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

Report 2

## R2.1
### Question

Given the two roots and the initial conditions:
    $\displaystyle \lambda _{1}=-2,\lambda _{2}=+5$ (1.0)
    $\displaystyle y(0)=1,y'(0)=0$ (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 $r(x).\!$ Now with no excitation and plot the solution:
    $\displaystyle r(x)=0$ (1.2)
Part 2.
Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the two values $\lambda _{1}=-2,\lambda _{2}=+5\!$ as the two roots of the corresponding characteristic equation.

### Solution

Part 1.
Characteristic Equation:
    $\displaystyle (\lambda -\lambda _{1})(\lambda -\lambda _{2})=0$ (1.3)
    $\displaystyle (\lambda -(-2))(\lambda -5)=0$ (1.4)
    $\displaystyle \lambda ^{2}-3\lambda -10=0$ (1.5)
Non-Homogeneous L2-ODE-CC:
    $\displaystyle y''-3y'-10y=r(x)\rightarrow standardform$ (1.6)
Homogenous Solution:
    $\displaystyle y_{h}(x)=c_{1}e^{-2x}+c_{2}e^{5x}$ (1.7)
Overall Solution:
    $\displaystyle y(x)=c_{1}e^{-2x}+c_{2}e^{5x}+y_{p}(x)$ (1.8)
    $\displaystyle y'(x)=-2c_{1}e^{-2x}+5c_{2}e^{5x}+y'_{p}(x)$ (1.9)
Satisfy Initial Conditions:
    $\displaystyle y(0)=1=c_{1}+c_{2}+y_{p}(0)$ (1.10)
    $\displaystyle y'(0)=0=-2c_{1}+5c_{2}+y'_{p}(0)$ (1.11)
No excitation:
    $\displaystyle r(x)=0\rightarrow y_{p}(x)=0\rightarrow y'_{p}(x)=0$ (1.12)
From (1.10):
    $\displaystyle c_{1}+c_{2}=1\rightarrow c_{1}=1-c_{2}$ (1.13)
From (1.11):
    $\displaystyle -2c_{1}+5c_{2}=0$ (1.14)
    $\displaystyle 5c_{2}=2c_{1}$ (1.15)
    $\displaystyle {\frac {5}{2}}c_{2}=c_{1}$ (1.16)
Plug (1.13) into (1.16):
    $\displaystyle {\frac {5}{2}}c_{2}=1-c_{2}$ (1.17)
Solve for $c_{2}\!$:
    $\displaystyle c_{2}={\frac {2}{7}}$ (1.18)
Plug (1.18) into (1.13) and solve for $c_{1}\!$:
    $\displaystyle c_{1}={\frac {5}{7}}$ (1.19)
Therefore, the final solution in terms of the initial conditions and the general excitation $r(x)\!$ is:
    $\displaystyle y(x)={\frac {5}{7}}e^{-2x}+{\frac {2}{7}}e^{5x}$ (1.20)
Plot of Solution:
Figure 1
Part 2.
Three non-standard and non-homogeneous solutions using the same roots given above:
1.
    $\displaystyle 2(\lambda -(-2))(\lambda -5)=0$ (1.21)
    $\displaystyle 2\lambda ^{2}-6\lambda -20=0$ (1.22)
2.
    $\displaystyle 3(\lambda -(-2))(\lambda -5)=0$ (1.23)
    $\displaystyle 3\lambda ^{2}-9\lambda -30=0$ (1.24)
3.
    $\displaystyle 4(\lambda -(-2))(\lambda -5)=0$ (1.25)
    $\displaystyle 4\lambda ^{2}-12\lambda -40=0$ (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 \ y''-10y'+25y=r(x)\$
For initial conditions:
$\displaystyle \ y(0)=1,y'(0)=0\$
Where there is no excitation, i.e. $\displaystyle \ r(x)=0\$

