View source for Intermediate engineering analysis/Problem set 2

== Problem R2.1 ==
<br>
===Part 1===
<br>
==== Problem Statement ====
<br>
Given the two roots and the initial conditions:<br>
<br>
<math>\lambda_{1}=-2,\lambda_{2}=5 \!</math><br>
<math>y(0)=1,y'(0)=0\!</math><br>
<br>
Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation <math>r(x)\!</math>.<br>
Consider no excitation: <br>
<math> r(x)=0\!</math><br>
Plot the solution<br>

==== Solution ====
<br>
=====Characteristic Equation:=====
<br>
<math> (\lambda-\lambda_1)(\lambda-\lambda_2)=0 \!</math> <br>
<math> (\lambda+2)(\lambda-5)=\lambda^2 +2\lambda-5\lambda-10=0\!</math><br>
  <math>\lambda^2 -3\lambda-10=0\!</math><br>
<br>
=====Non-Homogeneous L2-ODE-CC=====
<br>
  <math>y''-3y'-10=r(x)\!</math><br>
=====Homogeneous Solution:=====
<br>
<math> y_{h}(x)=c_1e^{-2x}+c_2e^{5x}\!</math><br>
<math> y(x)=c_1e^{-2x}+c_2e^{5x}+y_p(x)\!</math><br>
Since there is no excitation, <br>
<math>y_p(x)=0\!</math><br>
  <math> y(x)=c_1e^{-2x}+c_2e^{5x}\!</math><br>
=====Substituting the given initial conditions:=====
<br>
<math> y(0)=1\!</math><br>
  <math>1=c_1+c_2\!</math><br>
<math>y'(0)=0\!</math><br>
  <math>0=-2c_1+5c_2\!</math><br>
Solving these two equations for <math> c_1\!</math> and <math> c_2\!</math> yields:<br>
  <math> c_1=5/4, c_2=-1/4 \!</math>
=====Final Solution=====
<br>
  <math> y(x)=(5/4)e^{-2x}-(1/4)e^{5x}\!</math><br>
[[File:2.1fig1.jpg]]
===Part 2===
<br>
==== Problem Statement ====
<br>
Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the 2 values in (3a) p.3-7 as the 2 roots of the corresponding characteristic equation. <br>
====Solutions====
<br>
<math> 2(\lambda+2)(\lambda-5)=2\lambda^2 +4\lambda-10\lambda-20=0\!</math><br>
  <math>2\lambda^2 -6\lambda-20=0\!</math><br>
<br>
<math> 3(\lambda+2)(\lambda-5)=3\lambda^2 +6\lambda-15\lambda-30=0\!</math><br>
  <math>3\lambda^2 -9\lambda-30=0\!</math><br>
<br>
<math> 4(\lambda+2)(\lambda-5)=4\lambda^2 +8\lambda-20\lambda-40=0\!</math><br>
  <math>4\lambda^2 -12\lambda-40=0\!</math><br>
<br>
--[[User:Egm4313.s12.team11.gooding|Egm4313.s12.team11.gooding]] 02:01, 7 February 2012 (UTC)
==Report 2, Problem 2==
====Problem Statement====
Find and plot the solution for the homogeneous L2-ODE-CC 
:<math>y''(x) - 10y'(x) + 25y(x)=0\!</math>
with initial conditions <math>y(0)=1\!</math> ,and   <math>y'(0)=0\!</math>
====Characteristic Equation====
<math>\lambda^2-10\lambda+25=0\!</math>

<math>(\lambda-5)(\lambda-5)=0\!</math>

<math>\lambda=5\!</math>
====Homogeneous Solution====
The solution to a L2-ODE-CC with real double root is given by
 
<math>y(x)=c_1e^{\lambda x}+c_2xe^{\lambda x}\!</math>

First initial condition

<math>y(0)=1\!</math> 

<math>y(0)=c_1e^{5*0}+c_2*0*e^{5*0}=1\!</math>

<math>c_1=1\!</math>

Second initial condition 

<math>y'(0)=0\!</math>

<math>\frac{d}{dx}y(x)=y'(x)=5e^{5x}+c_2e^{5x}(5x+1)\!</math>

<math>y'(0)=5e^{5*0}+c_2e^{5*0}(5*0+1)=0\!</math>

<math>5+c_2=0\!</math>

<math>c_2=-5\!</math>

The solution to our L2-ODE-CC is 

                        <math>y(x)=e^{5x}(1-5x)\!</math>
====Plot====
<math>y(x)=e^{5x}(1-5x)\!</math>

[[File:PlotR2 2.jpg]]

