# University of Florida/Egm4313/s12.team5.R2

Report 2

## R2.1

### Question

Given the two roots and the initial conditions:

 ${\displaystyle \displaystyle \lambda _{1}=-2,\lambda _{2}=+5}$ (1.0)
 ${\displaystyle \displaystyle y(0)=1,y'(0)=0}$ (1.1)

Part 1.

Find the non-homogeneous L2-ODE-CC in standard form. Find the solution in terms of the initial conditions and the general excitation ${\displaystyle r(x).\!}$
Now with no excitation and plot the solution:

 ${\displaystyle \displaystyle r(x)=0}$ (1.2)

Part 2.

Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the two values ${\displaystyle \lambda _{1}=-2,\lambda _{2}=+5\!}$ as the two roots of the corresponding characteristic equation.

### Solution

Part 1.

Characteristic Equation:

 ${\displaystyle \displaystyle (\lambda -\lambda _{1})(\lambda -\lambda _{2})=0}$ (1.3)
 ${\displaystyle \displaystyle (\lambda -(-2))(\lambda -5)=0}$ (1.4)
 ${\displaystyle \displaystyle \lambda ^{2}-3\lambda -10=0}$ (1.5)

Non-Homogeneous L2-ODE-CC:

 ${\displaystyle \displaystyle y''-3y'-10y=r(x)\rightarrow standardform}$ (1.6)

Homogenous Solution:

 ${\displaystyle \displaystyle y_{h}(x)=c_{1}e^{-2x}+c_{2}e^{5x}}$ (1.7)

Overall Solution:

 ${\displaystyle \displaystyle y(x)=c_{1}e^{-2x}+c_{2}e^{5x}+y_{p}(x)}$ (1.8)
 ${\displaystyle \displaystyle y'(x)=-2c_{1}e^{-2x}+5c_{2}e^{5x}+y'_{p}(x)}$ (1.9)

Satisfy Initial Conditions:

 ${\displaystyle \displaystyle y(0)=1=c_{1}+c_{2}+y_{p}(0)}$ (1.10)
 ${\displaystyle \displaystyle y'(0)=0=-2c_{1}+5c_{2}+y'_{p}(0)}$ (1.11)

No excitation:

 ${\displaystyle \displaystyle r(x)=0\rightarrow y_{p}(x)=0\rightarrow y'_{p}(x)=0}$ (1.12)

From (1.10):

 ${\displaystyle \displaystyle c_{1}+c_{2}=1\rightarrow c_{1}=1-c_{2}}$ (1.13)

From (1.11):

 ${\displaystyle \displaystyle -2c_{1}+5c_{2}=0}$ (1.14)
 ${\displaystyle \displaystyle 5c_{2}=2c_{1}}$ (1.15)
 ${\displaystyle \displaystyle {\frac {5}{2}}c_{2}=c_{1}}$ (1.16)

Plug (1.13) into (1.16):

 ${\displaystyle \displaystyle {\frac {5}{2}}c_{2}=1-c_{2}}$ (1.17)

Solve for ${\displaystyle c_{2}\!}$:

 ${\displaystyle \displaystyle c_{2}={\frac {2}{7}}}$ (1.18)

Plug (1.18) into (1.13) and solve for ${\displaystyle c_{1}\!}$:

 ${\displaystyle \displaystyle c_{1}={\frac {5}{7}}}$ (1.19)

Therefore, the final solution in terms of the initial conditions and the general excitation ${\displaystyle r(x)\!}$ is:

 ${\displaystyle \displaystyle y(x)={\frac {5}{7}}e^{-2x}+{\frac {2}{7}}e^{5x}}$ (1.20)

Plot of Solution:

Figure 1

Part 2.

Three non-standard and non-homogeneous solutions using the same roots given above:

1.

 ${\displaystyle \displaystyle 2(\lambda -(-2))(\lambda -5)=0}$ (1.21)
 ${\displaystyle \displaystyle 2\lambda ^{2}-6\lambda -20=0}$ (1.22)

2.

