University of Florida/Egm4313/IEA-f13-team10/R4

Report 4

Problem 1: Basic Rule to Solve Non-Homogeneous ODE

Problem Statement

Solve the ODE:

$y^{″}+6y^{\prime }+9y=8x-2x^7$

With the initial conditions:

$y(0)=1$

$y^{\prime }(0)=.5$

Plot the homogeneous solution

Plot the particular solution

Plot the overall solution

Solution

Homogeneous solution: $y_h(x)=C_1{e}^{-3x}+C_{2}x{e}^{-3x}$

So that ${y_h}^{\prime }(x)=-3C_1{e}^{-3x}+C_2{e}^{-3x}-3C_{2}x{e}^{-3x}$

Solving for the initial conditions:

$y_h(0)=1=C_1+0$

${y_h}^{\prime }(0)=.5=-3C_1+C_2-0$

$C_1=1$
$C_2=3.5$

So the homogeneous solution is:

$y_h(x)=e^{-3x}+3.5x{e}^{-3x}$

Choose the particular solution to be: $y_p(x)=A{x}^7+B{x}^6+C{x}^5+D{x}^4+E{x}^3+F{x}^2+Gx+H$

So that:

${y_p}^{\prime }(x)=7A{x}^6+6B{x}^5+5C{x}^4+4D{x}^3+3E{x}^2+2Fx+G$

${y_p}^{″}(x)=42A{x}^5+30B{x}^4+20C{x}^3+12D{x}^2+6Ex+2F$

Substitute in the original equation $y^{″}+6y^{\prime }+9y=8x-2x^7$

$(42A{x}^5+30B{x}^4+20C{x}^3+12D{x}^2+6Ex+2F)+6(7A{x}^6+6B{x}^5+5C{x}^4+4D{x}^3+3E{x}^2+2Fx+G)+...$
$...9(A{x}^7+B{x}^6+C{x}^5+D{x}^4+E{x}^3+F{x}^2+Gx+H)=8x-2x^7$

Sorting by the x term gives:

$x^7(9A)+x^6(42A+9B)+x^5(42A+36B+9C)+x^4(30B+30C+9D)+...$
$...x^3(20C+24D+9E)+x^2(12D+18E+9F)+x(6E+12F+9G)+(2F+6G+9H)=8x-2x^7$

Giving us the system of equations:


$9A=-2$
$42A+9B=0$
$42A+36B+9C=0$
$30B+30C+9D=0$
$20C+24D+9E=0$
$12D+18E+9F=0$
$6E+12F+9G=8$
$2F+6G+9H=0$

The Matlab code we made to solve this was:

A = -2/9;

B = -(42*A)/9;

C = -(42*A + 36*B)/9;

D = -(30*B + 30*C)/9;

E = -(20*C + 24*D)/9;

F= -(12*D + 18*E)/9;

G = (8-(6*E + 12*F))/9;

H = -(2*F + 6*G)/9;

To get (rounded to the nearest tenth):

A= -0.2

B= 1.0

C= -3.1

D= 6.9

E= -11.5

F= 13.8

G= -9.9

H= 3.5

So now the particular solution is:

$y_p(x)=-0.2x^7+1.0x^6-3.1x^5+6.9x^4-11.5x^3+13.8x^2-9.9x+3.5$

The overall solution can be found by:

$y(x)=y_h(x)+y_p(x)$

$y(x)=C_1{e}^{-3x}+C_{2}x{e}^{-3x}-0.2x^7+1.0x^6-3.1x^5+6.9x^4-11.5x^3+13.8x^2-9.9x+3.5$

To solve for C_1 and C_2 (which are different from the homogeneous solution constants) we find:

$y^{\prime }(x)=-3C_1{e}^{-3x}+C_2{e}^{-3x}-3C_{2}x{e}^{-3x}-1.4x^6+6.0x^5-15.5x^4+27.6x^3-34.5x^2+27.6x-9.9$

$y(0)=1=C_1+3.5$ $y^{\prime }(0)=.5=-3C_1+C_2-9.9$

$C_1=-2.5$
$C_2=2.8$

The overall solution is:

$y(x)=-2.5e^{-3x}+2.8x{e}^{-3x}-0.2x^7+1.0x^6-3.1x^5+6.9x^4-11.5x^3+13.8x^2-9.9x+3.5$

Plot the homogeneous solution:
File:P1H.gif

Plot the particular solution:
File:P1P.gif

Plot the overall solution:
File:P1O.gif

Honor Pledge

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

Problem 2: Sum Rule to Find Particular Solution

Part 1

Use the Basic Rule 1 and the Sum Rule 3 to show that the appropriate particular solution for:
$y^{″}+6y^{\prime }+9y=8x-2x^7$
$y(0)=1,y^{\prime }(0)=1/2$

Part 2

Derive the Basic rule and the Sum rule, instead of just using them, based on the linearity
of the differential operator to obtain the expression (trial solution) for the particular solution $y_p(x)$

Solution

$y_p(x)=\sum _{j=0}^n{c}_j{x}^j$

${y_p}^{\prime }(x)=\sum _{j=1}^n{c}_{j}j{x}^{j-1}=\sum _{j=0}^{n-1}{c}_{j+1}(j+1)x^j$

${y_p}^{″}(x)=\sum _{j=2}^n{c}_{j}j(j-1)x^{j-2}=\sum _{j=0}^{n-2}{c}_{j+2}(j+2)(j+1)x^j$

With n =7 we get

$y_p(x)=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+c_5{x}^5+c_6{x}^6+c_7{x}^7$

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

${y_p}^{″}(x)=2c_2+6c_{3}x+12c_4{x}^2+20c_5{x}^3+30c_6{x}^4+42c_7{x}^5$

From report problem 1 it was already found that:

$c_0=3.5$
$c_1=-9.9$
$c_2=13.8$
$c_3=-11.5$
$c_4=6.9$
$c_5=-3.1$
$c_6=1.0$
$c_7=-0.2$

$y_p(x)=-0.2x^7+1.0x^6-3.1x^5+6.9x^4-11.5x^3+13.8x^2-9.9x+3.5$

Honor Pledge

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

Problem 3: Method of Undetermined Coefficients

Problem Set 2.7, problem 5

Find a real, general solution. State which rule you are using. Show each step of your work.

$y^{″}+6y^{\prime }+9y=e^{x}sinx$
$y(0)=1,y^{\prime }(0)=1/2$

plot on separate graphs:

(1) the homogeneous solution $y_h(x)$,

(2) the particular solution $y_p(x)$,

and (3) the overall solution $y(x)$.

Solution

We start by finding the general solution of the homogeneous ODE

$y^{″}+6y^{\prime }+9y=0$

The characteristic equation of the homogeneous ODE is

${\lambda }^2+6\lambda +9=0$
$(\lambda +3{)}^2=0$
$\lambda =-3,-3$

The roots are double real roots.
The general solution of the homogeneous ODE is $y_h=(c_1+x{c}_2)e^{-2x}$

Now we solve for the particular solution of the nonhomogeneous ODE

$y^{″}+6y^{\prime }+9y=e^{x}sinx$

We use the method of Undetermined Coefficients

$y_p=e^x(Kcosx+Msinx)$

$y_{p^{\prime }}=e^x(Kcosx+Msinx)+e^x(-Ksinx+Mcosx)$

$y_{p^{″}}=e^x(Kcosx+Msinx)+e^x(-Ksinx+Mcosx)+e^x(-Ksinx+Mcosx)+e^x(-Kcosx-Msinx)$
$y_{p^{″}}=e^x(Kcosx+Msinx)+2e^x(-Ksinx+Mcosx)+e^x(-Kcosx-Msinx)$

We now substitute the values of $y_p,y_{p^{\prime }},y_{p^{″}}$ into $y^{″}+6y^{\prime }+9y=e^{x}sinx$

$e^x(Kcosx+Msinx)+2e^x(-Ksinx+Mcosx)+e^x(-Kcosx-Msinx)$
$+6(e^x(Kcosx+Msinx)+e^x(-Ksinx+Mcosx))$
$+9(e^x(Kcosx+Msinx))=e^{x}sinx$