[[User:Egm4313.s12.team11.imponenti|Egm4313.s12.team11.imponenti]] 00:30, 8 February 2012 (UTC)

==Problem R2.3==

===Problem Statement (K 2011 p.59 pb. 3)===

Find a general solution. Check your answer by substitution.

===Given===

<math> y''+6y'+8.96y=0 \! </math>

===Solution===

We can write the above differential equation in the following form:

<math> \frac{d^2y}{dx^2}+6\frac{dy}{dx}+8.96y=0 </math>

Let <math> \frac{d}{dx} = \lambda. </math>

The characteristic equation of the given DE is

<math> \lambda^2+6\lambda+8.96=0 \! </math>

Now, in order to solve for <math> \lambda </math>, we can use the quadratic formula:

<math> \lambda=\frac{-(6)\pm \sqrt{(6)^2-4(1)(8.96)}}{2(1)} </math>

<math> \lambda=\frac{-6\pm \sqrt{36-35.84}}{2} </math>

<math> \lambda=\frac{-6\pm \sqrt{0.16}}{2} </math>

<math> \lambda=\frac{-6\pm 0.4}{2} </math>

Therefore, we have:

<math> \lambda_1=\frac{-6 + 0.4}{2}=\frac{-5.8}{2}=-2.8 </math>

and

<math> \lambda_1=\frac{-6 - 0.4}{2}=\frac{-6.4}{2}=-3.2 </math>

Thus, we have found that the general solution of the DE is actually:

<math> y=c_1 e^{-2.8x} + c_2 e^{-3.2x} \! </math>

'''Check:'''

To check if <math> y </math> is indeed the solution of the given DE, we can differentiate the 

what we found to be the general solution.

<math> \frac{dy}{dx}=y'=-2.8x c_1 e^{-2.8x} + -3.2x c_2 e^{-3.2x} \! </math>

<math> \frac{d^2y}{dx^2}=y''=7.84x c_1 e^{-2.8x} + 10.24 c_2 e^{-3.2x} \! </math>

Substituting the values of <math> y, y' </math> and <math> y'' </math> in the given equation, we get:

<math> 7.84c_1e^{-2.8x}+10.24c_2e^{-3.2x}+6(-2.8c_1e^{-2.8x}-3.2c_2e^{-3.2x})+ 8.96(c_1e^{-2.8x}+c_2e^{-3.2x}) = 0 \! </math>

<math> 7.84c_1e^{-2.8x}+10.24c_2e^{-3.2x}-16.8c_1e^{-2.8x}-19.2c_2e^{-3.2x}+ 8.96c_1e^{-2.8x}+8.96c_2e^{-3.2x} = 0 \! </math>

and thus:

<math> 0 \equiv 0. \! </math>

Therefore, the solution of the given DE is in fact:

<br> 
<math> y=c_1 e^{-2.8x} + c_2 e^{-3.2x} \! </math> 
<br>

===Problem Statement (K 2011 p.59 pb. 4)===

Find a general solution to the given ODE. Check your answer by substituting into the original equation.

===Given===

<math> {y}''+ 4{y}'+(\pi^2+4)y = 0\! </math>

===Solution===

The characteristic equation of this ODE is therefore:

<math>\lambda^2 + a\lambda  + b = \lambda^2 + 4\lambda  + (\pi^2+4 ) = 0\!</math>

Evaluating the discriminant: <br>

<br>

<math> a^2 - 4b = 4^2 - 4(\pi^2 + 4) = -4\pi^2 < 0\!</math><br>

<br>

Therefore the equation has two complex conjugate roots and a general homogenous solution of the form:

<br>

<br>

<math> y = e^{-ax/2}(Acos(\omega x) + Bsin(\omega x))\!</math><br>

<br>

Where: <math> \omega = \sqrt{b - \frac{1}{4}a^2} = \sqrt{\pi^2 + 4 - \frac{1}{4}(4)^2} = \sqrt{\pi^2} = \pi\!</math> <br 

/><br>

And finally we find the general homogenous solution:<br>

<br>
                                                   <math> y = e^{-2x}(Acos(\pi x )+ Bsin(\pi x))\!</math>
<br>

'''Check:'''<br><br>

We found that: <br>

<br>

<math> y = e^{-2x}(Acos(\pi x )+ Bsin(\pi x))\!</math> <br><br>

Differentiating <math> y\!</math> to obtain <math>{y}'\!</math> and <math>{y}''\!</math> respectively: <br> <br>

<math> {y}' = -2e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + e^{-2x}(\pi Acos(\pi x)-\pi Bsin(\pi x))\!</math> <br>
                                    <math> {y}' = -2e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + \pi e^{-2x}( Acos(\pi x)- Bsin(\pi x))\!</math>

<br>

<math> {y}'' = 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -2 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x))- 2 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!</math>

<br>
                  <math> {y}'' = 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -4 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!</math>
<br>

Substituting these equations into the original ODE yields: <br>
<br>
<math> 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -4 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!</math> <br>

<math> + 4(-2e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + \pi e^{-2x}( Acos(\pi x)- Bsin(\pi x))) + (\pi^2 + 4)(e^{-2x}(Acos(\pi x )+ Bsin(\pi x))) = 0\!</math> <br>
<br>

<math> 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -4 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!</math> <br>

<math> -8e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + 4\pi e^{-2x}( Acos(\pi x)- Bsin(\pi x)) + \pi^2 e^{-2x}(Acos(\pi x )+ Bsin(\pi x)) + 4e^{-2x}(Acos(\pi x )+ Bsin(\pi x)) = 0\!</math> <br> <br>
<math> (4-8+4)e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + (-4+4) \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) + (\pi^2-\pi^2) e^{-2x}(Acos(\pi x)+Bsin(\pi x)) = 0\!</math> <br> <br>

<math> 0 \equiv 0 \!</math> <br>
<br>
Therefore, the solution is correct.

--[[Egm4313.s12.team11.perez.gp|Gonzalo Perez]]
==Problem R2.4==
=== K 2011 p.59 pb. 5 ===
===== Problem Statement =====
Find a general solution to the given ODE. Check your answer by substituting into the original equation.<br><br>

===== Given =====
<math>{y}''+ 2\pi{y}'+\pi^2y = 0\!</math> <br>
<br>

===== Solution =====
The characteristic equation of this ODE is therefore:<br>
<br>
<math>\lambda^2 + a\lambda  + b = \lambda^2 + 2\pi\lambda  + \pi^2= 0\!</math><br>
<br>
Evaluating the discriminant:<br>
<br>
<math> a^2 - 4b = (2\pi)^2 - 4(\pi^2) = 4\pi^2 - 4\pi^2 = 0\!</math><br>
<br>
Therefore the equation has a real double root and a general homogenous solution of the form:
<br>
<br>
<math> y = (c_{1}+c_{2}x)e^{-ax/2}\!</math>  <ref>Kreyszig 2011, p.54-57.</ref><br>
<br>
And finally we find the general homogenous solution:<br>
<br>
                                                        <math> y = (c_{1}+c_{2}x)e^{-\pi x}\!</math>
<br>
<big>'''Checking:'''</big><br><br>
We found that: <br>
<br>
<math> y = (c_{1}+c_{2}x)e^{-\pi x}\!</math> <br><br>
Differentiating <math> y\!</math> to obtain <math>{y}'\!</math> and <math>{y}''\!</math> respectively: <br> <br>
                                                   <math> {y}' = c_{2}e^{-\pi x} - \pi (c_{1}+c_{2}x)e^{-\pi x}\!</math>
<br>
And,  <br>
<br>
<math> {y}'' =  -\pi c_{2} e^{-\pi x}- \pi c_{2}e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x}\!</math>
<br>
                                                  <math> {y}'' = -2\pi c_{2} e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x}\!</math>
<br>
Substituting these equations into the original ODE yields: <br>
<br>
<math> -2\pi c_{2} e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x} + 2\pi(c_{2}e^{-\pi x} - \pi (c_{1}+c_{2}x)e^{-\pi x}) + \pi^2((c_{1}+c_{2}x)e^{-\pi x}) = 0\!</math> <br>
<br>
<math> -2\pi c_{2} e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x} + 2\pi c_{2}e^{-\pi x} - 2\pi^2 (c_{1}+c_{2}x)e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x} = 0\!</math> <br>
<br>
<math> (-2\pi + 2\pi) c_{2} e^{-\pi x} + (\pi^2 - 2\pi^2 + \pi^2)(c_{1}+c_{2}x)e^{-\pi x}  = 0\!</math> <br> <br>

