University of Florida/Egm4313/s12.team5.R4

R4.1

Question

For the series shown in the notes on p. 7-20

$\sum _{j=0}^{n-2}[c_{j+2}(j+2)(j+1)+a{c}_{j+1}(j+1)+b{c}_j]x^j+a{c}_{n}n{x}^{n-1}+b[c_{n-1}{x}^{n-1}+c_n{x}^n]=\sum _{j=0}^n{d}_j{x}^j$

Obtain the equations for the coefficients of $x$, $x^2$, $x^{n-2}$ , $x^{n-1}$ , $x^n$

Then, set up the coefficient matrix A for the general case with the coefficients obtained.

Solution

For j=0:

$2c_2+a{c}_1+b{c}_0=d_0$  (1)

For j=1:

$6c_3+2a{c}_2+b{c}_1=d_1$  (2)

For j=2:

$12c_4+3a{c}_3+b{c}_2=d_2$ (3)

For j=n-2:

$[c_n(n)(n-1)+a{c}_{n-1}(n-1)+b{c}_{n-2}]=d_{n-2}$ (4)

For j=n-1:

$a{c}_{n}n+b{c}_{n-1}=d_{n-1}$ (5)

For j=n:

$b{c}_n=d_n$ (6)

Setting up the A matrix using equations (1)-(6):

$A=\left[\begin{array}{cccccc}b& a& 2& & & \\ & b& 2a& & & \\ & & b& & & \\ & & & b& a(n-1)& n(n-1)\\ & & & & b& an\\ & & & & & b\end{array}\right]$

Author

Solved and uploaded by Joshua House

Proofread by David Herrick

R 4.2

Question

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

$y^{″}-3y^{\prime }+2y=r(x)$

$r(x)=\sin x$

and with the initial conditions $y(0)=1,y^{\prime }(0)=0$

Part 1

Use the Taylor series for $\sin x$ in (1)p6-4 to reproduce the figure on p7-24

Part 2

Let $y_p(x)$ be the particular soln corresponding to the excitating $r_n(x)$ : $y^{\prime }{\text{'}}_p+ay{\text{'}}_p+b{y}_p=r_n(x)$

Let $r_n(x)$ be the truncated Taylor series of sin x :

$\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^k(t{)}^{2k+1}}{(2k+1)!}=t-\frac{t^3}{3!}+...+\frac{(-1{)}^n(t{)}^{2n+1}}{(2n+1)!}=:r_n(x)$

Let $y_n(x)$ be the overall soln for the L2-ODE-CC corresponding to (2)-(3) p.7-26 :

$y^{\prime }{\text{'}}_n+ay{\text{'}}_n+b{y}_n=r_n(x)$

with the same initial conditions (3b)p.3-7.

Find the $y_n(x)$ for n = 3, 5, 9; plot these solns for x in the interval [0,4π].

Part 3

Use the particular soln in K 2011 p.82 Table 2.1 to find the exact overall soln y(x) and plot it in the above figure to compare with $y_n(x)$ for n = 3, 5, 9.

Solution

Part 1

File:Partone.jpg

Part 2

To solve the non-homogeneous ODE we have to solve the homogeneous ODE and find any solution of $y_p(x)$. Using the method of undetermined coefficients, specifically the basic rule, where r(x) is in one of the functions in the first column in Table 2.1 K 2011 p. 82 and choose the $y_p(x)$ in the same line and determine its undetermined coefficients by substituting $y_p$ and its derivatives.

For an excitation $r_n(x)$

n = 3

$r_3(x)=t-(t^3)/3!$

n = 5

$r_5(x)=t-(t^3)/3!+(t^5)/5!$

n = 9

$r_9(x)=t-(t^3)/3!+(t^5)/5!-(t^7)/7!+(t^9)/9!$

---

The term in $r(x)$ that will be used will be $k{x}^n(n=3,5,9)$

For n = 3

$K_3{x}^3+K_2{x}^2+K_{1}x+K_0$

For n = 5

$K_5{x}^5+K_4{x}^4+K_3{x}^3+K_2{x}^2+K_{1}x+K_0$

For n = 9

$K_9{x}^9+K_8{x}^8+K_7{x}^7+K_6{x}^6+K_5{x}^5+K_4{x}^4+K_3{x}^3+K_2{x}^2+K_{1}x+K_0$

---

The homogeneous solution for y will be in the form of

$y_h(x)=C_1{e}^{2x}+C_2{e}^x$

Plugging ${y_p}^{″},{y_p}^{\prime },y_p$ into the original equation will give the following:

For n = 3

$(6K_{3}x+2K_2)-3(3K_3{x}^2+2K_{2}x+K_1)+2(K_3{x}^3+K_2{x}^2+K_{1}x+K_0)=x-(x^3)/6!$