### Solution

$\displaystyle \ \lambda ^{2}-10\lambda +25=0\$
Factoring:
$\displaystyle \ (\lambda -5)^{2}=0\$
Thus, there is a double root, where $\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 \ y_{h}(x)=y(x)=C_{1}e^{5x}+C_{2}xe^{5x}\$
Plugging in the initial condition,
$\displaystyle \ y(0)=1=C_{1}e^{0}+C_{2}(0)e^{0}\$
$\displaystyle \ C_{1}=1\$
In order to solve for the second constant, we need to take a derivative.
$\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 \ y'_{h}(0)=y'(0)=0=5(1)e^{0}+C_{2}e^{0}+5C_{2}(0)e^{0}\$
$\displaystyle \ 5+C_{2}=0\Rightarrow C_{2}=-5\$
So, the final solution is:
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

$\displaystyle 3.\;{y}''+6{y}'+8.96y=0$
We start by using the characteristic equation of this ODE in order to find the roots. We use:
$\displaystyle \lambda ^{2}+a\lambda +b=0$
where a = 6 and b = 8.96. Taking the discriminant, we find that
$\displaystyle 6^{2}-4(8.96)=0.16>0$
A positive discriminant implies that there exists two real roots, with a general solution of the form:
$\displaystyle y=c_{1}e^{\lambda _{1}x}+c_{2}e^{\lambda _{2}x}$
To find the unknowns, we find the roots of the above equation:
$\displaystyle \lambda ^{2}+6\lambda +8.96=0$
$\displaystyle \lambda _{1}=-3.2,\lambda _{2}=-2.8$
Therefore, the correct general solution is:
To check that this solution is correct, we take the first and second derivatives:
$\displaystyle {y}'=-3.2c_{1}e^{-3.2x}-2.8c_{2}e^{-2.8x}$
$\displaystyle {y}''=10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x}$
Now, we plug these values into the original equation:
$\displaystyle 10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x}-19.2c_{1}e^{-3.2x}-16.8c_{2}e^{-2.8x}+8.96c_{1}e^{-3.2x}+8.96c_{2}e^{-2.8x}=0$
Upon further inspection, all the terms on the left side of the equation cancel out and equal zero.

#### Problem 4

$\displaystyle 4.\;{y}''+4{y}'+(\pi ^{2}+4)y=0$
We start by using the characteristic equation of this ODE in order to find the roots. We use:
$\displaystyle \lambda ^{2}+a\lambda +b=0$
where a = 4 and b = (π^2 + 4). Taking the discriminant, we find that
$\displaystyle 4^{2}-4(\pi ^{2}+4)=-4\pi ^{2}<0$
A negative discriminant implies that there exists complex conjugate roots to this equation. The general solution for this type of equation is
$\displaystyle y=e^{-2x}(Acos(\omega x)+Bsin(\omega x))$
Where
$\displaystyle \omega ={\sqrt {b-{\frac {1}{4}}a^{2}}}={\sqrt {(\pi ^{2}+4)-({\frac {1}{4}})4^{2}}}=\pi$
This yields
To verify that this is the correct solution, we first find the first and second derivatives of the solution:
$\displaystyle y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))$
$\displaystyle {y}'=e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))$
$\displaystyle {y}''=e^{-2x}[-A\pi ^{2}cos(\pi x)-B\pi ^{2}sin(\pi x)+2A\pi sin(\pi x)-2B\pi sin(\pi x)]-2e^{-2x}[-A\pi sin(\pi x)+B\pi cos(\pi x)-2Acos(\pi x)-2Bsin(\pi x)]$
We then plug these values into the original problem to get this somewhat lengthy equation:
$\displaystyle e^{-2x}A\pi ^{2}cos(\pi x)-e^{-2x}B\pi ^{2}sin(\pi x)+2e^{-2x}A\pi sin(\pi x)-2e^{-2x}B\pi sin(\pi x)+2e^{-2x}A\pi sin(\pi x)-2e^{-2x}B\pi cos(\pi x)+$ $\displaystyle 4e^{-2x}Acos(\pi x)+4e^{-2x}Bsin(\pi x)-4e^{-2x}A\pi sin(\pi x)+4e^{-2x}B\pi cos(\pi x)-$ $\displaystyle 8e^{-2x}Acos(\pi x)-8e^{-2x}Bsin(\pi x)-e^{-2x}A\pi ^{2}cos(\pi x)+e^{-2x}B\pi ^{2}sin(\pi x)+4e^{-2x}Acos(\pi x)+4e^{-2x}Bsin(\pi x)=0$
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

