University of Florida/Egm4313/s12.team5.R5

R5.1

Question

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

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

$\gamma =\text{constant}$

Use (2)-(3)p.7-31 to find $R_c$ for the Taylor series of

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

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

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

Solution

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

Using L'Hospital's rule and (2)p.7-31, and setting $d_k=(k+1)k$:

$R_c={\left[\underset{k\to \mathrm{\infty }}{\lim }\left|\frac{(k+2)(k+1)}{(k+1)k}\right|\right]}^{-1}={\left[\underset{k\to \mathrm{\infty }}{\lim }\left|\frac{k+2}{k}\right|\right]}^{-1}={\left[\underset{k\to \mathrm{\infty }}{\lim }\left|\frac{1}{1}\right|\right]}^{-1}=1$

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

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

$R_c=\underset{k\to \mathrm{\infty }}{\lim }{\left[\sqrt[k]{\left|\frac{(-1{)}^k}{{\gamma }^k}\right|}\right]}^{-1}=\underset{k\to \mathrm{\infty }}{\lim }\frac{\sqrt[k]{|{\gamma }^k}|}{\sqrt[k]{|(-1{)}^k|}}=\underset{k\to \mathrm{\infty }}{\lim }\frac{\sqrt{|\gamma |}}{\sqrt{|-1|}}=\sqrt{|\gamma |}$

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

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

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

$R_c={\left[\underset{k\to \mathrm{\infty }}{\lim }\left|\frac{\frac{(-1{)}^{k+1}}{(2(k+1)+1)!}}{\frac{(-1{)}^k}{(2k+1)!}}\right|\right]}^{-1}={\left[\underset{k\to \mathrm{\infty }}{\lim }\left|\frac{-1}{(2k+3)(2k+2)}\right|\right]}^{-1}=\mathrm{\infty }$

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

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

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

$R_c={\left[\underset{k\to \mathrm{\infty }}{\lim }\left|\frac{\frac{(-1{)}^{k+1}}{(k+2)}}{\frac{(-1{)}^k}{(k+1)}}\right|\right]}^{-1}={\left[\underset{k\to \mathrm{\infty }}{\lim }\left|\frac{(-1)(k+1)}{(k+2)}\right|\right]}^{-1}={\left[\underset{k\to \mathrm{\infty }}{\lim }\left|\frac{-1}{1}\right|\right]}^{-1}=1$

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

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

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

$R_c={\left[\underset{k\to \mathrm{\infty }}{\lim }\left|\frac{\frac{(-1{)}^{k+2}}{(k+1)2^{k+1}}}{\frac{(-1{)}^{k+1}}{k{2}^k}}\right|\right]}^{-1}={\left[\underset{k\to \mathrm{\infty }}{\lim }\left|\frac{-1k}{2(k+1)}\right|\right]}^{-1}={\left[\underset{k\to \mathrm{\infty }}{\lim }\left|\frac{-1}{2}\right|\right]}^{-1}=2$

Author

This problem is solved and uploaded by Radina Dikova

R5.2

Question

Determine whether the following pairs of functions are linearly independent:

$1.f(x)=x^2,g(x)=x^4$

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

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

Solution

The Wronskian is defined as: $W(f,g):=\det \left[\begin{array}{cc}f& g\\ f^{\prime }& g^{\prime }\end{array}\right]=f{g}^{\prime }-g{f}^{\prime }$

If $W(f,g)\ne 0$ , then the functions f and g are linearly independent.

The Gramian is defined as: $\mathrm{\Gamma }(f,g):=\det \left[\begin{array}{cc}<f,f>& <f,g>\\ <g,f>& <g,g>\end{array}\right]$

Where $<f,g>:=\underset{a}{\overset{b}{\int }}f(x)g(x)dx$

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

1. Using Wronskian.

$W(f,g)=f{g}^{\prime }-g{f}^{\prime }=x^2(4x^3)-x^4(2x)=4x^5-2x^5=2x^5\ne 0$

Therefore, f and g are linearly independent.

2. Using Wronskian.

$W(f,g)=f{g}^{\prime }-g{f}^{\prime }=\cos (x)(3\cos (3x))-\sin (3x)(-\sin (x))\ne 0$

Therefore, f and g are linearly independent.

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

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

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

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

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

Therefore, f and g are linearly independent.

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

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

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

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

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

Therefore, f and g are linearly independent.

Author

This problem was solved and uploaded by David Herrick.