For n = 5

$(20K_5{x}^3+12K_4{x}^2+6K_{3}x+4K_{2}x)-3(5K_5{x}^4+4K_4{x}^3+3K_3{x}^2+2K_{2}x+K_1)+2(K_5{x}^5+K_4{x}^4+K_3{x}^3+K_2{x}^2+K_{1}x+K_0)$

$=x-(x^3)/3!+(x^5)/5!$

For n = 9

$(72K_9{x}^7+56K_8{x}^6+42K_7{x}^5+30K_6{x}^4+20K_5{x}^3+12K_4{x}^2+6K_{3}x+4K_{2}x)-3(9K_9{x}^8+8K_8{x}^7+7K_7{x}^6+6K_6{x}^5+5K_5{x}^4+4K_4{x}^3$

$+3K_3{x}^2+2K_{2}x+K_1+2(K_9{x}^9+K_8{x}^8+K_7{x}^7+K_6{x}^6+K_5{x}^5+K_4{x}^4+K_3{x}^3+K_2{x}^2+K_{1}x+K_0)=x-(x^3)/3!+(x^5)/5!-(x^7)/7!+(x^9)/9!$

---

Equating the coefficients of $x^n$ on both sides

For n = 3

$x^3:2K_3=-1/6$

$K_3=-1/12$

$x^2:-9K_3+2K_2=0$

$K_2=-3/8$

$x:6K_3-6K_2+2K_1=1$

$K_1=-3/8$

$c:2K_2-3K_1+2K_0=0$

$K_0=-3/16$

Now plugging in the particular and homogeneous equation $y=y_h+y_p$ and solving for the initial conditions.

$y=C_1{e}^{2x}+C_2{e}^x-(1/12)x^3-(3/8)x^2-(3/8)x-3/16$ $y(0)=C_1+C_2-3/16=1$ $y^{\prime }(0)=2C_1+C_2-3/8=0$

This gives constant values of $C_1=-13/16$ $C_2=2$

For n = 3 Final solution is:

$y_3=(-13/16)e^{2x}+2e^x-(1/12)x^3-(3/8)x^2-(3/8)x-3/16$

---

For n = 5

Equating the coefficients

$x^5:2K_5=1/120$

$K_5=1/240$

$x^4:-15K_5+2K_4=0$

$K_4=1/32$

$x^3:20K_5-12K_4+2K_3=-1/6$

$K_3=-11/24$

$x^2:12K_4-9K_3+2K_2=0$

$K_2=-9/4$

$x:6K_3-6K_2+2K_1=1$

$K_1=-39/8$

$c:4k_2-3K_1+2K_0=0$

$K_0=-45/16$

Now plugging in the particular and homogeneous equation $y=y_h+y_p$ and solving for the initial conditions.

$y=C_1{e}^{2x}+C_2{e}^x+(1/240)x^5+(1/32)x^4-(11/24)x^3-(9/4)x^2-(39/8)x-45/16$

$y(0)=C_1+C_2-45/16=1$

$y^{\prime }(0)=2C_1+C_2-39/8=0$

This gives constant values of $C_1=79/48$ $C_2=13/6$

For n = 5 the final solution is

$y_5=(79/48)e^{2x}+(13/6)e^x+(1/240)x^5+(1/32)x^4-(11/24)x^3-(9/4)x^2-(39/8)x-45/16$

---

For n = 9

Equating the coefficients

$x^9:2K_9=1/9!$

$K_9=1.3778e^{-6}$

$x^8:-27K_9+2K_8=0$

$k_8=1.86e^{-5}$

$x^7:72K_9-24K_8+2K_7=-1/7!$

$K_7=2.728e^{-4}$

$x^6:56K_8-21K_7+2K_6=0$

$K_6=0.0023$

$x^5:42K_7-18K_6+2K_5=1/5!$

$K_5=0.0111$

$x^4:30K_6-15K_5+2K_4=0$

$K_4=0.0488$

$x^3:20K_5-12K_4+2K_3=-1/3!$

$K_3=0.0976)$

$x^2:12K_4-9K_3+2K_2=0$

$K_2=0.1463$

$x:6K_3-6K_2+2K_1=1$

$K_1=0.6461$

$c:4K_2-3K_1+2K_0=0$

$K_0=0.6765$

Now plugging in the particular and homogeneous equation $y=y_h+y_p$ and solving for the initial conditions.