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

### Solution

Determine the characteristic equation and the nature of the determinate:
    $y=e^{-\lambda x}\$
    $y'=-\lambda e^{-\lambda x}\$
    $y''=\lambda ^{2}e^{-\lambda x}\$
Characteristic Equation:
    $\lambda ^{2}-2\pi \lambda +\pi ^{2}\lambda =0\$
    $a^{2}-4b=(2\pi )^{2}-4(\pi ^{2})=4\pi ^{2}-4\pi ^{2}=0\$
0 indicates a real double root, therefore:
    $y_{g}=e^{-{\frac {a}{2}}x}(C_{1}+C_{2}x)\$
$a=2\pi \$ from the original equation. Therefore:
    $y'_{g}=e^{-\pi x}(C_{2}-\pi C_{1}-\pi C_{2}x)\$
    $y''_{g}=e^{-\pi x}(C_{1}\pi ^{2}+C_{2}\pi ^{2}x-2C_{2}\pi )\$
Substituting into the original equation gives:
    $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\$
       $=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

    $10y''-32y'+25.6y=0\$

### Solution

Reduce original equation so that the first coefficient is 1:
    $y''-3.2y'+2.56y=0\$
Determine the characteristic equation and the nature of the determinate:
    $y=e^{-\lambda x}\$
    $y'=-\lambda e^{-\lambda x}\$
    $y''=\lambda ^{2}e^{-\lambda x}\$
Characteristic Equation:
    $\lambda ^{2}-3.2\lambda +2.56=0\$
    $a^{2}-4b=3.2^{2}-4(2.56)=10.24-10.24=0\$
0 indicates a real double root, therefore:
    $y_{g}=e^{-{\frac {a}{2}}x}(C_{1}+C_{2}x)\$
$a={\frac {-(-3.2)}{2}}=1.6$ from the original equation. Therefore:
    $y'_{g}=e^{1.6x}(1.6C_{1}+1.6C_{2}x+C_{2})\$
    $y''_{g}=e^{1.6x}(2.56C_{1}+2.56C_{2}x+3.2C_{2})\$
Substituting into the original equation gives:
    $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\$
       $=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
$\displaystyle {y}''+a{y}'+by=0$
for the given basis.

### Solution
#### Problem 16

$\displaystyle 16.\;e^{2.6x},e^{-4.3x}$
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: $\displaystyle (\lambda -2.6)(\lambda +4.3)=0\rightarrow \lambda ^{2}+1.7\lambda -11.18=0$
Then, you convert the characteristic equation to a regular ODE:

#### Problem 17

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

### 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 \lambda =-3\$, i.e., find values for parameters k,c,m.

### Solution

First we put the double root into the characteristic equation:
$\displaystyle (\lambda +3)^{2}=0$
$\displaystyle \lambda ^{2}+6\lambda +9=0$
The corresponding L2-ODE is: $\displaystyle {y}''+6{y}'+9y=0$
The equation of motion of the spring-dashpot-mass system is: $\displaystyle m\left(y_{k}''+{\frac {k}{c}}y_{k}'\right)+ky_{k}=0$
or: $\displaystyle my_{k}''+m{\frac {k}{c}}y_{k}'+ky_{k}=0$
Solving for the parameters k,c,m: $\displaystyle m=1$
$\displaystyle m{\frac {k}{c}}=6$
$\displaystyle k=9$
We get:

### Author

This problem was solved and uploaded by: (Radina Dikova)
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
    $f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f'''(a)}{3!}}(x-a)^{3}+...$
for $e^{t}$ at t=0:
    $e^{0}+{\frac {e^{0}}{1!}}(x)+{\frac {e^{0}}{2!}}(x)^{2}+{\frac {e^{0}}{3!}}(x)^{3}+...$
for $\cos(t)$ at t=0:
    $\cos(0)+{\frac {-\sin(0)}{1!}}(x)+{\frac {-\cos(0)}{2!}}(x)^{2}+{\frac {sin(0)}{3!}}(x)^{3}+{\frac {cos(0)}{4!}}(x)^{4}+...$
for $\sin(t)$ at t=0:
    $\sin(0)+{\frac {cos(0)}{1!}}(x)+{\frac {-\sin(0)}{2!}}(x)^{2}+{\frac {-cos(0)}{3!}}(x)^{3}+{\frac {\sin(x)}{4!}}(x)^{4}+{\frac {\cos(0)}{5!}}(x)^{5}+...$