<math> 0 \equiv 0 \!</math> <br>
<br>
Therefore this solution is correct. <br>
<references/>

=== K 2011 p.59 pb. 6 ===
===== Problem Statement =====
Find a general solution to the given ODE. Check your answer by substituting into the original equation.<br><br>

===== Given =====
<math>10{y}''-32{y}'+25.6y = 0\!</math> <br>
<br>

===== Solution =====
Initially we modify the original ODE to put it in the form of a second-order homogenous linear ODE with constant coefficients:<br>
<br>
<math>10{y}''-32{y}'+25.6y = 0\!</math> <br> <br>
Dividing both sides by 10: <br> <br>
<math>{y}''-3.2{y}+2.56y = 0\!</math> <br> <br>
The characteristic equation of this ODE is now therefore:<br>
<br>
<math>\lambda^2 + a\lambda  + b = \lambda^2 -3.2\lambda  + 2.56= 0\!</math><br>
<br>
Evaluating the discriminant:<br>
<br>
<math> a^2 - 4b = (-3.2)^2 - 4(2.56) = 10.24 - 10.24 = 0\!</math><br>
<br>
Therefore the equation has a real double root and a general homogenous solution of the form:
<br>
<br>
<math> y = (c_{1}+c_{2}x)e^{-ax/2}\!</math>  <ref>Kreyszig 2011, p.54-57.</ref><br>
<br>
And finally we find the general homogenous solution:<br>
<br>
                                                        <math> y = (c_{1}+c_{2}x)e^{1.6 x}\!</math>
<br>
<big>'''Checking:'''</big><br><br>
We found that: <br>
<br>
<math> y = (c_{1}+c_{2}x)e^{1.6x}\!</math> <br><br>
Differentiating <math> y\!</math> to obtain <math>{y}'\!</math> and <math>{y}''\!</math> respectively: <br> <br>
                                                   <math> {y}' = 1.6c_{1}e^{1.6x}+c_{2}e^{1.6x}+1.6c_{2}xe^{1.6x}\!</math>
<br>
And,  <br>
<br>
<math> {y}'' =  (1.6)^2 c_{1}e^{1.6x}+1.6c_{2}e^{1.6x}+1.6c_{2}e^{1.6x}+(1.6)^2c_{2}xe^{1.6x}\!</math>
<br>
                                                  <math> {y}'' = (1.6)^2 c_{1}e^{1.6x}+3.2c_{2}e^{1.6x}+(1.6)^2c_{2}xe^{1.6x}\!</math>
<br>
Substituting these equations into the original ODE yields: <br>
<br>
<math> 10[(1.6)^2 c_{1}e^{1.6x}+3.2c_{2}e^{1.6x}+(1.6)^2c_{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}+c_{2}x)e^{1.6 x}] = 0\!</math> <br>
<br>
<math> 25.6 c_{1}e^{1.6x}+32c_{2}e^{1.6x}+25.6c_{2}xe^{1.6x} -51.2c_{1}e^{1.6x}-32c_{2}e^{1.6x}-51.2c_{2}xe^{1.6x}+25.6c_{1}e^{1.6 x}+25.6c_{2}xe^{1.6 x} = 0\!</math> <br>
<br>
<math> (25.6-51.2+25.6) c_{1} e^{1.6x} + (32-32)c_{2}e^{1.6x} + (25.6-51.2+25.6)c_{2}xe^{1.6x}  = 0\!</math> <br> <br>

<math> 0 \equiv 0 \!</math> <br>
<br>
Therefore this solution is correct. <br>
<references/>

==Report 2, Problem 2.5==
=== Problem Statement ===
'''<u>Problem 2.5</u>''' Find an Ordinary Differential Equation.
Use <math>y''+ay'+by=0\!</math> for the given basis.

