# University of Florida/Egm4313/s12.team15.r2

<Team 15
REPORT 2

## R2.1
### Given

$\lambda _{1}=-2\!$
$\lambda _{2}=+5\!$
$y(0)=1\!$
$y'(0)=0\!$
$r(x)=0\!$

### Find

Find the non-homogeneous linear 2nd order ordinary differential equation with constant coefficients in standard form and the solution in terms of the initial conditions and the general excitation $r(x)$.
Then plot the solution.

### Solution

Some of the following standard equations were referenced from the textbook Advanced Engineering Mechanics 10th Ed. Kreyszig 2011.
Standard Form:
$\lambda ^{2}+a\lambda +b=0\!$
$(\lambda -\lambda _{1})(\lambda -\lambda _{2})=0\!$
$\lambda ^{2}-\lambda \lambda _{2}-\lambda \lambda _{1}+\lambda _{1}\lambda _{2}=0\!$
$\lambda ^{2}-5\lambda +2\lambda -10=0\!$
$\lambda ^{2}-3\lambda -10=0\!$
Therfore,
$a=-3\!$
$b=-10\!$
Standard Form:
$y''+ay'+by=r(x)=0\!$
$y''-3y'-10y=0\!$
Standard Form:
$y(x)=y_{h}(x)+y_{p}(x)\!$
$y_{h}(x)=c_{1}e^{-2x}+c_{2}e^{5x}\!$
$y_{p}(x)=0\!$
Therefore,
$y_{p}'(x)=0\!$
$y(x)=c_{1}e^{-2x}+c_{2}e^{5x}\!$
$y'(x)=-2c_{1}e^{-2x}+5c_{2}e^{5x}\!$
If
$y(0)=1\!$
then
$1=c_{1}e^{-2(0)}+c_{2}e^{5(0)}\!$
$1=c_{1}+c_{2}\!$
If
$y'(0)=0\!$
then
$0=c_{1}e^{-2(0)}+c_{2}e^{5(0)}\!$
$0=c_{1}+c_{2}\!$
$2c_{1}=5c_{2}\!$
$c_{1}={\frac {5}{2}}c_{2}\!$
$1={\frac {5}{2}}c_{2}+c_{2}\!$
$1={\frac {7}{2}}c_{2}\!$
$c_{2}={\frac {2}{7}}\!$
$1=c_{1}+c_{2}\!$
$c_{1}=1-c_{2}=1-{\frac {2}{7}}={\frac {5}{7}}\!$
Final Solution:
$y(x)={\frac {5}{7}}e^{-2x}+{\frac {2}{7}}e^{5x}\!$
Plot of Solution:

### Author

Solved and typed by Kristin Howe
Reviewed By -

## R2.2
### Author

Solved and typed by -
Reviewed By -

## R2.3

Problems referenced from Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.59 problems.3-4

### Given

3a) $y''+6y'+8.96y=0\!$
3b) $y''+4y'+(\pi ^{2}+4)y=0\!$

### Find

Find the general solution of the ODEs.

