# University of Florida/Egm4313/s12.team14.report2
## Problem 1
### Find

Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $\displaystyle r(x)$
Consider no excitation:
    $\displaystyle r(x)=0$ $\displaystyle (Eq.2)$
Plot the solution.

### Solution

Characteristic equation:
    $\displaystyle (\lambda -\lambda _{1})(\lambda -\lambda _{2})=0$ $\displaystyle (Eq.3)$
Substituting $\displaystyle (Eq.1a)$ into $\displaystyle (Eq.3)$:
    $\displaystyle (\lambda +2)(\lambda -5)=0$ $\displaystyle (Eq.4)$
    $\displaystyle \lambda ^{2}-3\lambda -10=0$ $\displaystyle (Eq.5)$
Non-homogeneous solution:
    $\displaystyle y''-3y'-10y=0$ $\displaystyle (Eq.6)$
Homogeneous solution:
    $\displaystyle y_{h}(x)=C_{1}e^{-2x}+C_{2}e^{5x}$ $\displaystyle (Eq.7)$
Overall solution:
    $\displaystyle y(x)=C_{1}e^{-2x}+C_{2}e^{5x}+y_{p}(x)$ $\displaystyle (Eq.8)$
No excitation:
    $\displaystyle r(x)=0=>y_{p}(x)=0$ $\displaystyle (Eq.9)$
From intitial conditions:
    $\displaystyle y(0)=1=C_{1}+C_{2}$ $\displaystyle (Eq.10)$
    $\displaystyle y'(0)=0=-2C_{1}+5C_{2}$ $\displaystyle (Eq.11)$
Solving for $\displaystyle C_{1}$ and $\displaystyle C_{2}$:
    $\displaystyle C_{1}={\frac {5}{7}}$ $\displaystyle$
and
    $\displaystyle C_{2}={\frac {2}{7}}$ $\displaystyle$
So, the final solution is:

## Problem 2
### Find

Find and plot the solution for $\displaystyle (Eq.1)$

### Solution

Due to no excitation, $\displaystyle (Eq.1)$ becomes:
    $\displaystyle y''-10y'+25y=0$ $\displaystyle (Eq.4)$
Substituting $\displaystyle \lambda$ into $\displaystyle (Eq.4)$:
    $\displaystyle \lambda ^{2}-10\lambda +25=0$ $\displaystyle (Eq.5)$
Factoring, and solving for $\displaystyle \lambda$:
    $\displaystyle (\lambda -5)^{2}=0$ $\displaystyle (Eq.6)$
    $\displaystyle \lambda _{2}=\lambda _{1}=\lambda =5$ $\displaystyle$
Since $\displaystyle \lambda$ is a double root, the general solution:
    $\displaystyle y=C_{1}e^{-5x}+C_{2}xe^{-5x}$ $\displaystyle (Eq.7)$
From intitial conditions:
    $\displaystyle y(0)=1=C_{1}$ $\displaystyle (Eq.8)$
    $\displaystyle y'(0)=0=5C_{1}+C_{2}$ $\displaystyle (Eq.9)$
Solving for $\displaystyle C_{1}$ and $\displaystyle C_{2}$:
    $\displaystyle C_{1}=1$ $\displaystyle$
and
    $\displaystyle C_{2}=-5$ $\displaystyle$
So, the final solution is:

## Problem 3
### Given

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

### Find

General solution to the differential equation

### Solution

(a)
Characteristic equation:
    $\displaystyle {\lambda ^{2}+6\lambda +8.96=0}$
Using the quadratic equation:
    $\displaystyle {x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\,}$
where:
    $\displaystyle {a=1,b=6,c=8.96}$
Place values into quadratic equation:
    $\displaystyle {x={\frac {-6\pm {\sqrt {6^{2}-4*1*8.96}}}{2*1}}\,}$
    $\displaystyle {x={\frac {-6\pm {\sqrt {36-35.84}}}{2}}\,}$
    $\displaystyle {x={\frac {-6\pm 0.4}{2}}\,}$
    $\displaystyle {x={\frac {-6.4}{2}}\,,{\frac {-5.6}{2}}}$
    $\displaystyle {x=-3.2,-2.8}$
(b)
Characteristic equation:
    $\displaystyle {\lambda ^{2}+4\lambda +(\pi ^{2}+4)=0}$
Using the quadratic equation:
    $\displaystyle {\lambda ={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\,}$
where:
$\displaystyle {a=1,b=4,c=(\pi ^{2}+4)}$
Place values into quadratic equation:
    $\displaystyle {\lambda ={\frac {-4\pm {\sqrt {4^{2}-4*1*(\pi ^{2}+4)}}}{2*1}}\,}$
    $\displaystyle {\lambda ={\frac {-4\pm {\sqrt {16-4\pi ^{2}-16)}}}{2}}\,}$
    $\displaystyle {\lambda ={\frac {-4\pm 2\pi i}{2}}\,}$
    $\displaystyle {\lambda =-2\pm \pi i}$