$y=C_1{e}^{2x}+C_2{e}^x+y_p$

$y(0)=C_1+C_2+0.6765=1$

$y^{\prime }(0)=2C_1+C_2+0.6461=0$

This gives constant values of

$C_1=-0.3226$

$C_2=0.6449$

For n = 9 the final solution is

$y_9=-0.3226e^{2x}+0.6449e^x+1.37e^{-6}{x}^9+1.86e^{-5}{x}^8+2.72e^{-4}{x}^7+0.0023x^6+0.0111x^5+0.0488x^4-0.0976x^3$

$-0.1463x^2-0.6461x-0.6765$

File:Parttwo.jpg

Part 3

The table 2.1 from K 2011 p.82 illustrates that for the method of undetermined coefficients for an r(x) term in the form of:

r(x) = k sin(ωx)

Then the choice for $y_p(x)$ would be:

K cos(ωx) + M sin(ωx)

The homogeneous solution for y will be in the form of

$y_h(x)=y(x)=C_1{e}^{2x}+C_2{e}^x$

with roots of 2 and 1 derived from $y^{″}-3y^{\prime }+2y=0$

Plugging in the derivatives for the particular solution into the original equation gives:

${y_p}^{″}-3{y_p}^{\prime }+2y_p=sin(x)$

where

     $y_p=Acos(x)+Bsin(x)$

     ${y_p}^{\prime }=-Asin(x)+Bcos(x)$

     ${y_p}^{″}=-Acos(x)-Bsin(x)$

     $-Acos(x)-Bsin(x)-3(-Asin(x)+Bcos(x))+2(Acos(x)+Bsin(x))=sin(x)$

     $(A-3B)cos(x)+(3A+B)sin(x)=sin(x)$

     $3A+B=1;A-3B=0;$

     $A=3/10;B=1/10$

Combining the particular and homogeneous solution of y gives the following:

$y=y_h+y_p=C_1{e}^{2x}+C_2{e}^x+(3/10)cos(x)+(1/10)sin(x)$

Solving for the initial conditions where $y(0)=1,and{y}^{\prime }(0)=0$

$y(0)=C_1+C_2=7/10$

$y^{\prime }(0)=2C_1+C_2+1/10$

$2C_1+7/10-C_1=-1/10$

$C_1=-8/10;C_2=-1/10$

This gives the final solution:

$y=(-8/10)e^{2x}-(1/10)e^x+(3/10)cos(x)+(1/10)sin(x)$

File:Ptthree.jpg

From the figure it is barely discernible between this figure and the figure above it in part two. The final solution for y in part three is very similar if not approximately identical to the lower curve in the figure in part two.

Author

This problem was solved and uploaded by Michael Wallace

R 4.3

Question

Consider the L2-ODE-CC:

$y^{″}-3y^{\prime }+2y=r(x)$

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

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

Part 1

Develop $\log (1+x)$ in Taylor series about x(hat) = 0 to reproduce the figure in the notes on p. 7-25.

Part 2

Let $r_n(x)$ be the truncated Taylor series with n terms -- which is also the highest degree of the Taylor (power) series -- of $\log (1+x)$.

Find $y_n(x)$ for n = 4, 7, and 11 such that:

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

Plot $y_n(x)$ for n = 4,7,11 for x in $[-\frac{3}{4},3]$

Part 3

Use the matlab command ODE45 to integrate numerically the same function with the same initial conditions to obtain $y(x)$ . Then plot $y(x)$ in the same figure with $y_n(x)$

Solution

Part 1

The formula to develop the Taylor Series about x(hat) = 0 is given by:

$f(0)+\frac{f^{\prime }(0)(x-0{)}^1}{1!}+\frac{f^{″}(0)(x-0{)}^2}{2!}+\frac{f^{‴}(0)(x-0{)}^3}{3!}+...\sum _{n=0}^{\mathrm{\infty }}\frac{f^n(0)(x-0{)}^n}{n!}$

Using the function $\log (1+x)$ we get:

$f^{\prime }(x)=\frac{1}{(x+1)},f^{″}(x)=\frac{-1}{(x+1{)}^2},f^{‴}(x)=\frac{2}{(x+1{)}^3},...$

Plugging in 0, the first few terms become:

$0+x-\frac{x^2}{2}+\frac{x^3}{3}$

The pattern of this expression can be represented by the Taylor series:

$\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(x{)}^{n+1}}{(n+1)}$

Part 2

First, the homogenous solution to the differential equation is given by:

$(\lambda -1)(\lambda -2)=0$

Therefore, $y_h=C_1{e}^x+C_2{e}^{2x}$

for $r_n(x)$ is given by the truncated Taylor series of 4 terms:

$r_4(x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$

Therefore, $y_{p4}=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4$

$y_{p^{\prime }4}=c_1+2c_{2}x+3c_3{x}^2+4c_4{x}^3,y_{p^{″}4}=2c_2+6c_{3}x+12c_4{x}^2$

For this solution, with n = 4, the coefficient matrix is set up like:

$A=\left[\begin{array}{ccccc}b& a(n-3)& (n-3)(n-2)& & \\ & b& a(n-2)& (n-1)(n-2)& \\ & & b& a(n-1)& n(n-1)\\ & & & b& an\\ & & & & b\end{array}\right]$