### Solution

3a)
$y''+6y'+8.96y=0\!$
$a=6\!$
$b=8.96\!$
$\lambda ^{2}+a\lambda +b=0\!$
$\lambda ^{2}+6\lambda +8.96=0\!$
Using the quadratic formula:
$\lambda _{1/2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2}}\,\!$ where $\lambda _{1/2},\!$ are the 2 different roots generated using quadratic formula.
Plug-in and solve using known variables (b=6, a=1, c=8.96)
$\lambda _{1/2}={\frac {-6\pm {\sqrt {0.16}}}{2}}\,\!$ $={\frac {-6\pm 0.4}{2}}\,\!$ $=-3\pm 0.2\,\!$
Therefore,
$\lambda _{1}=-2.8\,\!$
$\lambda _{2}=-3.2\,\!$
The General Solution formula for Distinct Real roots is $y(x)=C_{1}e^{\lambda _{1}x}+C_{2}e^{\lambda _{2}x}$
Now substitute the known roots into the General Solution formula:
Check using substitution. First, find the 1st and 2nd derivative of general solution formula:
${y}'=-3.2C_{1}e^{-3.2x}-2.8C_{2}e^{-2.8x}\!$
${y}''=10.24C_{1}e^{-3.2x}+7.84C_{2}e^{-2.8x}\!$
Then, plug in the known variables into the original equation:
$y''+6y'+8.96y=0\!$ (Original Equation)
$10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x}-19.2c_{1}e^{-3.2x}-16.8c_{2}e^{-2.8x}+8.96c_{1}e^{-3.2x}+8.96c_{2}e^{-2.8x}=0\!$
By inspection, all of the terms on the left side of the equation cancel out to 0, making the expression correct. Therefore, the general solution is correct.
3b)
$y''+4y'+(\pi ^{2}+4)y=0\!$
$a=4\!$
$b=(\pi ^{2}+4)\!$
$\lambda ^{2}+a\lambda +b=0\!$
$\lambda ^{2}+4\lambda +(\pi ^{2}+4)=0\!$
Using the quadratic formula:
$\lambda _{1/2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2}}\,\!$ where $\lambda _{1/2},\!$ are the 2 different roots generated using quadratic formula.
Plug-in and solve using known variables (b=4, a=1, c=[$\pi ^{2}$ + 4])
$\lambda _{1/2}={\frac {-4\pm {\sqrt {-4\pi ^{2}}}}{2}}\,\!$ $={\frac {-4\pm 2\pi i}{2}}\,\!$ $=-2\pm \pi i\,\!$
Therefore,
$\lambda _{1}=-2+\pi i\,\!$
$\lambda _{2}=-2-\pi i\,\!$
The General Solution formula for Complex Conjugate roots is $y(x)=e^{-{\frac {\alpha x}{2}}}(C_{1}\cos \beta x+C_{2}\sin \beta x)\!$
Now substitute the known roots into the General Solution formula:
Check using substitution. First, find the 1st and 2nd derivative of general solution formula:
${y}'=e^{-2x}(-C_{1}\pi sin(\pi x)+C_{2}\pi cos(\pi x))-2e^{-2x}(C_{1}cos(\pi x)+C_{2}sin(\pi x))\!$
${y}''=e^{-2x}(-C_{1}\pi ^{2}cos(\pi x)-C_{2}\pi ^{2}sin(\pi x)+2C_{1}\pi sin(\pi x)-2C_{2}\pi sin(\pi x))-2e^{-2x}(-C_{1}\pi sin(\pi x)+C_{2}\pi cos(\pi x)-2C_{1}cos(\pi x)-2C_{2}sin(\pi x))\!$
Then, plug in the known variables into the original equation:
$y''+4y'+(\pi ^{2}+4)y=0\!$ (Original Equation)
$e^{-2x}(-C_{1}\pi ^{2}cos(\pi x)-C_{2}\pi ^{2}sin(\pi x)+2C_{1}\pi sin(\pi x)-2C_{2}\pi sin(\pi x))-2e^{-2x}(-C_{1}\pi sin(\pi x)+C_{2}\pi cos(\pi x)-2C_{1}cos(\pi x)-2C_{2}sin(\pi x))\!$
$+4(e^{-2x}(-C_{1}\pi sin(\pi x)+C_{2}\pi cos(\pi x))-2e^{-2x}(C_{1}cos(\pi x)+C_{2}sin(\pi x)))\!$
$+(\pi ^{2}+4)(e^{-2x}(C_{1}\cos \pi x+C_{2}\sin \pi x))=0\!$
By inspection, all of the terms on the left side of the equation cancel out to 0, making the expression correct. Therefore, the general solution is correct.

### Author

Solved and typed by - Tim Pham 19:36, 07 February 2012 (UTC)
Reviewed By -

## R2.4

The following problems are referenced from Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.59.
For additional information and/or practice reference Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.53-59 and the lecture notes Lecture 5.
* Note the lecture notes and Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 textbook were used to solve the following.