## Problem 4
### Given

(5)
    $\displaystyle {y}''+2\pi {y}'+{\pi }^{2}y=0$ $\displaystyle (Eq.1)$
(6)
    $\displaystyle {10{y}''-32{y}'+25.6y=0}$ $\displaystyle (Eq.2)$
h

### Find

For both (5) and (6), find a general solution. Then, check answers by substitution.

### Solution

(5)
To obtain the general solution for (Eq. 1), we let
    $\displaystyle y=e^{\lambda x}$
This yields:
    $\displaystyle e^{\lambda x}(\lambda ^{2}+2\pi \lambda +\pi ^{2})=0$
Therefore, we have:
    $\displaystyle \lambda ^{2}+2\pi \lambda +\pi ^{2}=0$
Where $\displaystyle \lambda$ is a solution to the characteristic equation
    $\displaystyle \lambda ^{2}+a\lambda +b=0$
We can see that
    $\displaystyle a=2\pi$ and
    $\displaystyle b=\pi ^{2}$.
Next, we must examine the quadratic formula to determine whether the system is over-, under-, or critically-damped. The discriminant is:
    $\displaystyle \triangle =a^{2}-4b$
    $\displaystyle \triangle =(2\pi ^{2})^{2}-4(\pi ^{2})$
    $\displaystyle \triangle =0$
The general solution is therefore represented by:
    $\displaystyle y=c_{1}e^{\lambda _{1}x}+c_{2}xe^{\lambda _{2}x}$
Where $\displaystyle \lambda _{1}=\lambda _{2}=\lambda$.
We can determine the value of $\displaystyle \lambda$ by again using the quadratic equation, this time in full:
    $\displaystyle \lambda ={\frac {1}{2}}(-a+{\sqrt {a^{2}-4b}})$
for which we find that $\displaystyle \lambda =-\pi$.
Therefore, the general solution to the equation is:
Verification by substitution:
    $\displaystyle {y}'=-\pi c_{1}e^{-\pi x}+c_{2}e^{-\pi x}-\pi c_{2}xe^{-\pi x}$
    $\displaystyle {y}''={\pi }^{2}c_{1}e^{-\pi x}-\pi c_{2}e^{-\pi x}+{\pi }^{2}c_{2}xe^{-\pi x}-\pi c_{2}e^{-\pi x}$
Plugging into Eq. 1 and combining terms, it is found:
    $\displaystyle c_{1}e^{-\pi x}({\pi }^{2}-2{\pi }^{2}+{\pi }^{2})+c_{2}e^{-\pi x}(-\pi -\pi +2\pi )+c_{2}xe^{-\pi x}({\pi }^{2}-2{\pi }^{2}+{\pi }^{2})=0$
All the terms cancel, so the statement is true, and the solution is verified.
(6)
To obtain the general solution for (Eq. 2), let
    $\displaystyle y=e^{\lambda x}$
So then, we have
    $\displaystyle 10\lambda ^{2}-32\lambda +25.6=0$
as a solution to the characteristic equation. All terms must be divided by 10, to put the equation in standard form:
    $\displaystyle \lambda ^{2}-3.2\lambda +2.56=0$
Examining the discriminant, we find that
    $\displaystyle \triangle =a^{2}-4b=(-3.2)^{2}-4(2.56)=0$
The general solution is represented by:
    $\displaystyle y=c_{1}e^{\lambda _{1}x}+c_{2}xe^{\lambda _{2}x}$
and $\displaystyle \lambda _{1}=\lambda _{2}=\lambda$.
To determine the value of $\displaystyle \lambda$, we set:
    $\displaystyle \lambda ={\frac {1}{2}}(-a+{\sqrt {a^{2}-4b}})$
for which we find that $\displaystyle \lambda =1.6$.
Therefore, the general solution to the equation is:
Verification by substitution:
    $\displaystyle {y}'=1.6c_{1}e^{1.6x}+c_{2}e^{1.6x}+1.6c_{2}xe^{1.6x}$
    $\displaystyle {y}''=(1.6^{2})c_{1}e^{1.6x}+3.2c_{2}e^{1.6x}+(1.6^{2})c_{2}xe^{1.6x}$
Plugging into Eq. 1 and combining terms, it is found:
    $\displaystyle c_{1}e^{1.6x}(0)+c_{2}e^{1.6x}(0)+c_{2}xe^{1.6x}(0)=0$
All the terms cancel, so the statement is true, and the solution is verified.