$16e^x(Kcosx+Msinx)+8e^x(-Ksinx+Mcosx)+e^x(-Kcosx-Msinx)=e^{x}sinx$
$e^x(15(Kcosx+Msinx)+8(-Ksinx+Mcosx)=e^{x}sinx$
$e^x((15Kcosx+8Mcosx)+(15Msinx-8Ksinx))=e^{x}sinx$

Now we equate the coefficients of like terms on both sides
$15K+8M=0$
$15M-8K=1$

Now we solve these equations for the coefficients
$K=-8/15M$
$15M+64/15M=1$
$M=0.05190$

$15K+8(15/289)=0$
$K=0.02768$

These values are substituted into $y_p=e^x(Kcosx+Msinx)$ to get the particular solution of the ODE
$y_p=e^x(0.02768cosx+0.05190sinx)$

The general solution of the ODE is
$y=y_h+y_p$
$y=(c_1+x{c}_2)e^{-2x}+e^x(0.02768cosx+0.05190sinx)$

In order to determine the values of $c_1,c_2$ we use the initial conditions $y(0)=1,y^{\prime }(0)=1/2$
$y(0)=c_1+0.02768=1$
$c_1=0.97232$

$y^{\prime }=-2e^{-2x}(c_1+x{c}_2)+e^{-2x}(c_2)+e^x(0.02768cosx+0.05190sinx)+e^x(-0.02768sinx+0.05190cosx)$
$y^{\prime }=-2e^{-2x}(c_1+x{c}_2)+e^{-2x}(c_2)+e^x(0.07958cosx+0.02422sinx)$
$y(0{)}^{\prime }=-2(c_1)+c_2+0.07958=1/2$
$y(0{)}^{\prime }=-1.94464+c_2+0.07958=1/2$
$c_2=2.36506$

The general solution of the ODE is
$y=e^{-2x}(0.97232+2.36506x)+e^x(0.02768cosx+0.05190sinx)$

Plot the homogeneous solution:
File:Problem4-1.jpg
Plot the particular solution:
File:Problem4-2.jpg
Plot the overall solution:
File:Problem4-3.jpg

Honor Pledge

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

Problem 4: Method of Undetermined Coefficients

Problem Set 2.7, problem 5

Find a real, general solution. State which rule you are using. Show each step of your work.

$y^{″}+6y^{\prime }+9y=e^{-2x}\sinh x$
$y(0)=1,y^{\prime }(0)=1/2$

plot on separate graphs:

(1) the homogeneous solution $y_h(x)$,

(2) the particular solution $y_p(x)$,

and (3) the overall solution $y(x)$.

Solution

We start by finding the general solution of the homogeneous ODE

$y^{″}+6y^{\prime }+9y=0$

The characteristic equation of the homogeneous ODE is

${\lambda }^2+6\lambda +9=0$
$(\lambda +3{)}^2=0$
$\lambda =-3,-3$

The roots are double real roots.
The general solution of the homogeneous ODE is $y_h=e^{-2x}(c_1+x{c}_2)$

Now we solve for the particular solution of the non-homogeneous ODE

$y^{″}+6y^{\prime }+9y=e^{-2x}\sinh x$

By using the definition of hyperbolic trigonometric functions we can convert $\sinh x=1/2(e^x-e^{-x})$.
Our non-homogeneous ODE can now be written as:

$y^{″}+6y^{\prime }+9y=e^{-2x}(1/2(e^x-e^{-x})$

$y^{″}+6y^{\prime }+9y=1/2(e^{-x}-e^{-3x})$

Since the replacement of $\sinh x$ is the sum of two functions we can use the sum rule for the method of undetermined coefficients.