### Given

4a) $y''+2\piy'+\pi ^{2}y=0\!$
4b) $10y''-32y'+25.6y=0\!$

### Find

Find the general solution. Check answer by substitution.

### Solution

4a)
$y''+2\piy'+\pi ^{2}y=0\!$
$a=2\pi \!$
$b=\pi ^{2}\!$
$\lambda ^{2}+a\lambda +b=0\!$
$\lambda ^{2}+2\pi \lambda +\pi ^{2}=0\!$
Finding the roots:
$\lambda _{1}={\frac {1}{2}}(-2\pi +{\sqrt {(-2\pi )^{2}-4(\pi ^{2})}})=-\pi \!$
$\lambda _{2}={\frac {1}{2}}(-2\pi -{\sqrt {(-2\pi )^{2}-4(\pi ^{2})}})=-\pi \!$
Both solutions above show the double root characteristic so the following is applied:
$y_{1}=e^{-\pi {x}}\!$
$y_{2}=uy_{1}\!$
$u=c_{1}x+c_{2}\!$ This is due to double integration.
So,
$y_{2}=(c_{1}x+c_{2})y_{1}\!$
Choose $c_{1}=1c_{2}=0\!$
$y_{2}=(x+0)y_{1}\!$
$y_{2}=xy_{1}\!$
$y_{2}=xe^{-\pi {x}}\!$
$y=(c_{1}+c_{2}x)e^{-\pi {x}}\!$ General Solution
Check by substitution
$y=(c_{1}+c_{2}x)e^{-\pi {x}}\!$
$y'=-\pic_{1}e^{-\pi {x}}+c_{2}e^{-\pi {x}}-\pic_{2}e^{-\pi {x}}\!$
$y''=\pi ^{2}c_{1}e^{-\pi {x}}-\pic_{2}e^{-\pi {x}}+\pi ^{2}c_{2}xe^{-\pi {x}}-\pic_{2}e^{-\pi {x}}\!$
$\pi ^{2}c_{1}e^{-\pi {x}}-\pic_{2}e^{-\pi {x}}+\pi ^{2}c_{2}xe^{-\pi {x}}-\pic_{2}e^{-\pi {x}}+2\pi(-\pic_{1}e^{-\pi {x}}+c_{2}e^{-\pi {x}}-\pic_{2}e^{-\pi {x}})+\pi ^{2}(e^{-\pi {x}}c_{1}+e^{-\pi {x}}xc_{2})=0\!$
Since all terms cancel out then by substitution the general solution $y=(c_{1}+c_{2}x)e^{-\pi {x}}\!$ does hold true.
4b) $10y''-32y'+25.6y=0\!$
Divide by 10 to simplify equation :
$y''-3.2y'+2.56y=0\!$
$a=-3.2\!$
$b=2.56\!$
$\lambda ^{2}+a\lambda +b=0\!$
$\lambda ^{2}-3.2\lambda +2.56=0\!$
Solving for the roots:
$\lambda _{1}={\frac {1}{2}}(-(-3.2)+{\sqrt {(-3.2)^{2}-4(2.56)}})=1.6\!$
$\lambda _{2}={\frac {1}{2}}(-(-3.2)-{\sqrt {(-3.2)^{2}-4(2.56)}})=1.6\!$
$\lambda _{1}=\lambda _{2}\!$ This shows the double root characteristic.
So, the following is applied
$y_{1}=e^{1.6x}\!$
$y_{2}=uy_{1}\!$
$u=c_{1}x+c_{2}\!$ This is due to double integration
So
$y_{2}=(c_{1}x+c_{2})y_{1}\!$
Choose $c_{1}=1c_{2}=0\!$
$y_{2}=(x+0)y_{1}\!$
$y_{2}=xy_{1}\!$
$y_{2}=xe^{1.6x}\!$
$y=(c_{1}+c_{2}x)e^{1.6x}\!$ General Solution
Check by Substitution
$y''=2.56c_{1}e^{1.6x}+1.6c_{2}e^{1.6x}+1.6c_{2}xe^{1.6x}+2.56c_{2}xe^{1.6x}\!$
$y'=1.6c_{1}e^{1.6x}+c_{2}e^{1.6x}+1.6c_{2}xe^{1.6x}\!$
$y=c_{1}e^{1.6x}+c_{2}xe^{1.6x}\!$
$10(2.56c_{1}e^{1.6x}+1.6c_{2}e^{1.6x}+1.6c_{2}xe^{1.6x}+2.56c_{2}xe^{1.6x})-32(1.6c_{1}e^{1.6x}+c_{2}e^{1.6x}+1.6c_{2}xe^{1.6x})+25.6(c_{1}e^{1.6x}+c_{2}xe^{1.6x})=0\!$
Since all terms cancel out then through substitution it can be confirmed that the general solution
$y=(c_{1}+c_{2}x)e^{1.6x}\!$ does hold true.