## Problem 5
### Given

    $\displaystyle {y}''+a{y}'+by=0$
Given the basis:
(a)
    $\displaystyle e^{2.6x},e^{-4.3x}$
(b)
    $\displaystyle e^{-{\sqrt {5}}x},xe^{-{\sqrt {5}}x}$

### Find

For the given Information above, Find an ODE for both (a) and (b)

### Solution

(a)The general solution can be written as:
    $\displaystyle y=C_{1}e^{2.6x}+C_{2}e^{-4.3x}$
The characteristic equation can be written in the following way:
    $\displaystyle (\lambda -2.6)(\lambda +4.3)=\lambda ^{2}+1.7\lambda -11.18=0$
    $\displaystyle {y}''+1.7{y}'-11.18y=0$
(b) The general solution can be written as:
    $\displaystyle y=C_{1}e^{-{\sqrt {5}}x}+C_{2}xe^{-{\sqrt {5}}x}$
The characteristic equation can be written in the following way:
    $\displaystyle (\lambda +{\sqrt {5}})(\lambda +{\sqrt {5}})=\lambda ^{2}+2{\sqrt {5}}\lambda +5=0$
    $\displaystyle {y}''+2{\sqrt {5}}{y}'+5y=0$

## Problem 6
### Given

Spring-dashpot equation of motion from sec 1-5
    $\displaystyle {my{_{k}}''+m{\frac {k}{c}}y{_{k}}'+ky_{k}=f(t)}$

### Find

Spring-dashpot-mass system in series. Find the values for the parameters k,c,m with a double real root of $\lambda =-3$

### Solution

Consider the double real root
    $\displaystyle {\lambda =-3}$
Characteristic equation is
    $\displaystyle {(\lambda +3)^{2}=\lambda ^{2}+6\lambda +9=0}$ $\displaystyle (Eq.1)$
Recall
    $\displaystyle {my{_{k}}''+m{\frac {k}{c}}y{_{k}}'+ky_{k}=f(t)}$ $\displaystyle (Eq.2)$
Thus
    $\displaystyle {m=1}$
    $\displaystyle {m{\frac {k}{c}}=6}$
    $\displaystyle {k=9}$
Solve for c
    $\displaystyle {c={\frac {9}{6}}={\frac {3}{2}}}$
Therefore

## Problem 7
### Given

Taylor Series at t=0

### Find

Develop the MacLaurin Series for $e^{t},\cos t,\sin t\!$

### Solution

Taylor series is defined as
    $\sum _{n=0}^{\infty }{\frac {f^{(n)}(t)}{n!}}\,(x-t)^{n}$ $\displaystyle (Eq.1)$
The MacLaurin series occurs when t=0
    $\sum _{n=0}^{\infty }{\frac {f^{(n)}(0)}{n!}}\,(x)^{n}$ $\displaystyle (Eq.2)$
Development of MacLaurin series for $e^{t}$
    $e^{t}={\frac {1}{0!}}+{\frac {1}{1!}}t+{\frac {1}{2!}}t^{2}+{\frac {1}{3!}}t^{3}...$ $\displaystyle (Eq.3)$
Final Answer
Development of MacLaurin series for $cos(t)$
    $cos(t)={\frac {1}{0!}}+{\frac {0}{1!}}t-{\frac {1}{2!}}t^{2}+{\frac {0}{3!}}t^{3}+{\frac {1}{4!}}t^{4}...$ $\displaystyle (Eq.4)$
Final Answer
Development of MacLaurin series for $sin(t)$
    $sin(t)={\frac {0}{0!}}+{\frac {1}{1!}}t-{\frac {0}{2!}}t^{2}-{\frac {1}{3!}}t^{3}+{\frac {0}{4!}}t^{4}+{\frac {1}{5!}}t^{5}...$ $\displaystyle (Eq.5)$
Final Answer