### Author

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

## R2.8
### Question

Find a general solution. Check your answer by substitution.
8. $\displaystyle y^{''}+y^{'}+3.25y=0$
15. $\displaystyle y^{''}+0.54y^{'}+(0.0729+\pi )y=0$

### Solution

8. $\displaystyle y^{''}+y^{'}+3.25y=0$
Writing the characteristic equation:
    $\displaystyle \lambda ^{2}+\lambda +3.25=0$
Which is now in the form:
    $\displaystyle \lambda ^{2}+a\lambda +b=0$
Solving for the two roots:
    $\displaystyle \lambda _{1}={\frac {1}{2}}(-a+{\sqrt {a^{2}-4b}}),\lambda _{2}={\frac {1}{2}}(-a-{\sqrt {a^{2}-4b}})$
We will find the discriminant to be less than 0, leading to complex conjugate roots:
    $\displaystyle a^{2}-4b=1^{2}-4(3.25)=-12$
Which leads to a solution of the form:
    $\displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))$
Where $\displaystyle \omega ^{2}=b-{\frac {1}{4}}a^{2}$, therefore:
    $\displaystyle \omega ^{2}=3.25-{\frac {1}{4}}(1)^{2}=3\rightarrow \omega ={\sqrt {3}}$
Giving us a solution of:
Checking our solution:
    $\displaystyle y=e^{-x/2}(Acos({\sqrt {3}}x)+Bsin({\sqrt {3}}x))$
    $\displaystyle y^{'}=-{\frac {1}{2}}e^{-x/2}(Acos({\sqrt {3}}x)+Bsin({\sqrt {3}}x))+e^{-x/2}(-{\sqrt {3}}sin({\sqrt {3}}x)+{\sqrt {3}}Bcos({\sqrt {3}}x))$
    $\displaystyle y^{''}=-{\frac {1}{4}}e^{-x/2}(Acos({\sqrt {3}}x)+Bsin({\sqrt {3}}x))-{\frac {1}{2}}e^{-x/2}(-{\sqrt {3}}Asin({\sqrt {3}}x)+{\sqrt {3}}Bcos({\sqrt {3}}x))-{\frac {1}{2}}e^{-x/2}(-{\sqrt {3}}Asin({\sqrt {3}}x)+{\sqrt {3}}Bcos({\sqrt {3}}x))+e^{-x/2}(-3Acos({\sqrt {3}}x)-3Bsin({\sqrt {3}}x))$
Plugging all of this into the original differential equation and combining like terms gives us:
    $\displaystyle [{\frac {1}{4}}-3-{\frac {1}{2}}+3.25]e^{-x/2}Acos({\sqrt {3}}x)$
    $\displaystyle [{\frac {\sqrt {3}}{2}}+{\frac {\sqrt {3}}{2}}-{\sqrt {3}}]e^{-x/2}Asin({\sqrt {3}}x)$
    $\displaystyle [-{\frac {\sqrt {3}}{2}}-{\frac {\sqrt {3}}{2}}+{\sqrt {3}}]e^{-x/2}Bcos({\sqrt {3}}x)$
    $\displaystyle [{\frac {1}{4}}-3-{\frac {1}{2}}+3.25]e^{-x/2}Bsin({\sqrt {3}}x)$
The terms in the brackets all add up to zero, thus verifying we have a correct solution to our problem.
15. $\displaystyle y^{''}+0.54y{'}+(0.0729+\pi )y=0$
Writing the characteristic equation:
    $\displaystyle \lambda ^{2}+0.54\lambda +(0.0729+\pi )=0$
Which is now in the form:
    $\displaystyle \lambda ^{2}+a\lambda +b=0$
Solving for the two roots:
    $\displaystyle \lambda _{1}={\frac {1}{2}}(-a+{\sqrt {a^{2}-4b}}),\lambda _{2}={\frac {1}{2}}(-a-{\sqrt {a^{2}-4b}})$
We will find the discriminant to be less than 0, leading to complex conjugate roots:
    $\displaystyle a^{2}-4b=0.54^{2}-4(0.0729+\pi )=-4\pi$
Which leads to a solution of the form:
    $\displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))$
Where $\displaystyle \omega ^{2}=b-{\frac {1}{4}}a^{2}$, therefore:
    $\displaystyle \omega ^{2}=(0.0729-\pi )-{\frac {1}{4}}(0.54)^{2}=\pi \rightarrow \omega ={\sqrt {\pi }}$
Giving us a solution of:
Checking our solution:
    $\displaystyle y=e^{-0.27x}(Acos({\sqrt {\pi }}x)+Bsin({\sqrt {\pi }}x))$
    $\displaystyle y^{'}=-0.27e^{-0.27x}(Acos({\sqrt {\pi }}x)+Bsin({\sqrt {\pi }}x))+e^{-0.27x}(-{\sqrt {\pi }}Asin({\sqrt {\pi }}x)+{\sqrt {\pi }}Bcos({\sqrt {\pi }}x))$
    $\displaystyle y^{''}=0.0729e^{-0.27x}(Acos({\sqrt {\pi }}x)+Bsin({\sqrt {\pi }}x))-0.27e^{-0.27x}(-{\sqrt {\pi }}Asin({\sqrt {\pi }}x)+{\sqrt {\pi }}Bcos({\sqrt {\pi }}x))-0.27e^{-0.27x}(-{\sqrt {\pi }}Asin({\sqrt {\pi }}x)+{\sqrt {\pi }}Bcos({\sqrt {\pi }}x))+e^{-0.27x}(-\pi Acos({\sqrt {\pi }}x)-\pi Bsin({\sqrt {\pi }}x))$
Plugging all of this into the original differential equation and combining like terms gives us:
    $\displaystyle [0.0729-\pi -0.1458+0.0729+\pi ]e^{-0.27x}Acos{\sqrt {\pi }}x$
    $\displaystyle [0.27{\sqrt {\pi }}+0.27{\sqrt {\pi }}-0.54{\sqrt {\pi }}]e^{-0.27x}Asin{\sqrt {\pi }}x$
    $\displaystyle [-0.27{\sqrt {\pi }}-0.27{\sqrt {\pi }}+0.54{\sqrt {\pi }}]e^{-0.27x}Bcos{\sqrt {\pi }}x$
    $\displaystyle [0.0729-\pi -0.1458+0.0729+\pi ]e^{-0.27x}Bsin{\sqrt {\pi }}x$
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) This problem was proofread by: (Radina Dikova)

