2)(\lambda -1)=0\ }$

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

${\displaystyle \ Y_{h}=C_{1}e^{2x}+C_{2}e^{x}\ }$

${\displaystyle \ Y=Y_{h}+Y_{p}=C_{1}e^{2x}+C_{2}e^{x}-0.09399+0.591217x\ }$

Solving for the initial conditions

${\displaystyle \ y({\frac {3}{4}})=C_{1}e^{\frac {6}{4}}+C_{2}e^{\frac {3}{4}}-0.09399+0.591217({\frac {3}{4}})=1\ }$

${\displaystyle \ Y'==2C_{1}e^{2x}+C_{2}e^{x}+0.591217\ }$

${\displaystyle \ y'(-{\frac {3}{4}})=-{\frac {6}{4}}C_{1}e^{-{\frac {6}{4}}}+C_{2}e^{-{\frac {3}{4}}}+0.591217=0\ }$

This gives coefficients of:

${\displaystyle \ C_{1}=-9.5398,C_{2}=7.761\ }$

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For n = 1 the final solution will be;

${\displaystyle \ Y=-9.5398e^{2x}+7.761e^{x}+0.591217x-0.09399\ }$

This problem was solved and uploaded by Mike Wallace

Problem 2 was solved and Problem 5 was proofread by David Herrick

Problem 5 was solved and uploaded by Joshua House

Problems 1 and 7 were solved and uploaded by Radina Dikova

Problem 3 was solved and uploaded by Derik Bell

Problem 4 was solved and uploaded by John North

Problem 8 and 9 were solved and uploaded by Mike Wallace

Problem 6 was solved and uploaded by William Knapper