===ODE Solutions===
Two Real Roots: <math>y(x)=c_{1}e^{\lambda_{1}x}+c_{2}e^{\lambda_{2}x}\!</math></br>
Double Real Roots: <math>y(x)=c_{1}e^{\lambda_{1}x}+c_{2}xe^{\lambda_{2}x}\!</math>

===Reverse Engineering===
<math>e^{\lambda x}(\lambda^2+a\lambda+b)=0\!</math></br>
</br>
'''Case 1:''' <math>a^2-4b>0\!</math></br>
<math>\lambda^2-(\lambda_{1}+\lambda_{2})\lambda+\lambda_{1}\lambda_{2}\!</math></br>
<math>y''-(\lambda_{1}+\lambda_{2})y'+\lambda_{1}\lambda_{2}y=0\!</math></br>
</br>
'''Case 2:''' <math>a^2-4b=0\!</math></br>
<math>\lambda^2-(\lambda_{1}+\lambda_{2})\lambda+\lambda_{1}\lambda_{2}\!</math></br>
<math>y''-(\lambda_{1}+\lambda_{2})y'+\lambda_{1}\lambda_{2}y=0\!</math>

===Problem Solutions===
====R2.5 K2011 p.59 pbs.16====
<math>e^{2.6x}, e^{-4.3x}\!</math></br>
<math>(\lambda-2.6)(\lambda+4.3)\!</math></br>
<math>\lambda^2+1.7\lambda-11.18\!</math></br>


'''ODE Form:'''
                                                             <math>y''+1.7y'-11.18=0\!</math></br>

====R2.5 K2011 p.59 pbs.17====
<math>e^{-\sqrt{5}x}, xe^{-\sqrt{5}x}\!</math></br>
<math>(\lambda+\sqrt{5})^2\!</math></br>
<math>\lambda^2+2\sqrt{5}\lambda+5\!</math></br>


'''ODE Form:'''
                                                             <math>y''+2\sqrt{5}y'+5y=0\!</math></br>


Created by [[http://en.wikiversity.org/wiki/User:Egm4313.s12.team11.Suh| Daniel Suh]] 20:57, 7 February 2012 (UTC)

== Problem 2.6 ==
Solved by: [[user:Egm4313.s12.team11.vargas.aa | Andrea Vargas]]
===Problem Statement ===
For the following spring-dashpot-mass system (in series) find the values for the parameters <math> k, c ,m \!</math> knowing that the system has the double real root <math> \lambda = -3 \!</math>
==== Figure ====
[[File:R2.6.jpg]]
===Solution===
Previously, we have derived the following equation for such a system:<br>
(From [[media:iea.s12.sec1.djvu|Sec 1 (d)]], (3) p.1-5)<br>
<math> m\left ( y_k''+\frac{k}{c}y_k' \right )+ ky_k=f(t)\!</math><br>
We can write this equation in standard form by diving through by <math> m \!</math>:<br>
<math> y_k''+\frac{k}{c}y_k'+ \frac{k}{m}y_k=\frac{f(t)}{m} \!</math><br>
Here, we can take the coefficients of <math>y_k'\!</math> and  <math>y_k\!</math> as <math>a\!</math> and <math>b\!</math>:<br>

<math> y_k''+\underbrace{\frac{k}{c}}_ay_k'+ \underbrace{\frac{k}{m}}_by_k=\frac{f(t)}{m} \!</math><br>


Next,considering the double real root:<br>
<math> \lambda = -3 \!</math><br>
We can find the characteristic equation to be:<br>
<math> \left ( \lambda +3 \right )^{2} = \lambda^{2}+6\lambda+9=0 \!</math><br>
Which is in the form:<br>
<math>\lambda^{2}+a\lambda+b=0\!</math><br><br>
Then, we know that <math>a=6\!</math> and <math>b=9\!</math>:<br>

Setting <math>a\!</math> and <math>b\!</math> from the first equation equal to these, we obtain:
<br>

                                                              <math>\frac{k}{c}=6 \; and \;  \frac{k}{m}=9\!</math>
<br>


Clearly, there is an infinite amount of solutions to this problem because we have 2 equations but 3 unknowns. This can be solved by fixing one of the values and finding the other two.