R5.3

Question

Verify that $b_1$ and $b_2$ in (1)-(2) p.7-34 are linearly independent using the Gramian

Solution

The given Grammian:

$\mathrm{\Gamma }(b_1,b_2)=\left[\begin{array}{c}⟨b_1,b_1⟩⟨b_1,b_2⟩\\ ⟨b_2,b_1⟩⟨b_2,b_2⟩\end{array}\right]$

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

$⟨b_1,b_1⟩=(b_1\cdot b_2)$

Calculating the dot products yields:

$⟨b_1,b_1⟩=4+49=53$

$⟨b_1,b_2⟩=3+21=24$

$⟨b_2,b_1⟩=3+21=24$

$⟨b_2,b_2⟩=2.25+9=11.25$

Plugging the dot products into the Grammian yields:

$\mathrm{\Gamma }=596.25-576=20.25\ne 0$

Therefore, b1 and b2 are linearly independent.

Author

Solved and uploaded by Derik Bell

R5.4

Question

Show that: $y_p(x)=\sum _{i=0}^n(y_{p,i}(x))$
is the particular solution to: $y^{″}+p(x)y^{\prime }+q(x)y=r(x)$
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?

Solution

For a single excitation that satisfies $r_i(x)$,
$y_{p^{″},i}+p(x)y_{p^{\prime },i}+q(x)y_{p,i}=r_i(x)$
for example:
$y_{p^{″}0}+p(x)y_{p^{\prime }0}+q(x)y_{p0}=r_0(x)$
$y_{p^{″}1}+p(x)y_{p^{\prime }1}+q(x)y_{p1}=r_1(x)$
$y_{p^{″}2}+p(x)y_{p^{\prime }2}+q(x)y_{p2}=r_2(x)$
and so on until...
$y_{p^{″},n}+p(x)y_{p^{\prime },n}+q(x)y_{p,n}=r_n(x)$ where by linearity: $r(x)=\sum _{i=0}^n{r}_i(x)$
Since $y_{p,i}$ is the solution to a single iteration of r(x) and $r(x)=\sum _{i=1}^n{r}_i(x)$ , then by linearity, the solution to r(x) is: $y_p(x)=\sum _{i=1}^n{y}_{p,i}$
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:
$r(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}[a_{n}cos(nwx)+b_{n}sin(nwx)]$
r(x) can be further broken down as the sum of:
$r_a=a_0$
$r_b=\sum _{n=1}^{\mathrm{\infty }}{a}_{n}cos(nwx)$
$r_c=\sum _{n=1}^{\mathrm{\infty }}{b}_{n}sin(nwx)$
Where $r(x)=r_a+r_b+r_c$
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 $r_a$, $r_b$, and $r_c$, 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), $k{e}^{\alpha x}\cos wx$, and $k{e}^{\alpha x}\sin wx$ must all include both cos(x) and sin(x).

Author

This problem was solved and uploaded by John North.

R5.5

Question

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.

Solution

(1)

First, using Wronskian:

For 2 functions, f and g, the Wrosnkian is defined as $W(f,g):=det\left[\begin{array}{cc}f& g\\ f^{\prime }& g^{\prime }\end{array}\right]=f{g}^{\prime }-g{f}^{\prime }$

Where f and g are linearly independent if $W(f,g)\ne 0$

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

Then, $W(f,g):=det\left[\begin{array}{cc}cos(7x)& sin(7x)\\ -7sin(7x)& 7cos(7x)\end{array}\right]=7co{s}^2(7x)+7si{n}^2(7x)\ne 0$

Therefore, f and g are linearly independent

Second, using Gramian:

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

$<f,g>:=\underset{a}{\overset{b}{\int }}f(x)g(x)dx$

And the Gramian defined as

$\mathrm{\Gamma }(f,g):=det\left[\begin{array}{cc}<f,f>& <f,g>\\ <g,f>& <g,g>\end{array}\right]$

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

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

$<f,f>=\underset{0}{\overset{\frac{2\pi }{7}}{\int }}co{s}^2(7x)dx$

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

$<f,f>=\frac{1}{7}\underset{0}{\overset{2\pi }{\int }}co{s}^2(u)du=\frac{1}{7}[\frac{u}{2}+\frac{1}{4}sin2u{\mathbf{\mid }}_0^{2\pi }]=\frac{\pi }{7}$