## Problem 8
### Given

(8)
    $\displaystyle {y}''+y'+3.25y=0$ $\displaystyle (Eq.1a)$
(15)
    $\displaystyle {y}''+(0.54)y'+(0.0729+\pi )y=0$ $\displaystyle (Eq.1b)$

### Solution

(8)
Assume that the solution is of the form
    $\displaystyle y_{h}=e^{\lambda x}$ $\displaystyle (Eq.2)$
    $\displaystyle \therefore y'_{h}=\lambda e^{\lambda x}$ $\displaystyle (Eq.3)$
And
    $\displaystyle y''_{h}=\lambda ^{2}e^{\lambda x}$ $\displaystyle (Eq.4)$
Substituting equations 2, 3, and 4 into equation 1a yields
$\displaystyle \lambda ^{2}e^{\lambda x}+\lambda e^{\lambda x}+3.25e^{\lambda x}=0$
$\displaystyle e^{\lambda x}(\lambda ^{2}+\lambda +3.25)=0$
Since
$\displaystyle e^{\lambda x}\neq 0$
$\displaystyle \therefore (\lambda ^{2}+4\lambda +3.25)=0$
We must use the quadratic formula to obtain values for lambda
$\displaystyle \lambda ={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$
$\displaystyle \lambda ={\frac {-1\pm {\sqrt {(1)^{2}-4(1)(3.25)}}}{2(1)}}$
$\displaystyle \lambda ={\frac {-1\pm {\sqrt {1-13)}}}{2}}$
$\displaystyle \lambda ={\frac {-1\pm {\sqrt {-12}}}{2}}$
$\displaystyle \lambda =-0.5\pm j{\sqrt {3}}$
Where $\displaystyle j={\sqrt {-1}}$
$\displaystyle y_{h}={C_{1}}e^{(-0.5+j{\sqrt {3}})x}+{C_{2}}e^{(-0.5-j{\sqrt {3}})x}$
In order to verify solution, we must evaluate it's first and second derivative, then plug them into the equation
Let $\displaystyle a=-0.5,b={\sqrt {3}}$
$\displaystyle y_{h}={C_{1}}e^{(a+jb)x}+{C_{2}}e^{(a-jb)x}$
$\displaystyle y_{h}={C_{1}}e^{ax}e^{jbx}+{C_{2}}e^{ax}e^{-jbx}$
$\displaystyle {y_{h}}'={C_{1}}[e^{ax}({jb}e^{jbx})+e^{jbx}({a}e^{ax})]+{C_{2}}[e^{ax}(-{jb}e^{jbx})+e^{jbx}({a}e^{ax})]$
$\displaystyle {y_{h}}''={C_{1}}[e^{ax}((jb)^{2}e^{jbx})+({jb}e^{jbx})({a}e^{ax})+e^{jbx}((a)^{2}e^{ax})+({a}e^{ax})({jb}e^{jbx})]+{C_{2}}[e^{ax}((jb)^{2}e^{-jbx})+(-{jb}e^{-jbx})({a}e^{ax})+e^{-jbx}((a)^{2}e^{ax})+({a}e^{ax})(-{jb}e^{-jbx})]$
Plugging $\displaystyle y_{h},{y_{h}}',{y_{h}}''$ into Eq. 1 and collecting terms gives
$\displaystyle 0={C_{1}}e^{ax}e^{jbx}[-b^{2}+jab+a^{2}+jab+jb+a+3.25]+{C_{2}}e^{ax}e^{-jbx}[-b^{2}-jab+a^{2}-jab-jb+a+3.25]$
$\displaystyle 0={C_{1}}e^{-0.5x}e^{j{{\sqrt {3}}x}}[-3-j{\sqrt {3}}/2+1/4-j{\sqrt {3}}/2+j{\sqrt {3}}-1/2+3.25]+{C_{2}}e^{-0.5x}e^{-j{\sqrt {3}}x}[-3+j{\sqrt {3}}/2+1/4+j{\sqrt {3}}/2-j{\sqrt {3}}-1/2+3.25]$
$\displaystyle 0={C_{1}}e^{-0.5x}e^{j{\sqrt {3}}x}(0)+{C_{2}}e^{-0.5x}e^{-j{\sqrt {3}}x}(0)$