 ${\displaystyle \displaystyle 3(\lambda -(-2))(\lambda -5)=0}$ (1.23)
 ${\displaystyle \displaystyle 3\lambda ^{2}-9\lambda -30=0}$ (1.24)

3.

 ${\displaystyle \displaystyle 4(\lambda -(-2))(\lambda -5)=0}$ (1.25)
 ${\displaystyle \displaystyle 4\lambda ^{2}-12\lambda -40=0}$ (1.26)

### Author

This problem was solved and uploaded by [Derik Bell]

This problem was proofread by: David Herrick

## R2.2

### Question

From Section 5 in the notes, pg. 5-6

Solve the L2-ODE-CC from pg. 5-5, equation (4)

${\displaystyle \displaystyle \ y''-10y'+25y=r(x)\ }$

For initial conditions:

${\displaystyle \displaystyle \ y(0)=1,y'(0)=0\ }$

Where there is no excitation, i.e. ${\displaystyle \displaystyle \ r(x)=0\ }$

### Solution

${\displaystyle \displaystyle \ \lambda ^{2}-10\lambda +25=0\ }$

Factoring:

${\displaystyle \displaystyle \ (\lambda -5)^{2}=0\ }$

Thus, there is a double root, where ${\displaystyle \displaystyle \ \lambda =5\ }$

Because there is no excitation, the homogenous solution will be the same as the final solution. Thus, the solution will be:

${\displaystyle \displaystyle \ y_{h}(x)=y(x)=C_{1}e^{5x}+C_{2}xe^{5x}\ }$

Plugging in the initial condition,

${\displaystyle \displaystyle \ y(0)=1=C_{1}e^{0}+C_{2}(0)e^{0}\ }$

${\displaystyle \displaystyle \ C_{1}=1\ }$

In order to solve for the second constant, we need to take a derivative.

${\displaystyle \displaystyle \ y'_{h}(x)=y'(x)=5C_{1}e^{5x}+C_{2}e^{5x}+5C_{2}xe^{5x}\ }$

Plugging in C1 and the other initial condition:

${\displaystyle \displaystyle \ y'_{h}(0)=y'(0)=0=5(1)e^{0}+C_{2}e^{0}+5C_{2}(0)e^{0}\ }$

${\displaystyle \displaystyle \ 5+C_{2}=0\Rightarrow C_{2}=-5\ }$

So, the final solution is:

${\displaystyle \displaystyle \ y(x)=e^{5x}-5xe^{5x}\ }$


Matlab Code:

x = 0:0.001:1;

f = exp(5*x) - 5*x.*exp(5*x);

plot(x,f)

xlabel ('x')

ylabel ('y(x)')

### Author

This problem was solved and uploaded by [David Herrick]
This problem was checked by [William Knapper]

## R2.3

### Question

Complete problems 3 and 4 from p.59 of K 2011.

### Solution

The two problems have the same instructions: "Find a general solution. Check your answer by substitution."

#### Problem 3

${\displaystyle \displaystyle 3.\;{y}''+6{y}'+8.96y=0}$

We start by using the characteristic equation of this ODE in order to find the roots. We use:

${\displaystyle \displaystyle \lambda ^{2}+a\lambda +b=0}$

where a = 6 and b = 8.96. Taking the discriminant, we find that

${\displaystyle \displaystyle 6^{2}-4(8.96)=0.16>0}$

A positive discriminant implies that there exists two real roots, with a general solution of the form:

${\displaystyle \displaystyle y=c_{1}e^{\lambda _{1}x}+c_{2}e^{\lambda _{2}x}}$

To find the unknowns, we find the roots of the above equation:

${\displaystyle \displaystyle \lambda ^{2}+6\lambda +8.96=0}$

${\displaystyle \displaystyle \lambda _{1}=-3.2,\lambda _{2}=-2.8}$

Therefore, the correct general solution is:

${\displaystyle \displaystyle y=c_{1}e^{-3.2x}+c_{2}e^{-2.8x}}$


To check that this solution is correct, we take the first and second derivatives:

${\displaystyle \displaystyle {y}'=-3.2c_{1}e^{-3.2x}-2.8c_{2}e^{-2.8x}}$

${\displaystyle \displaystyle {y}''=10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x}}$

Now, we plug these values into the original equation:

${\displaystyle \displaystyle 10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x}-19.2c_{1}e^{-3.2x}-16.8c_{2}e^{-2.8x}+8.96c_{1}e^{-3.2x}+8.96c_{2}e^{-2.8x}=0}$

Upon further inspection, all the terms on the left side of the equation cancel out and equal zero.

#### Problem 4

${\displaystyle \displaystyle 4.\;{y}''+4{y}'+(\pi ^{2}+4)y=0}$

We start by using the characteristic equation of this ODE in order to find the roots. We use:

${\displaystyle \displaystyle \lambda ^{2}+a\lambda +b=0}$

where a = 4 and b = (π^2 + 4). Taking the discriminant, we find that

${\displaystyle \displaystyle 4^{2}-4(\pi ^{2}+4)=-4\pi ^{2}<0}$

A negative discriminant implies that there exists complex conjugate roots to this equation. The general solution for this type of equation is

${\displaystyle \displaystyle y=e^{-2x}(Acos(\omega x)+Bsin(\omega x))}$

Where

${\displaystyle \displaystyle \omega ={\sqrt {b-{\frac {1}{4}}a^{2}}}={\sqrt {(\pi ^{2}+4)-({\frac {1}{4}})4^{2}}}=\pi }$

This yields

${\displaystyle \displaystyle y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))}$


To verify that this is the correct solution, we first find the first and second derivatives of the solution:

${\displaystyle \displaystyle y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))}$
${\displaystyle \displaystyle {y}'=e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))}$
${\displaystyle \displaystyle {y}''=e^{-2x}[-A\pi ^{2}cos(\pi x)-B\pi ^{2}sin(\pi x)+2A\pi sin(\pi x)-2B\pi sin(\pi x)]-2e^{-2x}[-A\pi sin(\pi x)+B\pi cos(\pi x)-2Acos(\pi x)-2Bsin(\pi x)]}$

We then plug these values into the original problem to get this somewhat lengthy equation:

${\displaystyle \displaystyle e^{-2x}A\pi ^{2}cos(\pi x)-e^{-2x}B\pi ^{2}sin(\pi x)+2e^{-2x}A\pi sin(\pi x)-2e^{-2x}B\pi sin(\pi x)+2e^{-2x}A\pi sin(\pi x)-2e^{-2x}B\pi cos(\pi x)+}$
${\displaystyle \displaystyle 4e^{-2x}Acos(\pi x)+4e^{-2x}Bsin(\pi x)-4e^{-2x}A\pi sin(\pi x)+4e^{-2x}B\pi cos(\pi x)-}$
${\displaystyle \displaystyle 8e^{-2x}Acos(\pi x)-8e^{-2x}Bsin(\pi x)-e^{-2x}A\pi ^{2}cos(\pi x)+e^{-2x}B\pi ^{2}sin(\pi x)+4e^{-2x}Acos(\pi x)+4e^{-2x}Bsin(\pi x)=0}$

Upon further inspection, all the terms on the left side of the equation cancel out to 0.

### Author

This problem was solved and uploaded by: (Will Knapper)

This problem was proofread by: David Herrick

## R2.4

### Question

K 2011 p.59 pbs.5,6. Find a general solution. Check your answer by substitution.