$<g,g>=\underset{0}{\overset{\frac{2\pi }{7}}{\int }}si{n}^2(7x)dx$

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

$<g,g>=\frac{1}{7}\underset{0}{\overset{2\pi }{\int }}si{n}^2(u)du=\frac{1}{7}[\frac{u}{2}-\frac{1}{4}sin2u\underset{0}{\overset{2\pi }{\mid }}]=\frac{\pi }{7}$

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

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

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

Plugging in the results of each integral into the Gramian

$\mathrm{\Gamma }(f,g)=det\left[\begin{array}{cc}<f,f>& <f,g>\\ <g,f>& <g,g>\end{array}\right]=det\left[\begin{array}{cc}\frac{\pi }{7}& 0\\ 0& \frac{\pi }{7}\end{array}\right]=\frac{{\pi }^2}{49}\ne 0$

Therefore, f and g are linearly independent

(2)

Given $y^{″}-3y^{\prime }-10y=3cos(7x)$

And particular solutions of the form $y_p(x)=Mcos(7x)+Nsin(7x),y{\text{'}}_p(x)=-M7sin(7x)+N7cos(7x),y^{\prime }{\text{'}}_p(x)=-M49cos(7x)-N49sin(7x)$

Plug particular solutions back into original ODE and collect like terms

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

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

Equating coefficients

$-59M=3\to M=-\frac{3}{59}$

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

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

(3)

The overall solution $y(x)=y_p(x)+y_h(x)$ consists of the particular solution and homogeneous soloution

Homogeneous solotuion

$y^{″}-3y^{\prime }-10y=0$

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

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

$y_h(x)=c_1{e}^{5x}+c_2{e}^{-2x}$

Using initial conditions $y(0)=1,y^{\prime }(0)=1$ and $y{\text{'}}_h(x)=5c_1{e}^{5x}-2c_2{e}^{-2x}$

$5c_1-2c_2=0$

$c_1+c_2=1$

Solving the two equations for the two unknowns yields $c_1=\frac{2}{7},c_2=\frac{5}{7}$

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

Particular solution

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

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

Giving us an overall solution of

$y(x)=y_h(x)+y_p(x)=\frac{2}{7}{e}^{5x}+\frac{5}{7}{e}^{-2x}-\frac{3}{59}cos(7x)-\frac{1}{7}sin(7x)$

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

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)

Author

Solved and uploaded by Joshua House

R5.6

Question

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

$y^{″}+4y^{\prime }+13y=2e^{-2x}cos(3x)$
$y_h(x)=e^{-2x}(Acos(3x)+Bsin(3x))$
$y_p(x)=x{e}^{-2x}(Mcos(3x)+Nsin(3x))$
$y(0)=1,y^{\prime }(0)=0$

Solution

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

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

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

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

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

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

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

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

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

Therefore, the particular solution is:

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

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

$y_h(x)=e^{-2x}[Acos(3x)+Bsin(3x)]$

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:

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

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

$y_h(0)=e^{-2x}[Acos(3x)+Bsin(3x)]=1$
$y{\text{'}}_h(x)=e^{-2x}[(-3A-2B)sin(3x)+(-2A+3B)cos(3x)]=0$

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

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

Therefore, the homoegenous solution is:

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

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

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

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)

File:R5.6.png

Author

This problem was solved and uploaded by Will Knapper

R 5.7

Question

$v=4e_1+2e_2=c_1{b}_1+c_2{b}_2$
The oblique basis vectors are:
$b_1=2e_1+7e_2$
$b_2=1.5e_1+3e_2$

1. Find the components $c_1,c_2$ 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 $b_1,b_2$ relative to the basis $e_1,e_2$, as discussed on p.7c-34.

Solution

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

$\left[\begin{array}{cc}<b_1,b_1>& <b_1,b_2>\\ <b_2,b_1>& <b_2,b_2>\end{array}\right]$ $\left\{\begin{array}{c}{c}_1\\ c_2\end{array}\right\}$ = $\left\{\begin{array}{c}<b_1,v>\\ <c_2,v>\end{array}\right\}$

From (3)p.8-9 we know that $<b_i,b_j>=b_i\cdot b_j$. Solving the various components we get:

$\left[\begin{array}{cc}53& 24\\ 24& 11.25\end{array}\right]$ $\left\{\begin{array}{c}{c}_1\\ c_2\end{array}\right\}$ = $\left\{\begin{array}{c}22\\ 12\end{array}\right\}$

In order to solve for $c_1,c_2$ we need to calculate the inverse of $\left[\begin{array}{cc}53& 24\\ 24& 11.25\end{array}\right]$.