Therefore, the matrix solution with n = 4, a = -3, and b = 2 is:

$A=\left[\begin{array}{ccccc}2& -3& 2& & \\ & 2& -6& 6& \\ & & 2& -9& 12\\ & & & 2& -12\\ & & & & 2\end{array}\right]\left[\begin{array}{c}{c}_0\\ c_1\\ c_2\\ c_3\\ c_4\end{array}\right]=\left[\begin{array}{c}0\\ 1\\ -\frac{1}{2}\\ \frac{1}{3}\\ -\frac{1}{4}\end{array}\right]$

Solving by back substitution we get:

$y_{p4}=-\frac{65}{16}-\frac{33}{8}x-\frac{17}{8}{x}^2-\frac{7}{12}{x}^3-\frac{1}{8}{x}^4$

$y_4(x)=C_1{e}^x+C_2{e}^{2x}-\frac{65}{16}-\frac{33}{8}x-\frac{17}{8}{x}^2-\frac{7}{12}{x}^3-\frac{1}{8}{x}^4$

Plugging in the initial conditions we get:

$y_4(x)=8.90e^x-5.587e^{2x}-\frac{65}{16}-\frac{33}{8}x-\frac{17}{8}{x}^2-\frac{7}{12}{x}^3-\frac{1}{8}{x}^4$

for $r_n(x)$ is given by the truncated Taylor series of 7 terms:

$r_7(x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^7}{7}$

Therefore: $y_{p7}=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+c_5{x}^5+c_6{x}^6+c_7{x}^7$

$y_{p^{\prime }7}=c_1+2c_{2}x+3c_3{x}^2+4c_4{x}^3+5c_5{x}^4+6c_6{x}^5+7c_7{x}^6$

$y_{p^{″}7}=2c_2+6c_{3}x+12c_4{x}^2+20c_5{x}^3+30c_6{x}^4+42c_7{x}^5$

For this solution, with n = 7, the coefficient matrix is set up like:

$A=\left[\begin{array}{cccccccc}b& a(n-6)& (n-6)(n-5)& & & & & \\ & b& a(n-5)& (n-5)(n-4)& & & & \\ & & b& a(n-4)& (n-4)(n-3)& & & \\ & & & b& a(n-3)& (n-3)(n-2)& & \\ & & & & b& a(n-2)& (n-2)(n-1)& \\ & & & & & b& a(n-1)& (n-1)(n)\\ & & & & & & b& an\\ & & & & & & & b\end{array}\right]$

Therefore, the matrix solution with n = 7, a = -3, and b = 2 is:

$A=\left[\begin{array}{cccccccc}2& -3& 2& & & & & \\ & 2& -6& 6& & & & \\ & & 2& -9& 12& & & \\ & & & 2& -12& 20& & \\ & & & & 2& -15& 30& \\ & & & & & 2& -18& 42\\ & & & & & & 2& -21\\ & & & & & & & 2\end{array}\right]\left[\begin{array}{c}{c}_0\\ c_1\\ c_2\\ c_3\\ c_4\\ c_5\\ c_6\\ c_7\end{array}\right]=\left[\begin{array}{c}0\\ 1\\ -\frac{1}{2}\\ \frac{1}{3}\\ -\frac{1}{4}\\ \frac{1}{5}\\ -\frac{1}{6}\\ \frac{1}{7}\end{array}\right]$

Solving by back substitution we get:

$y_{p7}=\frac{49317}{80}+\frac{24573}{40}x+\frac{12201}{40}{x}^2+\frac{1205}{12}{x}^3+\frac{195}{8}{x}^4+\frac{23}{5}{x}^5+\frac{2}{3}{x}^6+\frac{1}{14}{x}^7$

Since the homogeneous solution remains the same:

$y_7(x)=C_1{e}^x+C_2{e}^{2x}+\frac{49317}{80}+\frac{24573}{40}x+\frac{12201}{40}{x}^2+\frac{1205}{12}{x}^3+\frac{195}{8}{x}^4+\frac{23}{5}{x}^5+\frac{2}{3}{x}^6+\frac{1}{14}{x}^7$

Plugging in the initial conditions we get:

$y_7(x)=-613.504e^x-3.868e^{2x}+\frac{49317}{80}+\frac{24573}{40}x+\frac{12201}{40}{x}^2+\frac{1205}{12}{x}^3+\frac{195}{8}{x}^4+\frac{23}{5}{x}^5+\frac{2}{3}{x}^6+\frac{1}{14}{x}^7$

The image below demonstrates the general matrix for n=11 and the matrix with values filled in: File:EML HW4 matricies1.png

for $r_n(x)$ is given by the truncated Taylor series of 11 terms:

$r_{11}(x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^7}{7}-\frac{x^8}{8}+\frac{x^9}{9}-\frac{x^{10}}{10}+\frac{x^{11}}{11}$

Therefore, $y_{p11}=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+c_5{x}^5+c_6{x}^6+c_7{x}^7+c_8{x}^8+c_9{x}^9+c_{10}{x}^{10}+c_{11}{x}^{11}$

Solving by back substitution we get:

$y_{p11}=3301079.406+3300338.812x+1649428.813x^2+549316.042x^3+137082.188x^4+27317.725x^5+4520.042x^6$

$+636.321x^7+77.188x^8+8.056x^9+0.7x^{10}+0.0455x^{11}$

Since the homogeneous solution remains the same:

$y_{11}(x)=C_1{e}^x+C_2{e}^{2x}+3301079.406+3300338.812x+1649428.813x^2+549316.042x^3+137082.188x^4+27317.725x^5+4520.042x^6$

$+636.321x^7+77.188x^8+8.056x^9+0.7x^{10}+0.0455x^{11}$

Plugging in the initial conditions we get:

$y_{11}(x)=-3391815.1e^x+734.909e^{2x}+3301079.406+3300338.812x+1649428.813x^2+549316.042x^3+137082.188x^4+27317.725x^5$

$+4520.042x^6+636.321x^7+77.188x^8+8.056x^9+0.7x^{10}+0.0455x^{11}$

File:R4.3.png

Part 3

Matlab Code:

function yp = F(t,y)

yp = zeros(2,1);

yp(1) = y(2);

yp(2) = log(t+1) + 3*y(2)-2*y(1);

end

[t,y] = ode45('F',[-.75,3],[1,0]);

plot(x,y4,'g',x,y7,'r',x,y11,'y',t,y(:,1),'b')

axis([-0.75 3 -4 2])

File:Part3.png

Author

Part 2 n=7 and n =11 were solved and uploaded by Cameron North.

Part 1 and Part 2 n = 4 was solved and uploaded by David Herrick

R 4.4

Question

Part 1

Find n sufficiently high so that $y_n(x_1),y{\text{'}}_n(x_1)$ do not differ from the numerical solution by more than $10^{-5}$ at $x_1=0.9$

Part 2

Develop log(1+x) in Taylor series about $\hat{x}=1$. Plot the results.

Part 3

Find $y_n(x)$ for n=4,7,11, such that ${y_n}^{″}+a{y_n}^{\prime }+b{y}_n=r_n(x)$ for x in [0.9,3] with the initial conditions found in Part 1. Plot the results.

Part 4

Use the matlab command 'ode45' to integrate numerically (5) p.7b-7 with (1) p.7-28 and the initial conditions $y_n(x_1),y{\text{'}}_n(x_1)$ to obtain the numerical solution for $y(x)$.
Plot $y(x)$ in the same figure with $y_n(x)$

Solution

Part 1

With MATLAB, a program was used to iteratively add terms into the taylor series of $log(1+x)$. Until the error between the exact answer and the series was less than $10^{-5}$., more terms were added.

File:IEA 4 4 1a.jpg

File:IEA 4 4 1asolution.jpg

$n=39$ for the error to be of a magnitude of $10^{-5}$. The error found: 9.7422e-005


With a very similar process for $y{\text{'}}_n(x_1)$.

File:IEA 4 4 1b.jpg

File:IEA 4 4 1bsolution.jpg

$n=74$ for the error to be of a magnitude of $10^{-5}$. The error found: 9.3967e-005

Part 2

The formula to develop the Taylor Series about $\hat{x}=1$ is given by:

$f(1)+\frac{f^{\prime }(1)(x-1{)}^1}{1!}+\frac{f^{″}(1)(x-1{)}^2}{2!}+\frac{f^{‴}(1)(x-1{)}^3}{3!}+...\sum _{n=0}^{\mathrm{\infty }}\frac{f^n(1)(x-1{)}^n}{n!}$

Using the function $\log (1+x)$ for n=11 we get:

$f^{\prime }(x)=\frac{1}{(x+1)},f^{″}(x)=\frac{-1}{(x+1{)}^2},f^{‴}(x)=\frac{2}{(x+1{)}^3},f^{IV}(x)=\frac{-6}{(x+1{)}^4},$
$f^V(x)=\frac{24}{(x+1{)}^5},f^{VI}(x)=\frac{-120}{(x+1{)}^6},f^{VII}(x)=\frac{720}{(x+1{)}^7},f^{VIII}(x)=\frac{-5040}{(x+1{)}^8},$
$f{\text{'}}^X(x)=\frac{40320}{(x+1{)}^9},f^X(x)=\frac{-362880}{(x+1{)}^{10}},f^{XI}(x)=\frac{3628800}{(x+1{)}^{11}}$