====Example of Solution====
An example of fixing one of the constants to find the other two is provided here. By solving the simple equations above, we can illustrate how to find <math> k, c ,m \!</math> . We had:<br>
<math> \frac{k}{c}=6 \; and \;  \frac{k}{m}=9\!</math><br>

If we fix the mass to <math> m=10 \; kg \!</math>. We find:<br>
<math>\frac{k}{10}=9\!</math><br><br>

<math>k=90\!</math><br>

Then,<br>
<math>\frac{90}{c}=6\!</math><br><br>

<math>c=15\!</math><br>

Finally, we obtain:
<br>

                                                              <math>m=10, \; k=90, \; and \; c=15 \!</math>
<br>
--[[User:Egm4313.s12.team11.vargas.aa|Andrea Vargas]] 21:44, 7 February 2012 (UTC)


==Problem 2.7: McLaurin Series==

====Problem Statement====

:Develop the McLaurin Series (Taylor Series at t=0) for <math> e^t , \cos t , \sin t \!</math>

====Solution====

:<math> e^t= \sum_{n=0}^{\infty }\frac{t^n}{n!} \!</math>

::<math>=\frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+...\frac{t^n}{n!} \!</math>

                                       <math>e^t=1+t+\frac{t^2}{4}+\frac{t^3}{6}+...\frac{t^n}{n!}\!</math>


:<math> \cos t= \sum_{k=0}^{\infty }\frac{(-1)^kt^{2k}}{(2k)!} \!</math>

::<math>=\frac{(-1)^0t^{2(0)}}{(2(0))!}+\frac{(-1)^1t^{2(1)}}{(2(1))!}+\frac{(-1)^2t^{2(2)}}{(2(2))!}+\frac{(-1)^3t^{2(3)}}{(2(3))!}+...\frac{(-1)^kt^{2k}}{(2k)!} \!</math>

                                       <math>\cos t=1-\frac{t^2}{2}+\frac{t^4}{24}+\frac{t^6}{720}...\frac{(-1)t^{2k}}{(2k)!} \!</math>


:<math> \sin t= \sum_{k=0}^{\infty }\frac{(-1)^kt^{2k+1}}{(2k+1)!} \!</math>

::<math>=\frac{(-1)^0t^{2(0)+1}}{(2(0)+1)!}+\frac{(-1)^1t^{2(1)+1}}{(2(1)+1)!}+\frac{(-1)^2t^{2(2)+1}}{(2(2)+1)!}+\frac{(-1)^3t^{2(3)+1}}{(2(3)+1)!}+...\frac{(-1)^kt^{2k+1}}{(2k+1)!} \!</math>

                                       <math>\sin t=t-\frac{t^3}{6}+\frac{t^5}{120}-\frac{t^7}{5040}+...\frac{(-1)^kt^{2k+1}}{(2k+1)!} \!</math>

--[[User:Egm4313.s12.team11.arrieta|Egm4313.s12.team11.arrieta]] 17:07, 6 February 2012 (UTC)



== Problem R2.8 ==
<br>
====Problem Statement====
<br>
Find a general solution. Check your answer by substitution.<br>

===Problem 8===
<br>
<math>y''+ y'+3.25y=0\!</math><br>

Let:<br>
<math> \lambda=d/dx\!</math><br>
=====Characteristic Equation=====
<br>
<math> \lambda^2+\lambda+3.25=0\!</math><br>
Using the quadratic equation to find roots we get:<br>
<math>\lambda_1=\frac{-1+i\sqrt(12)}{2}\!</math><br>
<math>\lambda_2=\frac{-1-i\sqrt(12)}{2}\!</math><br>
Therefore:<br>
 <math>y_h(x)=e^{-\frac{1}{2}x}(c_1\cos(x\sqrt{3})+c_2\sin(x\sqrt{3}))\!</math><br>