#### Problem 5

${\displaystyle y''+2\pi y'+\pi ^{2}y=0\ }$

### Solution

Determine the characteristic equation and the nature of the determinate:

${\displaystyle y=e^{-\lambda x}\ }$
${\displaystyle y'=-\lambda e^{-\lambda x}\ }$
${\displaystyle y''=\lambda ^{2}e^{-\lambda x}\ }$

Characteristic Equation:

${\displaystyle \lambda ^{2}-2\pi \lambda +\pi ^{2}\lambda =0\ }$
${\displaystyle a^{2}-4b=(2\pi )^{2}-4(\pi ^{2})=4\pi ^{2}-4\pi ^{2}=0\ }$

0 indicates a real double root, therefore:

${\displaystyle y_{g}=e^{-{\frac {a}{2}}x}(C_{1}+C_{2}x)\ }$

${\displaystyle a=2\pi \ }$ from the original equation. Therefore:

${\displaystyle y_{g}=e^{-\pi x}(C_{1}+C_{2}x)\ }$

${\displaystyle y'_{g}=e^{-\pi x}(C_{2}-\pi C_{1}-\pi C_{2}x)\ }$
${\displaystyle y''_{g}=e^{-\pi x}(C_{1}\pi ^{2}+C_{2}\pi ^{2}x-2C_{2}\pi )\ }$

Substituting into the original equation gives:

${\displaystyle e^{-\pi x}[(C_{1}\pi ^{2}+C_{2}\pi ^{2}x-2C_{2}\pi )+2\pi (C_{2}-\pi C_{1}-\pi C_{2}x)+\pi ^{2}(C_{1}+C_{2}x)]=0\ }$
${\displaystyle =e^{-\pi x}[C_{1}\pi ^{2}+C_{2}\pi ^{2}x-2C_{2}\pi +2C_{2}\pi -2C_{1}\pi ^{2}-2C_{2}\pi ^{2}x+C_{1}\pi ^{2}+C_{2}\pi ^{2}x]\ }$

Canceling gives 0 = 0, confirming the general solution found above.

### Author

This problem was solved and uploaded by: (John North)

This problem was proofread by: David Herrick

### Question 2

#### Problem 5

${\displaystyle 10y''-32y'+25.6y=0\ }$

### Solution

Reduce original equation so that the first coefficient is 1:

${\displaystyle y''-3.2y'+2.56y=0\ }$

Determine the characteristic equation and the nature of the determinate:

${\displaystyle y=e^{-\lambda x}\ }$
${\displaystyle y'=-\lambda e^{-\lambda x}\ }$
${\displaystyle y''=\lambda ^{2}e^{-\lambda x}\ }$

Characteristic Equation:

${\displaystyle \lambda ^{2}-3.2\lambda +2.56=0\ }$
${\displaystyle a^{2}-4b=3.2^{2}-4(2.56)=10.24-10.24=0\ }$

0 indicates a real double root, therefore:

${\displaystyle y_{g}=e^{-{\frac {a}{2}}x}(C_{1}+C_{2}x)\ }$

${\displaystyle a={\frac {-(-3.2)}{2}}=1.6}$ from the original equation. Therefore:

${\displaystyle y_{g}=e^{1.6x}(C_{1}+C_{2}x)\ }$

${\displaystyle y'_{g}=e^{1.6x}(1.6C_{1}+1.6C_{2}x+C_{2})\ }$
${\displaystyle y''_{g}=e^{1.6x}(2.56C_{1}+2.56C_{2}x+3.2C_{2})\ }$

Substituting into the original equation gives:

${\displaystyle e^{1.6x}[(2.56C_{1}+2.56C_{2}x+3.2C_{2})-3.2(1.6C_{1}+1.6C_{2}x+C_{2})+2.56(C_{1}+C_{2}x)]=0\ }$
${\displaystyle =e^{1.6x}[(2.56C_{1}+2.56C_{2}x+3.2C_{2}-5.12C_{1}-5.12C_{2}x-3.2C_{2}+2.56C_{1}+2.56C_{2}x]=0\ }$

Canceling gives 0 = 0, confirming that general solution found above.

### Author

This problem was solved and uploaded by: (John North)

This problem was proofread by: David Herrick

## R2.5

### Question

Problems 16 and 17 from pg. 59 of the textbook. The two questions have the same instructions: Find an ODE of the form
${\displaystyle \displaystyle {y}''+a{y}'+by=0}$
for the given basis.