$y_p=y_{p1}+y_{p2}=J{e}^{-x}-K{e}^{-3x}$

$y_p=J{e}^{-x}-K{e}^{-3x}$

$y{\text{'}}_p=-J{e}^{-x}+3K{e}^{-3x}$

$y^{\prime }{\text{'}}_p=J{e}^{-x}-9K{e}^{-3x}$

We now substitute the values of $y_p,y_{p^{\prime }},y_{p^{″}}$ into $y^{″}+6y^{\prime }+9y=1/2(e^{-x}-e^{-3x})$

$J{e}^{-x}-9K{e}^{-3x}$
$+6(-J{e}^{-x}+3K{e}^{-3x})$
$+9(J{e}^{-x}-K{e}^{-3x})=1/2(e^{-x}-e^{-3x})$

$J{e}^{-x}-9K{e}^{-3x}-6J{e}^{-x}+18K{e}^{-3x}+9J{e}^{-x}-9K{e}^{-3x}=1/2(e^{-x}-e^{-3x})$

$4J{e}^{-x}=1/2(e^{-x}-e^{-3x})$

Now we equate the coefficients of like terms on both sides and solve for the coefficients
$4J=1/2$
$J=1/8$
$K=0$

These values are substituted into $y_p=J{e}^{-x}-K{e}^{-3x}$ to get the particular solution of the ODE

$y_p=1/8e^{-x}$

The general solution of the ODE is:

$y=y_h+y_p$
$y=e^{-2x}(c_1+x{c}_2)+1/8e^{-x}$

In order to determine the values of $c_1,c_2$ we use the initial conditions $y(0)=1,y^{\prime }(0)=1/2$
$y(0)=c_1+1/8=1$
$c_1=7/8$

$y^{\prime }=-2e^{-2x}(c_1+x{c}_2)+e^{-2x}{c}_2$
$y^{\prime }=-2e^{-2x}(7/8+x{c}_2)+e^{-2x}{c}_2$
$y(0{)}^{\prime }=-14/8+c_2=1/2$
$c_2=9/4$

The general solution of the ODE is
$y=e^{-2x}(7/8+9/4x)+1/8e^{-x}$

Plot the homogeneous solution:
File:Problem5-1.jpg
Plot the particular solution:
File:Problem5-2.jpg
Plot the general solution:
File:Problem5-3.jpg

Honor Pledge

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

Problem 5: Display of Equality by Series Expansion

Problem Statement

Expand the series on both sides of (1)-(2) p.7-12b to verify these equalities
$(1)$

$\sum _{j=2}^5{c}_j*j*(j-1)*x^{j-2}=\sum _{j=0}^3{c}_{j+2}*(j+2)*(j+1)*x^j$

$(2)$

$\sum _{j=1}^5{c}_j*j*x^{j-1}=\sum _{j=0}^4{c}_{j+1}*(j+1)*x^j$

Solution

Evaluating the right-hand side of (1):

$c_{2}2(2-1)x^{2-2}+c_{3}3(3-1)x^{3-2}+c_{4}4(4-1)x^{4-2}+c_{5}5(5-1)x^{5-2}$

$=2c_2+6c_{3}x+12c_4{x}^2+20c_5{x}^3$

Now evaluating the left-hand side of (1):

$c_{0+2}(0+2)(0+1)x^0+c_{1+2}(1+2)(1+1)x^1+c_{2+2}(2+2)(2+1)x^2+c_{3+2}(3+2)(3+1)x^3$

$=2c_2+6c_{3}x+12c_4{x}^2+20c_5{x}^3$

So both sides are equal.

Now evaluating the right-hand side of (2):

$c_1(1)x^{1-1}+c_2(2)x^{2-1}+c_3(3)x^{3-1}+c_4(4)x^{4-1}+c_5(5)x^{5-1}$

$=c_1+2c_{2}x+3c_3{x}^2+4c_4{x}^3+5c_5{x}^4$

The left-hand side of (2) expands into the following:

$c_{0+1}(0+1)x^0+c_{1+1}(1+1)x^1+c_{1+1}(1+1)x^1+c_{2+1}(2+1)x^2+c_{3+1}(3+1)x^3+c_{4+1}(4+1)x^4$

$=c_1+2c_{2}x+3c_3{x}^2+4c_4{x}^3+5c_5{x}^4$

So both sides are equal.