Therefore the solution holds for any values of $\displaystyle C_{1},C_{2},x$
(15)
Assume that the solution is of the form
    $\displaystyle y_{h}=e^{\lambda x}$ $\displaystyle (Eq.2)$
    $\displaystyle \therefore y'_{h}=\lambda e^{\lambda x}$ $\displaystyle (Eq.3)$
And
    $\displaystyle y''_{h}=\lambda ^{2}e^{\lambda x}$ $\displaystyle (Eq.4)$
Substituting equations 2, 3, and 4 into equation 1b yields
$\displaystyle \lambda ^{2}e^{\lambda x}+(0.54)\lambda e^{\lambda x}+(0.0729+\pi )e^{\lambda x}=0$
$\displaystyle e^{\lambda x}(\lambda ^{2}+(0.54)\lambda +(0.0729+\pi ))=0$
Since
$\displaystyle e^{\lambda x}\neq 0$
$\displaystyle \therefore (\lambda ^{2}+(0.54)\lambda +(0.0729+\pi ))=0$
We must use the quadratic formula to obtain values for lambda
$\displaystyle \lambda ={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$
$\displaystyle \lambda ={\frac {-(0.54)\pm {\sqrt {(0.54)^{2}-4(1)(0.0729+\pi )}}}{2(1)}}$
$\displaystyle \lambda ={\frac {-0.54\pm j{3.545}}{2}}$
$\displaystyle \lambda =-0.27\pm j(1.772)$
Where $\displaystyle j={\sqrt {-1}}$
$\displaystyle y_{h}={C_{1}}e^{(-0.27+j(1.772))x}+{C_{2}}e^{(-0.27-j(1.772))x}$
In order to verify solution, we must evaluate it's first and second derivative, then plug them into the equation
Let $\displaystyle a=-0.27,b=1.772$
$\displaystyle y_{h}={C_{1}}e^{(a+jb)x}+{C_{2}}e^{(a-jb)x}$
$\displaystyle y_{h}={C_{1}}e^{ax}e^{jbx}+{C_{2}}e^{ax}e^{-jbx}$
$\displaystyle {y_{h}}'={C_{1}}[e^{ax}({jb}e^{jbx})+e^{jbx}({a}e^{ax})]+{C_{2}}[e^{ax}(-{jb}e^{jbx})+e^{jbx}({a}e^{ax})]$
$\displaystyle {y_{h}}''={C_{1}}[e^{ax}((jb)^{2}e^{jbx})+({jb}e^{jbx})({a}e^{ax})+e^{jbx}((a)^{2}e^{ax})+({a}e^{ax})({jb}e^{jbx})]+{C_{2}}[e^{ax}((jb)^{2}e^{-jbx})+(-{jb}e^{-jbx})({a}e^{ax})+e^{-jbx}((a)^{2}e^{ax})+({a}e^{ax})(-{jb}e^{-jbx})]$
Plugging $\displaystyle y_{h},{y_{h}}',{y_{h}}''$ into Eq. 1 and collecting terms gives
$\displaystyle 0={C_{1}}e^{ax}e^{jbx}[-b^{2}+jab+a^{2}+jab+(0.54)(jb+a)+0.0729+\pi ]+{C_{2}}e^{ax}e^{-jbx}[-b^{2}-jab+a^{2}-jab+(0.54)(-jb+a)+0.0729+\pi ]$
$\displaystyle 0={C_{1}}e^{-0.27x}e^{j{1.772x}}[-(1.772)^{2}+j(-0.27)(1.772)+(0.27)^{2}+j(-0.27)(1.772)+(0.54)(j(1.772)-0.27)+0.0729+\pi ]+{C_{2}}e^{-0.27x}e^{-j1.772x}[-(1.772)^{2}-j(-0.27)(1.772)+(0.27)^{2}-j(-0.27)(1.772)+(0.54)(-j(1.772)-0.27)+0.0729+\pi ]$
$\displaystyle 0={C_{1}}e^{-0.27x}e^{j1.772x}(0)+{C_{2}}e^{-0.27x}e^{-j1.772x}(0)$
Therefore the solution holds for any values of $\displaystyle C_{1},C_{2},x$