### Author

Solved and typed by - Cynthia Hernandez
Reviewed By -

## R2.5
### Given

Problem 16, P 59 Kreyszig:
$e^{2.6x},e^{-4.3x}\!$
Problem 17, P 59 Kreyszig:
$e^{-{\sqrt {5}}x},xe^{-{\sqrt {5}}x}\!$

### Find

Problem 16, P 59 Kreyszig:
ODE in the form:
${y}''+a{y}'+by=0\!$
Problem 17, P 59 Kreyszig:
ODE in the form:
${y}''+a{y}'+by=0\!$

### Solution

Problem 16, P 59 Kreyszig:
For two distinct real-roots:
$y=c_{1}e^{\lambda _{1}x}+c_{2}e^{\lambda _{2}x}\!$
Where:
$\lambda _{1}=2.6\!$ and $\lambda _{2}=-4.3\!$
The characteristic equation is:
$\lambda ^{2}+a\lambda +b=0\!$
Multiplying the roots of the equation together:
$(\lambda -2.6)(\lambda +4.3)=\lambda ^{2}+1.7\lambda -11.18\!$
Therefore the ODE is:
Problem 17, P 59 Kreyszig:
For real double root:
$y=(c_{1}+c_{2}x)e^{\lambda x}\!$
Where:
$\lambda =-{\sqrt {5}}\!$
The characteristic equation is:
$\lambda ^{2}+a\lambda +b=0\!$
Multiplying the roots of the equation together:
$(\lambda +{\sqrt {5}})^{2}=\lambda ^{2}+2{\sqrt {5}}\lambda +5\!$
Therefore the ODE is:

### Author

Solved and typed by - Neil Tidwell 1:55, 06 February 2012 (UTC)
Reviewed By -

## R2.6
### Given

Spring-Dashpot-Mass System from Sec 1 lecture notes.
Has double real root $\lambda =-3\!$

### Find

$k,c,m\!$

### Solution

For spring-mass-dashpot system:
$m(y_{k}''+{\frac {k}{c}}y_{k}')+ky_{k}=f(t)\!$
This simplifies to:
$y_{k}''+{\frac {k}{mc}}y_{k}'+{\frac {k}{m}}y_{k}=f(t)\!$
Multiply roots together to get characteristic equation:
$(\lambda +3)^{2}=\lambda ^{2}+6\lambda +9\!$
Set coefficients of simplified spring-mass-dashpot equation equal to those of characteristic equation:
${\frac {k}{mc}}=6\!$
${\frac {k}{m}}=9\!$
Have one degree of freedom, so arbitrarily choose $m=1\!$
Solve system:

### Author

Solved and typed by - Neil Tidwell 2:14, 06 February 2012 (UTC)
Reviewed By -

## R2.7
### Given

Develop the MacLaurin Series (Taylor Series at $t=0\!$) for $e^{t},\cos t,\sin t\!$
$L[y]andL[w]thenL[y+w]\!$
$L[y+w]=L[y]+L[w]\!$
$L[cy]=cL[y]$
$L[kw]=kL[w]\!$

### Author

Solved and typed by - Jenny Schulze
Reviewed By -

## R2.8
### Given

Find a general solution to the equations, and check answers by substitution.
Problem 8
$y''+y'+3.25y=0\!$
(8.0)
Problem 15
$y''+0.54y'+(0.0729+\pi )y=0\!$
(8.1)

### Find

Factor as in the text and solve.