### Solution

#### Problem 16

${\displaystyle \displaystyle 16.\;e^{2.6x},e^{-4.3x}}$
This basis implies that there are two distinct, real roots: 2.6 and -4.3. We put these roots into the characteristic equation to get: ${\displaystyle \displaystyle (\lambda -2.6)(\lambda +4.3)=0\rightarrow \lambda ^{2}+1.7\lambda -11.18=0}$
Then, you convert the characteristic equation to a regular ODE:

${\displaystyle \displaystyle {y}''+1.7{y}'-11.18y=0}$


#### Problem 17

${\displaystyle \displaystyle 17.\;e^{-{\sqrt {5}}x}\;,xe^{-{\sqrt {5}}x}}$
A basis of this form implies that the solution is of the form:
${\displaystyle \displaystyle y=(c_{1}+c_{2}x)e^{-ax/2}}$
The real double root can be found by this equation:
${\displaystyle \displaystyle \lambda =-a/2=-{\sqrt {5}}}$
That means that the characteristic equation can be found by solving:
${\displaystyle \displaystyle (\lambda +{\sqrt {5}})^{2}=\lambda ^{2}+2{\sqrt {5}}\lambda +5}$
Converting the characteristic equation to the ODE, we get:

${\displaystyle \displaystyle {y}''+2{\sqrt {5}}{y}'+5y=0}$


### Author

This problem was solved and uploaded by: (Will Knapper)
This problem was proofread by (Josh House)

## R2.6

### Question

Realize spring-dashpot-mass system in series as shown in Fig. p.1-4 with the similar characteristic as in (3)p.5-5, but with the double real root ${\displaystyle \displaystyle \lambda =-3\ }$, i.e., find values for parameters k,c,m.

### Solution

First we put the double root into the characteristic equation:
${\displaystyle \displaystyle (\lambda +3)^{2}=0}$
${\displaystyle \displaystyle \lambda ^{2}+6\lambda +9=0}$

The corresponding L2-ODE is:
${\displaystyle \displaystyle {y}''+6{y}'+9y=0}$

The equation of motion of the spring-dashpot-mass system is:
${\displaystyle \displaystyle m\left(y_{k}''+{\frac {k}{c}}y_{k}'\right)+ky_{k}=0}$ [1]
or:
${\displaystyle \displaystyle my_{k}''+m{\frac {k}{c}}y_{k}'+ky_{k}=0}$

Solving for the parameters k,c,m:
${\displaystyle \displaystyle m=1}$

${\displaystyle \displaystyle m{\frac {k}{c}}=6}$

${\displaystyle \displaystyle k=9}$

We get:

${\displaystyle \displaystyle m=1}$
${\displaystyle \displaystyle k=9}$
${\displaystyle \displaystyle c={\frac {3}{2}}}$


### Author

This problem was proofread by: Mike Wallace

## R2.7

### Question

Develop the McLauren series (Taylor series at t=0) for e^t, cos(t), and sin(t).

### Solution

The general form of the Taylor series expansion at point a is

${\displaystyle f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f'''(a)}{3!}}(x-a)^{3}+...}$

for ${\displaystyle e^{t}}$ at t=0:

${\displaystyle e^{0}+{\frac {e^{0}}{1!}}(x)+{\frac {e^{0}}{2!}}(x)^{2}+{\frac {e^{0}}{3!}}(x)^{3}+...}$

${\displaystyle =1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+...}$


for ${\displaystyle \cos(t)}$ at t=0:

${\displaystyle \cos(0)+{\frac {-\sin(0)}{1!}}(x)+{\frac {-\cos(0)}{2!}}(x)^{2}+{\frac {sin(0)}{3!}}(x)^{3}+{\frac {cos(0)}{4!}}(x)^{4}+...}$

${\displaystyle =1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+...}$


for ${\displaystyle \sin(t)}$ at t=0:

${\displaystyle \sin(0)+{\frac {cos(0)}{1!}}(x)+{\frac {-\sin(0)}{2!}}(x)^{2}+{\frac {-cos(0)}{3!}}(x)^{3}+{\frac {\sin(x)}{4!}}(x)^{4}+{\frac {\cos(0)}{5!}}(x)^{5}+...}$

${\displaystyle =x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+...}$


### Author

This problem was solved and uploaded by: (John North)
This problem was checked by: William Knapper

## R2.8

### Question

8. ${\displaystyle \displaystyle y^{''}+y^{'}+3.25y=0}$

15. ${\displaystyle \displaystyle y^{''}+0.54y^{'}+(0.0729+\pi )y=0}$

### Solution

8. ${\displaystyle \displaystyle y^{''}+y^{'}+3.25y=0}$

Writing the characteristic equation:

${\displaystyle \displaystyle \lambda ^{2}+\lambda +3.25=0}$

Which is now in the form:

${\displaystyle \displaystyle \lambda ^{2}+a\lambda +b=0}$

Solving for the two roots:

${\displaystyle \displaystyle \lambda _{1}={\frac {1}{2}}(-a+{\sqrt {a^{2}-4b}}),\lambda _{2}={\frac {1}{2}}(-a-{\sqrt {a^{2}-4b}})}$

We will find the discriminant to be less than 0, leading to complex conjugate roots:

${\displaystyle \displaystyle a^{2}-4b=1^{2}-4(3.25)=-12}$

Which leads to a solution of the form:

${\displaystyle \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))}$

Where ${\displaystyle \displaystyle \omega ^{2}=b-{\frac {1}{4}}a^{2}}$, therefore:

${\displaystyle \displaystyle \omega ^{2}=3.25-{\frac {1}{4}}(1)^{2}=3\rightarrow \omega ={\sqrt {3}}}$

Giving us a solution of:

${\displaystyle \displaystyle y=e^{-x/2}(Acos({\sqrt {3}}x)+Bsin({\sqrt {3}}x))}$


Checking our solution:

${\displaystyle \displaystyle y=e^{-x/2}(Acos({\sqrt {3}}x)+Bsin({\sqrt {3}}x))}$
${\displaystyle \displaystyle y^{'}=-{\frac {1}{2}}e^{-x/2}(Acos({\sqrt {3}}x)+Bsin({\sqrt {3}}x))+e^{-x/2}(-{\sqrt {3}}sin({\sqrt {3}}x)+{\sqrt {3}}Bcos({\sqrt {3}}x))}$
${\displaystyle \displaystyle y^{''}=-{\frac {1}{4}}e^{-x/2}(Acos({\sqrt {3}}x)+Bsin({\sqrt {3}}x))-{\frac {1}{2}}e^{-x/2}(-{\sqrt {3}}Asin({\sqrt {3}}x)+{\sqrt {3}}Bcos({\sqrt {3}}x))-{\frac {1}{2}}e^{-x/2}(-{\sqrt {3}}Asin({\sqrt {3}}x)+{\sqrt {3}}Bcos({\sqrt {3}}x))+e^{-x/2}(-3Acos({\sqrt {3}}x)-3Bsin({\sqrt {3}}x))}$

Plugging all of this into the original differential equation and combining like terms gives us:

${\displaystyle \displaystyle [{\frac {1}{4}}-3-{\frac {1}{2}}+3.25]e^{-x/2}Acos({\sqrt {3}}x)}$
${\displaystyle \displaystyle [{\frac {\sqrt {3}}{2}}+{\frac {\sqrt {3}}{2}}-{\sqrt {3}}]e^{-x/2}Asin({\sqrt {3}}x)}$
${\displaystyle \displaystyle [-{\frac {\sqrt {3}}{2}}-{\frac {\sqrt {3}}{2}}+{\sqrt {3}}]e^{-x/2}Bcos({\sqrt {3}}x)}$
${\displaystyle \displaystyle [{\frac {1}{4}}-3-{\frac {1}{2}}+3.25]e^{-x/2}Bsin({\sqrt {3}}x)}$

The terms in the brackets all add up to zero, thus verifying we have a correct solution to our problem.