## Problem 10
### Given

Differential equation, initial conditions, and forcing function as shown:
    $\displaystyle y''-10y'+25y=r(x),$ $\displaystyle (Eq.1)$
    $\displaystyle y(0)=4,y'(0)=-5,r(x)=7e^{5x)-2x^{2}.$ $\displaystyle (Eq.1)$

### Find

The solution to Eq. 1

### Solution

    $\displaystyle y_{p}=C_{1}x^{2}e^{5x}+C_{2}x^{2}+C_{3}x+C_{4}$ $\displaystyle (Eq.2)$
    $\displaystyle y_{p}'=5C_{1}x^{2}e^{5x}+2C_{1}xe^{5x}+2C_{2}x+C_{3}$ $\displaystyle (Eq.3)$
    $\displaystyle y_{p}''=25C_{1}x^{2}e^{5x}+20C_{1}xe^{5x}+2C_{1}e^{5x}+2C_{2}$ $\displaystyle (Eq.4)$
Substituting Equations 2, 3, and 4 into Equation 1 yields
    $\displaystyle 7e^{5x}-2x^{2}=2C_{1}e^{5x}+25C_{2}x^{2}+25C_{3}x-20C_{2}x+2C_{2}-10C_{3}+25C_{4}$ $\displaystyle (Eq.5)$
To solve for the constants:
    $\displaystyle 7=2C_{1}$ $\displaystyle (Eq.6)$
    $\displaystyle -2=25C_{2}$ $\displaystyle (Eq.7)$
    $\displaystyle 0=25C_{3}-20C_{2}$ $\displaystyle (Eq.8)$
    $\displaystyle 0=2C_{2}-10C_{3}+25C_{4}$ $\displaystyle (Eq.9)$
Solving Equations 6-9 yields
    $\displaystyle C_{1}={\frac {7}{2}},C_{2}=-{\frac {2}{25}},C_{3}=-{\frac {8}{125}},C_{4}=-{\frac {12}{625}}$ $\displaystyle$
The homogeneous solution:
    $\displaystyle y_{h}=C_{5}e^{5x}+C_{6}xe^{5x}$ $\displaystyle (Eq.10)$
The total solution:
    $\displaystyle y=C_{5}e^{5x}+C_{6}xe^{5x}+2{\frac {7}{2}}x^{2}e^{5x}-{\frac {2}{25}}x^{2}-{\frac {8}{125}}x-{\frac {12}{625}}$ $\displaystyle (Eq.11)$
Setting $\displaystyle x=0$ yields
    $\displaystyle y(0)=C_{5}-{\frac {12}{625}}=4$ $\displaystyle (Eq.12)$
Solving for $\displaystyle C_{5}$ yields
    $\displaystyle C_{5}={\frac {2512}{625}}$ $\displaystyle (Eq.13)$
Finding $\displaystyle y'$ yields
    $\displaystyle y'={\frac {35}{2}}x^{2}e^{5x}+5C_{6}xe^{5x}+7xe^{5x}+{\frac {2512}{125}}e^{5x}+C_{6}e^{5x}-{\frac {4}{25}}x^{2}-{\frac {8}{125}}$ $\displaystyle (Eq.14)$
Setting $\displaystyle x=0$ yields
    $\displaystyle y'(0)={\frac {2512}{125}}+C_{6}-{\frac {8}{125}}=-5$ $\displaystyle (Eq.15)$
Solving for $\displaystyle C_{6}$ yields
    $\displaystyle C_{6}=-{\frac {3129}{125}}$ $\displaystyle (Eq.16)$
Therefore, the total solution is
    $\displaystyle y={\frac {2512}{625}}e^{5x}-{\frac {3129}{125}}xe^{5x}+2{\frac {7}{2}}x^{2}e^{5x}-{\frac {2}{25}}x^{2}-{\frac {8}{125}}x-{\frac {12}{625}}$ $\displaystyle (ANSEq.17)$

## Team Member Tasks

| Name | Responsibilities | Checked by |
| Bo Turano | Problem R2. | -- |
| David Parsons | Problem R2.4 | -- |
| Dean Pickett | Problem R2.8 | -- |
| Giacomo Savardi | Problem R2. | -- |
| Isaac Kimiagarov | Problem R2.3 | -- |
| Kyle Steiner | Problem R2. | -- |
| Tony Han | Problem R2.6, R2.7 | -- |
All team members contributed to the coding of this page.