### Author

Solved and typed by - Jenny Schulze
Reviewed By -

## R2.9
### Given

Initial Conditions:
$y(0)=1,y'(0)=0\$
No excitation:
$r(x)=0\$

### Find

Find and plot the solution for the L2-ODE-CC corresponding to:
$\lambda ^{2}+4\lambda +13=0\!$
In another Fig., superimpose 3 Figs.: (a) this Fig., (b) the Fig. in R2.6 p.5-6, (c) the Fig. in R2.1 p.3-7.

### Solution

The corresponding ODE in standard form:
$y''+4y'+13y=r(x)\!$
To find the roots of the ODE, convert ODE to this form:
$a\lambda ^{2}+b\lambda +c=0\!$
which in this case is:
$\lambda ^{2}+4\lambda +13=0\!$
use the formula to find the roots:
$\lambda _{1,2}={\frac {-b\pm i{\sqrt {b^{2}-4ac}}}{2a}}=-\alpha \pm i\omega$
$\lambda _{1,2}={\frac {-4\pm i{\sqrt {4^{2}-4*1*13}}}{2*1}}=-2\pm 3i$
because the roots are complex conjugates, the general solution can be given as:
$y(t)=e^{-\alpha t}(Acos\omega t+bsin\omega t)=e^{-2t}(Acos3t+Bsin3t)\!$
to find A and B, we use the initial conditions and take the first derivative of y(t):
$y(0)=1=e^{-2*0}(Acos(3*0)+Bsin(3*0)=1*A=A\!$
$A=1\!$
$y'(t)=e^{-2t}((-2A+3B)cos(3t)-(3A+2B)sin(3t))\!$
$y'(0)=0=e^{-2*0}((-2*1+3B)cos(3*0)-(3*1+2B)sin(3*0))=1*(-2+3B)=-2+3B\!$
$B={\frac {2}{3}}\!$
thus:
$y(t)=e^{-2t}(cos3t+{\frac {2}{3}}sin3t)\!$
The graph of the solved equation is shown below in Fig. 1.
The three graphed equations are shown together below in Fig. 2
Equation 1 (R2.9):
$y_{1}(t)=e^{-2t}(cos3t+{\frac {2}{3}}sin3t)\!$
Equation 2 (R2.6):
$y_{2}(t)=e^{-3t}+3te^{-3t}\!$
Equation 3 (R2.1):
$y_{3}(t)={\frac {5}{7}}e^{-2t}+{\frac {2}{7}}e^{5t}\!$

### Author

Solved and typed by - --James Moncrief 01:42, 8 February 2012 (UTC)
Reviewed By -

### Contributing Members

Team Contribution Table :
| Problem Number | Lecture | Assigned To | Solved By | Typed By | Proofread By |
| 2.1 | Lecture 3 | Kristin Howe | Kristin Howe | Kristin Howe | Name |
| 2.2 | Lecture 5 | Jerry Shugart | Jerry Shugart | Jerry Shugart | Name |
| 2.3 | Lecture 5 | Tim Pham | Tim Pham | Tim Pham | Name |
| 2.4 | Lecture 5 | Cynthia Hernandez | Cynthia Hernandez | Cynthia Hernandez | Name |
| 2.5 | Lecture 5 | Neil Tidwell | Neil Tidwell | Neil Tidwell | Name |
| 2.6 | Lecture 5 | Neil Tidwell | Neil Tidwell | Neil Tidwell | Name |
| 2.7 | [Lecture link] | Jenny Schulze | Jenny Schulze | Jenny Schulze | Name |
| 2.8 | [Lecture link] | Jenny Schulze | Jenny Schulze | Jenny Schulze | Name |
| 2.9 | Lecture 6 | James Moncrief | James Moncrief | James Moncrief | Name |