=====Check By Substitution=====
<br>
<math>y'(x)=-\frac{1}{2}e^{-\frac{1}{2}x}(c_1\cos(x\sqrt{3})+c_2\sin(x\sqrt{3})+e^{-\frac{1}{2}x}(-\sqrt{3}c_1\sin{\sqrt{3}x}+\sqrt{3}c_1\cos{\sqrt{3}x})\!</math><br>
<math>y''(x)=\frac{1}{4}e^{-\frac{1}{2}x}(c_1\cos(x\sqrt{3})+c_2\sin(x\sqrt{3})-\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{3}c_1\sin{\sqrt{3}x}+\sqrt{3}c_1\cos{\sqrt{3}x})-\!</math><br>
<math>\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{3}c_1\sin{\sqrt{3}x}+\sqrt{3}c_1\cos{\sqrt{3}x})e^{-\frac{1}{2}x}(-3c_1\cos(x\sqrt{3})-3c_2\sin(x\sqrt{3})\!</math><br>
Substituting <math>y,y',y''\!</math> into the original equation, the result is <br>
  <math>y''+ y'+3.25y=0\!</math><br>
===Problem 15===
<br>
<math>y''+0.54y'+(0.0729+\pi)y=0\!</math><br>
Let: <math>\frac{d}{dx}=\lambda\!</math><br>
====Characteristic Equation====
<br>
<math> \lambda^2+0.54\lambda+(0.0729+\pi)=0\!</math><br>
Using the quadratic equation to find roots we get:<br>
<math>\lambda_1=\frac{-0.27+i\sqrt(\pi)}{2}\!</math><br>
<math>\lambda_2=\frac{-0.27-i\sqrt(\pi)}{2}\!</math><br>
Therefore:<br>
 <math>y_h(x)=e^{-0.27x}(c_1\cos(x\sqrt{\pi})+c_2\sin(x\sqrt{\pi})\!</math><br>
=====Check By Substitution=====
<br>
<math>y'(x)=-0.27e^{-0.27x}(c_1\cos(x\sqrt{\pi})+c_2\sin(x\sqrt{\pi})+e^{-0.27x}(-\sqrt{\pi}c_1\sin{\sqrt{\pi}x}+\sqrt{\pi}c_1\cos{\sqrt{\pi}x})\!</math><br>

<math>y''(x)=0.0729e^{-0.27x}(c_1\cos(x\sqrt{\pi})+c_2\sin(x\sqrt{\pi})-0.27e^{-0.27x}(-\sqrt{\pi}c_1\sin{\sqrt{\pi}x}+\sqrt{\pi}c_1\cos{\sqrt{\pi}x})-\!</math><br>

<math>0.27e^{-0.27x}(\sqrt{\pi}c_1\sin{\sqrt{\pi}x}+\sqrt{\pi}c_1\cos{\sqrt{\pi}x})+e^{-0.27x}(-\pi(c_1\cos(x\sqrt{\pi}))-\pi(c_2\sin(x\sqrt{\pi})))\!</math><br>

Substituting <math>y,y',y''\!</math> into the original equation, the result is <br>
  <math>y''+0.54y'+(0.0729+\pi)y=0\!</math><br>
[[User:Egm4313.s12.team11.gooding|Egm4313.s12.team11.gooding]] 03:41, 7 February 2012 (UTC)
==Report 2, Problem 9==
====Problem Statement====
Find and plot the solution for the L2-ODE-CC corresponding to 

<math>\lambda^2 + 4 \lambda + 13\!</math>

with <math>r(x)=0\!</math> 

and initial conditions <math>y(0)=1\!</math>,  <math>y'(0)=0\!</math>

In another figure, superimpose 3 figs.:(a)this fig. (b) the fig. in R2.6 p.5-6, and (c) the fig. in R2.1 p.3-7
====Quadratic Equation====
<math>\lambda = \frac{-b \pm \sqrt(b^2-4ac)}{2a}\!</math>
with <math>a=1,  b=4,  c=13\!</math>

