# University of Florida/Egm4313/f13-team9-R1

## Problem 1.1 (Pb-10.1 in sec.10.)

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

## Problem 1.2 (Sec. 1, Pb 1-2)

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

### Problem Statement

Derive the equation of motion of the mass-spring-dashpot in Fig. 53 in K2011 p.85 with applied force r(t) on the ball.

## Problem 1.4 ( Sec. 2, Pb 2-1)

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

## Problem 1.5 ( P 2.2.5, P 2.2.12, Kreyszig, 2011)

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

### Problem 2.2.12

#### Problem Statement

Solve the initial value problem and graph the solution over the intervals $\displaystyle 0\leq x\leq 1\,,\,0\leq x\leq 5$ $y''+9y'+20=0$ (1)

#### Given

$\displaystyle y(0)=1$ $\displaystyle y'(0)={\frac {1}{2}}$ #### Solution

##### Step 1: Find a General Solution

The ODE is a linear, second-order, homogeneous differential equation with constant coefficients. So, the following equation was chosen as a solution.

 $y=e^{\lambda x}$ (2)

The first and second derivatives are as follows:

 $y'=\lambda e^{\lambda x}$ (3)
 $y''=\lambda ^{2}e^{\lambda x}$ (4)

Plugging the solution and its derivatives back into the original ODE, we receive

 $(\lambda ^{2}+9\lambda +20)e^{\lambda x}=0$ (5)

and the characteristic equation

 $\lambda ^{2}+9\lambda +20=0$ (6)

This gives us 2 real solutions from the quadratic formula, $\displaystyle \lambda _{1}=-4\,,\,\lambda _{2}=-5$ and the general solution:

 $y=c_{1}e^{-4x}+c_{2}e^{-5x}$ (7)
##### Step 2: Solve the IVP

Equation (7) and its derivative

 $y'=-4c_{1}e^{-4x}-5c_{2}e^{-5x}$ (8)

can be set equal to the initial values given

 $y(0)=1=c_{1}+c_{2}$ (9)
 $y'(0)={\frac {1}{2}}=-4c_{1}-5c_{2}$ (10)

Solving (9) and (10) simultaneously gives us the c-values and the solution to the IVP

 $y=5.5e^{-4x}-4.5e^{-5x}$ (11)
##### Step 3: Check Answer with Substitution

Our solution and its first two derivatives can be substituted into the original ODE

 $y''+8y'+20y=0$ (12)
 $y=5.5e^{-4x}-4.5e^{-5x}$ (13)
 $y'=-22e^{-4x}+22.5e^{-5x}$ (14)
 $y''=88e^{-4x}-112.5e^{-5x}$ (15)
 $[88+9(-22)+20(5.5)]e^{-4x}+[-112.5+9(22.5)+20(-4.5)]e^{-5x}=0$ (16)

Which is true.

## Problem 1.6 (P3.17, Beer2012)

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.