15. ${\displaystyle \displaystyle y^{''}+0.54y{'}+(0.0729+\pi )y=0}$

Writing the characteristic equation:

${\displaystyle \displaystyle \lambda ^{2}+0.54\lambda +(0.0729+\pi )=0}$

Which is now in the form:

${\displaystyle \displaystyle \lambda ^{2}+a\lambda +b=0}$

Solving for the two roots:

${\displaystyle \displaystyle \lambda _{1}={\frac {1}{2}}(-a+{\sqrt {a^{2}-4b}}),\lambda _{2}={\frac {1}{2}}(-a-{\sqrt {a^{2}-4b}})}$

We will find the discriminant to be less than 0, leading to complex conjugate roots:

${\displaystyle \displaystyle a^{2}-4b=0.54^{2}-4(0.0729+\pi )=-4\pi }$

Which leads to a solution of the form:

${\displaystyle \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))}$

Where ${\displaystyle \displaystyle \omega ^{2}=b-{\frac {1}{4}}a^{2}}$, therefore:

${\displaystyle \displaystyle \omega ^{2}=(0.0729-\pi )-{\frac {1}{4}}(0.54)^{2}=\pi \rightarrow \omega ={\sqrt {\pi }}}$

Giving us a solution of:

${\displaystyle \displaystyle y=e^{-0.27x}(Acos({\sqrt {\pi }}x)+Bsin({\sqrt {\pi }}x))}$


Checking our solution:

${\displaystyle \displaystyle y=e^{-0.27x}(Acos({\sqrt {\pi }}x)+Bsin({\sqrt {\pi }}x))}$
${\displaystyle \displaystyle y^{'}=-0.27e^{-0.27x}(Acos({\sqrt {\pi }}x)+Bsin({\sqrt {\pi }}x))+e^{-0.27x}(-{\sqrt {\pi }}Asin({\sqrt {\pi }}x)+{\sqrt {\pi }}Bcos({\sqrt {\pi }}x))}$
${\displaystyle \displaystyle y^{''}=0.0729e^{-0.27x}(Acos({\sqrt {\pi }}x)+Bsin({\sqrt {\pi }}x))-0.27e^{-0.27x}(-{\sqrt {\pi }}Asin({\sqrt {\pi }}x)+{\sqrt {\pi }}Bcos({\sqrt {\pi }}x))-0.27e^{-0.27x}(-{\sqrt {\pi }}Asin({\sqrt {\pi }}x)+{\sqrt {\pi }}Bcos({\sqrt {\pi }}x))+e^{-0.27x}(-\pi Acos({\sqrt {\pi }}x)-\pi Bsin({\sqrt {\pi }}x))}$

Plugging all of this into the original differential equation and combining like terms gives us:

${\displaystyle \displaystyle [0.0729-\pi -0.1458+0.0729+\pi ]e^{-0.27x}Acos{\sqrt {\pi }}x}$
${\displaystyle \displaystyle [0.27{\sqrt {\pi }}+0.27{\sqrt {\pi }}-0.54{\sqrt {\pi }}]e^{-0.27x}Asin{\sqrt {\pi }}x}$
${\displaystyle \displaystyle [-0.27{\sqrt {\pi }}-0.27{\sqrt {\pi }}+0.54{\sqrt {\pi }}]e^{-0.27x}Bcos{\sqrt {\pi }}x}$
${\displaystyle \displaystyle [0.0729-\pi -0.1458+0.0729+\pi ]e^{-0.27x}Bsin{\sqrt {\pi }}x}$

The terms in the brackets all add up to zero, thus verifying we have a correct solution to our problem.