<math>\lambda=\frac{-4 \pm \sqrt(4^2-4*1*13)}{2*1}=-2 \pm 3i\!</math>

<math>\lambda=-2 \pm 3i\!</math>
====Homogeneous Solution====
The solution to a L2-ODE-CC with two complex roots is given by

<math>y(x)=e^{-\frac{a}{2}x}[Acos(\omega x)+Bsin(\omega x)]\!</math>

where <math>\lambda=-\frac{a}{2} \pm \omega i=-2 \pm 3i\!</math>

<math>y(x)=e^{-2x}[Acos(3x)+Bsin(3x)]\!</math>
====Solving for A and B====
first initial condition <math>y(0)=1\!</math>

<math>y(x)=e^{-2x}[Acos(3x)+Bsin(3x)]\!</math>

<math>y(0)=e^{-2*0}[Acos(3*0)+Bsin(3*0)]=1\!</math>

<math>A=1\!</math>

second initial condition <math>y'(0)=0\!</math>

<math>y'(x)=\frac{d}{dx}y(x)=\frac{d}{dx}e^{-2x}[cos(3x)+Bsin(3x)]\!</math>

<math>y'(x)=e^{-2x}[(-2B-3)sin(3x)+(3B-2)cos(3x)]\!</math>

<math>y'(0)=e^{-2*0}[(-2B-3)sin(3*0)+(3B-2)cos(3*0)]\!</math>

<math>0=3B-2\!</math>

<math>B=\frac{2}{3}\!</math>

so the solution to our L2-ODE-CC is 
                       <math>y(x)=e^{-2x}[cos(3x)+\frac{2}{3}sin(3x)]\!</math>
====Solution to R2.6====
After solving for the constants <math>\frac{k}{c}</math> and <math>\frac{k}{m}</math> we have the following homogeneous equation

<math>y''(x)+6y'(x)+9y=0\!</math>
=====Characteristic Equation and Roots=====
<math>\lambda^2 + 6\lambda + 9=0\!</math>

<math>(\lambda+3)(\lambda+3)=0\!</math>

We have a real double root <math>\lambda=-3\!</math>
=====Homogeneous Solution=====
We know the homogeneous solution to a L2-ODE-CC with a double real root to be

<math>y(x)=c_1e^{\lambda x}+c_2xe^{\lambda x}\!</math>

Assuming object starts from rest

<math>y(0)=1\!</math>,  <math>y'(0)=0\!</math>

Plugging in <math>\lambda</math> and applying our first initial condition

<math>y(0)=c_1e^{-3*0}+c_2*0*e^{-3*0}=1\!</math>

<math>c_1=1\!</math>

Taking the derivative and applying our second condition

<math>y'(x)=\frac{d}{dx}y(x)=\frac{d}{dx}e^{-3x}+c_2xe^{-3x}\!</math>

<math>y'(x)=-3e^{-3x}+c_2e^{-3x}-3c_2xe^{-3x}\!</math>

<math>y'(0)=-3e^{-3*0}+c_2e^{-3*0}-3c_2*0*e^{-3*0}=0\!</math>

<math>c_2=3\!</math>

Giving us the final solution
                  <math>y(x)=e^{-3x}+3xe^{-3x}\!</math>
====Plots====
=====Solution to this Equation=====
<math>y(x)=e^{-2x}[cos(3x)+\frac{2}{3}sin(3x)]\!</math>

[[File:Plotr2_9.jpg]]
=====Superimposed Graph=====
Our solution: <math>y(x)=e^{-2x}[cos(3x)+\frac{2}{3}sin(3x)]\!</math> shown in blue

Equation for fig. in R2.1 p.3-7: <math>y(x)=\frac{5}{4}e^{-2x}-\frac{1}{4}e^{5x}\!</math> shown in red

Equation for fig. in R2.6 p.5-6:<math>y(x)=e^{-3x}+3xe^{-3x}\!</math> shown in green

[[File:R2superposed.jpg]]