Honor Pledge

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

Problem 6: Taylor Series to Solve ODE

Problem Statement

$y^{″}+6y^{\prime }+9y=cos(2x)$
$y(0)=1$ and $y^{\prime }(0)=1/2$

Solution

The taylor series for the excitation is $\sum \frac{(-4{)}^n*x^{2n}}{(2n)!}$ from n=0 to infinity
For n=3, this equals $1-2x^2+2x^4/3-4x^6/45$
For n=5, this equals $1-2x^2+2x^4/3-4x^6/45+2x^8/315-4x^{10}/14175$
For n=9, this equals $1-2x^2+2x^4/3-4x^6/45+2x^8/315-4x^{10}/14175+4x^{12}/467775-8x^{14}/42567525+2x^{16}/638512875-4x^{18}/97692469875$

For n=3,
$y_p=A{x}^6+B{x}^5+C{x}^4+D{x}^3+E{x}^2+Fx+G$
$y{\text{'}}_p=6A{x}^5+5B{x}^4+4C{x}^3+3D{x}^2+2Ex+F$
$y^{\prime }{\text{'}}_p=30A{x}^4+20B{x}^3+12C{x}^2+6Dx+2E$

Plugging these into the original equation using the taylor series approximation as the excitation,
$(A{x}^6+B{x}^5+C{x}^4+D{x}^3+E{x}^2+Fx+G)+6(6A{x}^5+5B{x}^4+4C{x}^3+3D{x}^2+2Ex+F)+9(30A{x}^4+20B{x}^3+12C{x}^2+6Dx+2E)$
$=1-2x^2+2x^4/3-4x^6/45$
Rearranging the coefficients,
$A{x}^6+(B+36A)x^5+(C+30B+270A)x^4+(D+24C+180B)x^3+(E+18D+108C)x^2+(F+12E+54D)x+(G+6F+18E)$
$=1+0x-2x^2+0x^3+2x^4/3+0x^5-4x^6/45$


Equating x^6 coefficients, A=-45/4
Equating x^5 coefficients, B=405
Equating x^4 coefficients, C=-118461.833
Equating x^3 coefficients, D=2770183.992
Equating x^2 coefficients, E=-37069430.492
Equating x^1 coefficients, F=-594423101.472
Equating x^0 coefficients, G=4233788358.69
$y_p=-45/4x^6+405x^5+-118461.833x^4+2770183.992x^3-37069430.492x^2-594423101x+4233788358.69$

The graph shown is the taylor series for cos(2x) for the 0th through 3rd order.

File:Cos(2x) graph.png

Honor Pledge

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

Problem 7: Taylor Series Expansion of the log Function

Problem Statement

$y^{″}+6y^{\prime }+9y=log(3+4x)$
$y(0)=1,y^{\prime }(0)=.5$
Use the point
$\hat{x}=-1/2$
$\log x={\log }_{e}x=\ln x$

Solution

The taylor series expansion for $log(3+4x)$ around $\hat{x}=-1/2$ up to 16 terms is
$4(x+.5)-8(x+.5{)}^2+\frac{64}{3}(x+.5{)}^3-64(x+.5{)}^4+\frac{1024}{5}(x+.5{)}^5-\frac{2048}{3}(x+.5{)}^6+\frac{16384}{7}(x+.5{)}^7-8192(x+.5{)}^8$
$+\frac{262144}{9}(x+.5{)}^9-\frac{524288}{5}(x+.5{)}^{10}+\frac{4194304}{11}(x+.5{)}^{11}-\frac{4194304}{3}(x+.5{)}^{12}+\frac{67108864}{13}(x+.5{)}^{13}$
$-\frac{134217728}{7}(x+.5{)}^{14}+\frac{1073741824}{15}(x+.5{)}^{15}-\frac{268435456}{3}(x+.5{)}^{16}$
Plots of taylor series expansion: Up to order 4
File:Taylor1.JPG

Up to order 7
File:Taylor2.JPG

Up to order 11
File:Taylor3.JPG

Up to order 16
File:Taylor4.JPG

The visually estimated domain of convergence is from .8 to .2.
Now use the transformation of variable
$x⟶t\text{ such that }3+4x=1+t$
$x=\frac{t-2}{4}$

If $log(1+t)$ has a domain of convergence from $[-1,+1]$ then $log(3+4x)$ converges from $[\frac{-3}{4},\frac{-1}{4}]$

Honor Pledge

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