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

$\left[\begin{array}{cc}0.55556& -1.18519\\ -1.18519& 2.61728\end{array}\right]$ $\left\{\begin{array}{c}22\\ 12\end{array}\right\}$ = $\left\{\begin{array}{c}{c}_1\\ c_2\end{array}\right\}$

Thus:
$c_1=-2;c_2=5.333=\frac{16}{3}$

2. Plugging in $b_1,b_2$ into $v$ we get:

$c_1(2e_1+7e_2)+c_2(1.5e_1+3e_2)=4e_1+2e_2$

$2c_1{e}_1+7c_1{e}_2+1.5c_2{e}_1+3c_2{e}_2=4e_1+2e_2$

$e_1(2c_1+1.5c_2)+e_2(7c_1+3c_2)=4e_1+2e_2$

Separating into components:

$2c_1+1.5c_2=4$ and $7c_1+3c_2=2$

Solving the two linearly independent equations we get:

$$
$c_1=-2;c_2=\frac{16}{3}$

Author

This problem was solved and uploaded by Radina Dikova

R 5.8

Question

Find the integral (see R5.9)

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

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

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

$(\genfrac{}{}{0ex}{}{n}{k})=\frac{n!}{k!(n-k)!}=\frac{n(n-1)...(n-k+1)}{k!}$

Solution

The indefinite integral of $x^n$ is

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

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

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

And the indefinite integral of $log(1+x)$ is

$(x+1)log(1+x)-x+C$

Integration by parts

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

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

Author

Solved and uploaded by Mike Wallace

R 5.9

Question

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

$y^{″}-3y^{\prime }+2y=r(x)$

$r(x)=log(1+x)$

and the initial conditions

$y(-\frac{3}{4})=1,y^{\prime }(-\frac{3}{4})=0$

Part One

Project the excitation r(x) on the polynomial basis $[b_j(x)=x^j,j=0,1,...,n]$

i.e., find $d_j$ such that

r(x) ≈ $r_n(x)=\sum _{j=0}^n{d}_j{x}^j$

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

Plot $r(x),r_n(x)$ to show uniform approximation and convergence

Part Two

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 $y_n(x)$ such that:

${y_n}^{″}+a{y_n}^{\prime }+b{y}_n=r_n(x)$

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

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

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.

Solution

Using $y^{″}-3y^{\prime }+2y=log(1+x)$

For n = 0

$b_0=x^0$

$r=log(1+x)=C_0{x}^0$

$d_0=<x^0,log(1+x)>=\underset{-\frac{3}{4}}{\overset{3}{\int }}log(1+x)dx=2.14175$

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

$d_0=C_0<x^0,x^0>=C_0(\frac{15}{4})=2.14175$

$C_0=0.57113$

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$Y=Y_h+Y_p=C_1{e}^{2x}+C_2{e}^x+0.57113$

$Y^{\prime }=2C_1{e}^{2x}+C_2{e}^x$

$y(\frac{3}{4})=C_1{e}^{\frac{6}{4}}+C_2{e}^{\frac{3}{4}}+0.57113=1$

$y^{\prime }(-\frac{3}{4})=-\frac{6}{4}{C}_1{e}^{-\frac{6}{4}}+C_2{e}^{-\frac{3}{4}}=0$

Subtracting Y from Y' in order to find the coefficients

$C_1{e}^{-\frac{6}{4}}-0.57113=-1$

$C_1=--1.92205$

$C_2=1.81582$

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For n = 0 the final solution will be

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

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For n = 1

$C_0=-.09399,C_1=0.591217$

We need to find the homogeneous Y

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

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

$Y_h=C_1{e}^{2x}+C_2{e}^x$

$Y=Y_h+Y_p=C_1{e}^{2x}+C_2{e}^x-0.09399+0.591217x$

Solving for the initial conditions

$y(\frac{3}{4})=C_1{e}^{\frac{6}{4}}+C_2{e}^{\frac{3}{4}}-0.09399+0.591217(\frac{3}{4})=1$

$Y^{\prime }==2C_1{e}^{2x}+C_2{e}^x+0.591217$

$y^{\prime }(-\frac{3}{4})=-\frac{6}{4}{C}_1{e}^{-\frac{6}{4}}+C_2{e}^{-\frac{3}{4}}+0.591217=0$

This gives coefficients of:

$C_1=-9.5398,C_2=7.761$

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For n = 1 the final solution will be;

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

Author

This problem was solved and uploaded by Mike Wallace

Contribution Summary

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