Plugging in x=1 up to n=4, and using the formula for developing the Taylor Series, we get $y_{n4}$ :

$y_{n4}=log(2)+\frac{(x-1)}{2}-\frac{(x-1{)}^2}{8}+\frac{(x-1{)}^3}{24}-\frac{(x-1{)}^4}{64}$,

Then we do the same for n=7 terms to get $y_{n7}$ :

$y_{n7}=log(2)+\frac{(x-1)}{2}-\frac{(x-1{)}^2}{8}+\frac{(x-1{)}^3}{24}-\frac{(x-1{)}^4}{64}+\frac{(x-1{)}^5}{160}-\frac{(x-1{)}^6}{384}+\frac{(x-1{)}^7}{896}$

And finally for $y_{n11}$:

$y_{n11}=log(2)+\frac{(x-1)}{2}-\frac{(x-1{)}^2}{8}+\frac{(x-1{)}^3}{24}-\frac{(x-1{)}^4}{64}+\frac{(x-1{)}^5}{160}-\frac{(x-1{)}^6}{384}+\frac{(x-1{)}^7}{896}-\frac{(x-1{)}^8}{2048}+\frac{(x-1{)}^9}{4608}-\frac{(x-1{)}^{10}}{10240}+\frac{(x-1{)}^{11}}{22528}$

Plotting the functions:
Matlab code:

x=-3:0.2:6;
y = log(1+x);
y4 = log(2) + (x-1)/2 - (x-1).^2/8 + (x-1).^3/24 - (x-1).^4/64;
y7 = log(2) + (x-1)/2 - (x-1).^2/8 + (x-1).^3/24 - (x-1).^4/64 + (x-1).^5/160 - (x-1).^6/384 + (x-1).^7/896;
y11 = log(2) + (x-1)/2 - (x-1).^2/8 + (x-1).^3/24 - (x-1).^4/64 + (x-1).^5/160 - (x-1).^6/384 + (x-1).^7/896 - (x-1).^8/2048 + (x-1).^9/4608 - (x-1).^(10)/10240 + (x-1).^(11)/22528;
plot(x,y, x,y4, x,y7, x,y11)
hleg1 = legend('log(1+x)','n=4','n=7','n=11');

Graph:

Convergence is in the domain [-1.5,4].

Part3

For n=4 $r_x$ is:

$r_x=log(2)+\frac{(x-1)}{2}-\frac{(x-1{)}^2}{8}+\frac{(x-1{)}^3}{24}-\frac{(x-1{)}^4}{64}$ 

$y_{p4}=c_0+c_1(x-1)-c_2(x-1{)}^2+c_3(x-1{)}^3-c_4(x-1{)}^4$
$y_{n^{\prime }4}=c_1-2c_2(x-1)+3c_3(x-1{)}^2-4c_4(x-1{)}^3$
$y_{n^{″}4}=-2c_2+6c_3(x-1)-12c_4(x-1{)}^2$,

Plugging the equations into the above ODE will give matrices like the following:

$\left[\begin{array}{ccccc}2& 0& 0& 0& 0\\ -12& 2& 0& 0& 0\\ 12& -9& 2& 0& 0\\ 0& 6& -6& 2& 0\\ 0& 0& 2& -3& 2\end{array}\right]*\left[\begin{array}{c}{c}_4\\ c_3\\ c_2\\ c_1\\ c_0\end{array}\right]=\left[\begin{array}{c}-\frac{1}{64ln(10)}\\ \frac{1}{24ln(10)}\\ -\frac{1}{8ln(10)}\\ \frac{1}{2ln(10)}\\ log(2)\end{array}\right]$

The unknown vector $c$ can be solved by forward substitution, with MATLAB to do the calculations:

$c_4=-.0034,c_3=-.0113,c_2=-.0577,c_1=-.1624,c_0=.1624$

Particular and general solution $y_{p4}$ , $y_4$:

$y_{p4}=0.1624-0.1624*(x-1)-.0577*(x-1{)}^2-.0113*(x-1{)}^3-.0034*(x-1{)}^4$

$y_4(x)=y_h(x)+y_{p4}(x)$

$y_4(x)=C_1{e}^x+C_2{e}^{2x}+0.1624-0.1624*(x-1)$ $-.0577*(x-1{)}^2-.0113*(x-1{)}^3-.0034*(x-1{)}^4$

Then with the Initial Conditions,

$y_4(x)=.0595e^x-.0076e^{2x}+0.1624-0.1624*(x-1)$
$-.0577*(x-1{)}^2-.0113*(x-1{)}^3-.0034*(x-1{)}^4$

For n=7 $r_x$ is:

$r_x=log(2)+\frac{(x-1)}{2}-\frac{(x-1{)}^2}{8}+\frac{(x-1{)}^3}{24}-\frac{(x-1{)}^4}{64}+\frac{(x-1{)}^5}{160}-\frac{(x-1{)}^6}{384}+\frac{(x-1{)}^7}{896}$

$y_{p7}=c_0+c_1(x-1)-c_2(x-1{)}^2+c_3(x-1{)}^3-c_4(x-1{)}^4+c_5(x-1{)}^5-c_6(x-1{)}^6+c_7(x-1{)}^7$

$y_{p^{\prime }7}=c_1-2c_2(x-1)+3c_3(x-1{)}^2-4c_4(x-1{)}^3+5c_5(x-1{)}^4-6c_6(x-1{)}^5+7c_7(x-1{)}^6$

$y_{p^{″}7}=-2c_2+6c_3(x-1)-12c_4(x-1{)}^2+20c_5(x-1{)}^3-30c_6(x-1{)}^4+42c_7(x-1{)}^5$

The same process used when solving when n=4 is used to construct a matrix equation for n=7:

$\left[\begin{array}{cccccccc}2& 0& 0& 0& 0& 0& 0& 0\\ -21& 2& 0& 0& 0& 0& 0& 0\\ 42& -18& 2& 0& 0& 0& 0& 0\\ 0& 30& -15& 2& 0& 0& 0& 0\\ 0& 0& 20& -12& 2& 0& 0& 0\\ 0& 0& 0& 12& -9& 2& 0& 0\\ 0& 0& 0& 0& 6& -6& 2& 0\\ 0& 0& 0& 0& 0& 2& -3& 2\end{array}\right]*\left[\begin{array}{c}{c}_7\\ c_6\\ c_5\\ c_4\\ c_3\\ c_2\\ c_1\\ c_0\end{array}\right]=\left[\begin{array}{c}\frac{1}{896ln(10)}\\ -\frac{1}{384ln(10)}\\ \frac{1}{160ln(10)}\\ -\frac{1}{64ln(10)}\\ \frac{1}{24ln(10)}\\ -\frac{1}{8ln(10)}\\ \frac{1}{2ln(10)}\\ log(2)\end{array}\right]$

Using MATLAB to solve for the unknown vector $c$:

$c_7=.0002,c_6=.0020,c_5=.0141,c_4=.0725,c_3=.3034,c_2=.9029,c_1=1.9072,c_0=2.1084$

Particular and general solutions $y_{p7}$ , $y_7$:

$y_{p7}=2.1084+1.9072*(x-1)+.9029*(x-1{)}^2+.3034*(x-1{)}^3+.0725*(x-1{)}^4+.0141*(x-1{)}^5+.0020*(x-1{)}^6+.0002*(x-1{)}^7$

$y_7(x)=y_h(x)+y_{p7}(x)$

$y_7(x)=C_1{e}^x+C_2{e}^{2x}+2.1084+1.9072*(x-1)+.9029*(x-1{)}^2+.3034*(x-1{)}^3+.0725*(x-1{)}^4+.0141*(x-1{)}^5$

$+.0020*(x-1{)}^6+.0002*(x-1{)}^7$

With the initial conditions:

$y_7(x)=-.7271e^x+.0233e^{2x}+2.1084+1.9072*(x-1)+.9029*(x-1{)}^2+.3034*(x-1{)}^3+.0725*(x-1{)}^4+.0141*(x-1{)}^5!$

$+.0020*(x-1{)}^6+.0002*(x-1{)}^7$

For n=11 $r_x$ is:

$r_x=log(2)+\frac{(x-1)}{2}-\frac{(x-1{)}^2}{8}+\frac{(x-1{)}^3}{24}-\frac{(x-1{)}^4}{64}+\frac{(x-1{)}^5}{160}-\frac{(x-1{)}^6}{384}+\frac{(x-1{)}^7}{896}-\frac{(x-1{)}^8}{2048}+\frac{(x-1{)}^9}{4608}-\frac{(x-1{)}^{10}}{10240}+\frac{(x-1{)}^{11}}{22528}$

$y_{p11}=c_0+c_1(x-1)-c_2(x-1{)}^2+c_3(x-1{)}^3-c_4(x-1{)}^4+c_5(x-1{)}^5-c_6(x-1{)}^6+c_7(x-1{)}^7-(x-1{)}^8+(x-1{)}^9-(x-1{)}^{10}+(x-1{)}^{11}$ $y_{p^{\prime }11}=c_1-2c_2(x-1)+3c_3(x-1{)}^2-4c_4(x-1{)}^3+5c_5(x-1{)}^4-6c_6(x-1{)}^5+7c_7(x-1{)}^6$