### Author

This problem was solved and uploaded by: (Josh House)

## R2.9

### Question

For the given ODE, find and plot the solution for the L2-ODE-CC corresponding to

${\displaystyle \displaystyle \ \lambda ^{2}+4\lambda +13=0\ }$

In another figure superpose this figure, Fig. from R2.6 p.5-6, and the Fig. from R2-1 p.3-7.

This corresponds to the L2-ODE-CC in standard form

${\displaystyle \displaystyle \ y''+4y'+13y=r(x)\ }$

### Given

Initial conditions

${\displaystyle \displaystyle \ y(0)=1,y'(0)=0\ }$

Where there is no excitation

${\displaystyle \displaystyle \ r(x)=0\ }$

### Solution

Solving the quadratic equation for ${\displaystyle \displaystyle \ \lambda \ }$ determines the roots of the ODE

${\displaystyle \displaystyle \ \lambda _{1}={\frac {-b+{\sqrt {b^{2}-(4ac)}}}{2a}}\!={\frac {-4+{\sqrt {(4)^{2}-4(1)(13)}}}{2(1)}}\!}$

${\displaystyle \displaystyle \ \lambda _{2}={\frac {-b-{\sqrt {b^{2}-(4ac)}}}{2a}}\!={\frac {-4-{\sqrt {(4)^{2}-4(1)(13)}}}{2(1)}}\!}$

The roots of the quadratic equation come out to be:

${\displaystyle \displaystyle \ \lambda =(-2\pm 3i)\ }$

The general homogeneous solution of the ODE will resemble the following form.

${\displaystyle \displaystyle \ y=C_{1}e^{a_{1}x}cosb_{1}x+C_{2}e^{a_{2}x}sinb_{2}x\ }$

Plugging in the roots into the equation we get

${\displaystyle \displaystyle \ y=C_{1}e^{-2x}cos3x+C_{2}e^{-2x}sin3x\ }$

Implementing initial conditions where: ${\displaystyle \displaystyle \ y(0)=1,y'(0)=0\ }$

${\displaystyle \displaystyle \ y(0)=C_{1}e^{-2(0)}cos3(0)+C_{2}e^{-2(0)}sin3(0)=1\ }$

${\displaystyle \displaystyle \ y'(0)=-2C_{1}e^{-2x}cos3x-3C_{1}e^{-2x}sin3x-2C_{2}e^{-2x}sin3x+3C_{2}e^{-2x}cos3x=0\ }$

${\displaystyle \displaystyle \ y'(0)=-2(1)e^{-2(0)}cos3(0)+3C_{2}e^{-2(0)}cos3(0)=-2+3C_{2}=0;\ }$

So from our initial conditions the constants are determined to be

${\displaystyle \displaystyle \ C_{1}=1,C_{2}=2/3\ }$

The final solution becomes

${\displaystyle \displaystyle \ y(x)=e^{-2x}cos3x+{\frac {2}{3}}e^{-2x}sin3x\ }$


### Author

This problem was solved and uploaded by: Mike Wallace
This problem was proofread by John North

## Contribution Summary

Problem 2 was solved and Problems 1, 3, and 4 were proofread by David Herrick 15:56, 6 February 2012 (UTC)

Problem 8 was solved and Problem 5 was proofread by Josh House 23:41, 7 February 2012 (UTC)

Problem 6 was solved and Problem 8 was proofread by Radina Dikova 16:26, 7 February 2012 (UTC)

Problem 9 was solved and Problem 6 was proofread by Mike Wallace 23:35, 7 February 2012 (UTC)

Problems 3 and 5 were solved and Problems 2 and 7 were proofread by William Knapper 05:04, 8 February 2012 (UTC)

Problem 1 was solved by Derik Bell 15:44, 8 February 2012 (UTC)

Problems 4 and 7 were solved and Problem 9 was proofread by John North 15:47, 8 February 2012 (UTC)

## References

1. Vu-Quoc , Loc. 2012. IEA S12 Lecture notes, audios, videos. Section 1(d) https://docs.google.com/Doc?id=dc82nfgb_858gmw855hj