## R2.9
### Question

For the given ODE, find and plot the solution for the L2-ODE-CC corresponding to
$\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 \ y''+4y'+13y=r(x)\$

### Given

Initial conditions
$\displaystyle \ y(0)=1,y'(0)=0\$
Where there is no excitation
$\displaystyle \ r(x)=0\$

### Solution

Solving the quadratic equation for $\displaystyle \ \lambda \$ determines the roots of the ODE
$\displaystyle \ \lambda _{1}={\frac {-b+{\sqrt {b^{2}-(4ac)}}}{2a}}\!={\frac {-4+{\sqrt {(4)^{2}-4(1)(13)}}}{2(1)}}\!$
$\displaystyle \ \lambda _{2}={\frac {-b-{\sqrt {b^{2}-(4ac)}}}{2a}}\!={\frac {-4-{\sqrt {(4)^{2}-4(1)(13)}}}{2(1)}}\!$
The roots of the quadratic equation come out to be:
$\displaystyle \ \lambda =(-2\pm 3i)\$
The general homogeneous solution of the ODE will resemble the following form.
$\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 \ y=C_{1}e^{-2x}cos3x+C_{2}e^{-2x}sin3x\$
Implementing initial conditions where: $\displaystyle \ y(0)=1,y'(0)=0\$
$\displaystyle \ y(0)=C_{1}e^{-2(0)}cos3(0)+C_{2}e^{-2(0)}sin3(0)=1\$
$\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 \ 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 \ C_{1}=1,C_{2}=2/3\$
The final solution becomes

### 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)