$-8c_8(x-1{)}^7+9c_9(x-1{)}^8-10c_{10}(x-1{)}^9+11c_{11}(x-1{)}^{10}$ $y_{p^{″}11}=-2c_2+6c_3(x-1)-12c_4(x-1{)}^2+20c_5(x-1{)}^3-30c_6(x-1{)}^4+42c_7(x-1{)}^5-56c_8(x-1{)}^6$

$+72c_9(x-1{)}^7-90c_{10}(x-1{)}^8+110c_{11}(x-1{)}^9$

Creating another matrix system and solving for the unknown vector $c$:

$c_{11}=0,c_{10}=.0002,c_9=.0019,c_8=.0181,c_7=.15,c_6=1.0675,c_5=6.4597,$

$c_4=32.4318,c_3=130.0033,c_2=390.3968,c_1=781.289,c_0=781.6873$

Particular and general solutions $y_{p11}$ , $y_{1}1(x)$:

$y_{p11}=781.6873+781.289*(x-1)+390.3968*(x-1{)}^2+130.0033*(x-1{)}^3+32.4318*(x-1{)}^4+$

$6.4597*(x-1{)}^5+1.0675*(x-1{)}^6+.15*(x-1{)}^7+.0181*(x-1{)}^8+.0019*(x-1{)}^9+.0002*(x-1{)}^{10}$

$y_{11}(x)=y_h(x)+y_{p11}(x)$

$y_{11}(x)=C_1{e}^x+C_2{e}^{2x}+781.6873+781.289*(x-1)$

$+390.3968*(x-1{)}^2+130.0033*(x-1{)}^3+32.4318*(x-1{)}^4$

$6.4597*(x-1{)}^5+1.0675*(x-1{)}^6+.15*(x-1{)}^7+.0181*(x-1{)}^8+.0019*(x-1{)}^9+.0002*(x-1{)}^{10}$

With initial conditions:

$y_{11}(x)=-287.5907e^x+.05e^{2x}+781.6873+781.289*(x-1)+390.3968*(x-1{)}^2+130.0033*(x-1{)}^3+32.4318*(x-1{)}^4$

$6.4597*(x-1{)}^5+1.0675*(x-1{)}^6+.15*(x-1{)}^7+.0181*(x-1{)}^8+.0019*(x-1{)}^9+.0002*(x-1{)}^{10}$

plot:

Part 4

The initial condition equation is:

$y^{″}-3y^{\prime }+2y=ln(1+x)$

The initial conditions are:

$y(0)=39,y^{\prime }(0)=74$

To solve the problem, we first convert the 2nd order differential equation into two 1st order differential equations. The initial 2nd order then turns into these two equations:

$y_1=y,y_2=y^{\prime }$

Taking the derivatives, we find that:

$y{\text{'}}_1=y_2,y{\text{'}}_2=-2y_1-3y_2+ln(1+x)$

with the new initial conditions as:

$y_1(t_0)=1,y_2(t_0)=0,t_0=\frac{-3}{4}$

Now, we create a MATLAB function, "F", that will return a vector-valued function. We do this with the following function:

function yp=F(x,y)
yp=zeros(2,1);
yp(1)=y(2);
yp(2)=-2*y(1)-3*y(2)+log(1+x);

We now incorporate the function into a simple MATLAB program which will calculate the results:

EDU>> [x,y]=ode45('F',[0,5],[39,74]);
EDU>> plot(y(:,1),y(:,2))

This generates the following graph of $y_n(x)$:



Now we solve the initial value problem by hand to find y(x):

$y^{″}-3y^{\prime }+2y=ln(1+x)$

First, convert to the characteristic equation:

${\lambda }^2-3\lambda +2=0$
${\lambda }_1=2,{\lambda }_2=1$
$y_h(x)=C_1{e}^x+C_2{e}^{2}x$

The excitation factor of ln(1+x) does not yield any useful value for $y_p(x)$ since the integral is nonelementary. Therefore, the final solution for y(x) with what is know is simply:

$y(x)=C_1{e}^x+C_2{e}^{2}x,y^{\prime }(x)=C_1{e}^x+2C_2{e}^{2}x$ $y(0)=39,C_1+C_2=39$
$y^{\prime }(0)=74,C_1+2C_2=74$
$C_1=4,C_2=35$
$y(x)=4e^x+35e^{2x}$

Graphing both solutions on the same graph yields the following:

MATLAB code: EDU>> z = 4*exp(x) + 25*exp(2*x)
plot(y(:,1),y(:,2),y(:,1),z)
File:Graph2.jpg

Contribution Summary

Problem 1 was solved and uploaded by Joshua House 15:34, 13 March 2012 (UTC)

Problem 2 was solved by Mike Wallace

Problem 3 part 2 was solved by John North and parts 1 and 3 were solved by David Herrick

Problem 4 part 1 and part of part3 were solved and uploaded by Derik Bell, parts 2 and 3 were solved by Radina Dikova, and part 4 was solved by William Knapper