# University of Florida/Egm4313/IEA-f13-team10/R3

## Report 3

### Problem 1

#### Problem Statement

Find the complete homogeneous solution using variation of parameters
${\displaystyle 16y''+8y'+y=r(x)}$

#### Solution

${\displaystyle 16y''+8y'+y=0}$

The solution is ${\displaystyle y=e^{rx}}$
Therefore, ${\displaystyle y'=re^{rx}}$ and ${\displaystyle y''=r^{2}e^{rx}}$

Plugging this back into the original homogeneous equation, ${\displaystyle e^{rx}(16r^{2}+8r+1)=0}$
${\displaystyle 16r^{2}+8r+1=0}$
${\displaystyle (4r+1)(4r+1)=0}$ so ${\displaystyle r=-1/4}$
${\displaystyle y_{h}=y_{1,h}+y_{2,h}=y_{1,h}+xy_{1,h}}$
${\displaystyle y_{h}(x)=cbar_{1}e^{-x/4}+cbar_{2}xe^{-x/4}}$

${\displaystyle y_{2}=u(x)y_{1}}$
${\displaystyle y=e^{-x/4}}$
${\displaystyle y=-1/4*e^{-x/4}}$
${\displaystyle y=1/16*e^{-x/4}}$
${\displaystyle y'_{2}=u'y_{1}+uy'_{1}}$
${\displaystyle y''_{2}=u''y_{1}+2u'y'_{1}+uy''}$

Plugging this into the original homogeneous equation
${\displaystyle 16(u''y_{1}+2u'y'_{1}+uy''_{1})+8(u'y_{1}+uy'_{1})+(uy)=0}$

Plugging in values for y and its derivatives, everything cancels out to zero.

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### Problem 2

#### Problem Statement

Find and plot the solution for the L2-ODE_CC
${\displaystyle y''-10y'+25y=r(x)}$ ${\displaystyle y(0)=1,y'(0)=1,r(x)=0}$

#### Solution

${\displaystyle y''-10y'+25y=r(x)}$
${\displaystyle y''-10y'+25y=0}$
This is a linear, first order ODE with constant coefficients.

To find the general solution to this ODE set ${\displaystyle y=e^{rx}}$

so that ${\displaystyle y'=re^{rx}}$ and ${\displaystyle y''=r^{2}e^{rx}}$

Substituting in y to the ODE and factoring out ${\displaystyle e^{rx}}$ we get:

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

Using the quadratic formula to solve for r we get

${\displaystyle r={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$ where ${\displaystyle a=1,b=-10,}$ and ${\displaystyle c=25}$

Solving to get ${\displaystyle r=5}$

Since we have a repeated root, we need to find v(x) so that y2(x) = v(x)y1(x)

Taking the first and second derivative of y2(x) we get:

${\displaystyle y_{2}'(x)=v'(x)y_{1}(x)+v(x)y_{1}'(x)}$
${\displaystyle y_{2}''(x)=v''(x)y_{1}(x)+2v'(x)y_{1}(x)+v(x)y_{1}''(x)}$

Substituting into the original ODE, we get:

${\displaystyle (v''(x)y_{1}(x)+2v'(x)y_{1}(x)+v(x)y_{1}''(x))-10(v'(x)y_{1}(x)+v(x)y_{1}'(x))+25(v(x)y1(x))=0}$
solving for ${\displaystyle v''(x)=0}$ so v(x) = kx + c

So y2(x) = x y1(x)

We get the general solution ${\displaystyle y(x)=C_{1}e^{5x}+C_{2}xe^{5x}}$

Now with the initial values y(0) = 1 and y'(0) = 0

${\displaystyle y(0)=C_{1}=1}$

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

${\displaystyle C_{1}=1}$ , ${\displaystyle C_{2}=-5}$

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

#### Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

### Problem 3

#### Problem Statement

##### Problem Sec 2.4 problem 3

How does the frequency of the harmonic oscillation change if we (i) double the mass (ii) take a spring of twice the modulus?

##### Problem Sec 2.4 problem 4

Could you make a harmonic oscillation move faster by giving the body a greater push?

#### Solution

##### Problem Sec 2.4 problem 3
###### Part 1

${\displaystyle \omega ={\sqrt {\frac {k}{m}}}}$
${\displaystyle f={\frac {\omega }{2\pi }}={\frac {\sqrt {\frac {k}{m}}}{2\pi }}}$
Now double the mass
${\displaystyle f={\frac {\omega }{2\pi }}={\frac {\sqrt {\frac {k}{2m}}}{2\pi }}}$
${\displaystyle f={\frac {\sqrt {\frac {k}{m}}}{2{\sqrt {2}}\pi }}}$
The frequency is decreased by ${\displaystyle {\sqrt {2}}}$.

###### Part 2

Multiply k by 2
${\displaystyle f={\frac {{\sqrt {2}}{\sqrt {\frac {k}{m}}}}{2\pi }}}$
The frequency is increased by ${\displaystyle {\sqrt {2}}}$.

##### Problem Sec 2.4 problem 4

No because frequency depends on the ratio of the spring modulus and mass.

#### Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

### Problem 4

#### Problem Statement

##### Section 2.4 Problem 16

Show the maxima of an underdamped motion occur at equidistant t-values and find the distance.

##### Section 2.4 Problem 17

Determine the values of t corresponding to the maxima and minima of the oscillation ${\displaystyle y(t)=e^{-t}sin(t)}$. Check your result by graphing y(t).

#### Part 1=

The general solution of underdamped motion is
${\displaystyle y(t)=Ce^{\alpha t}cos(\omega ^{*}t-\delta )}$
The maximas occur at ${\displaystyle y(t)=Ce^{\alpha t}}$
Set the two equations equal to each other a solve for t.
${\displaystyle Ce^{\alpha t}=Ce^{\alpha t}cos(\omega ^{*}t-\delta )}$
${\displaystyle 1=cos(\omega ^{*}t-\delta )}$
${\displaystyle \cos ^{-1}(1)=\omega ^{*}t-\delta }$
${\displaystyle t={\frac {2\pi n+\delta }{\omega ^{*}}}}$ where n=0,1,2,3.....

###### Part 2

${\displaystyle tan(\delta )={\frac {B}{A}}}$ shows delta is a constant.
The periodic distance between maximas is ${\displaystyle {\frac {2\pi }{\omega }}}$

##### Section 2.4 Problem 17

${\displaystyle y(t)=e^{-t}sin(t)}$
${\displaystyle y'(t)=e^{-t}[-sin(t)+cos(t)]}$

To find critical points, set y'(t)=0

${\displaystyle e^{-t}[-sin(t)+cos(t)]=0}$
${\displaystyle [-sin(t)+cos(t)]=0}$
${\displaystyle sin(t)=cos(t)}$
${\displaystyle {\frac {sin(t)}{cos(t)}}=1}$
${\displaystyle tan(t)=1}$
${\displaystyle t=\tan ^{-1}(1)}$
${\displaystyle t=n\pi +{\frac {\pi }{4}}}$where n=0,1,2,3...

As seen in the graph, the maximum of t was at ${\displaystyle {\frac {\pi }{4}}}$ and the minimum was at ${\displaystyle {\frac {5\pi }{4}}}$.

#### Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

### Problem 5

#### Problem Statement

Using the formula for Taylor series at x = 0 (the origin, i.e., McLaurin series), develop into Taylor series at the origin x = 0 for the following functions: cos x, sin x, exp(x), tan x, and write these series in compact form with the summation sign and a single summand.

#### Solution

Part 1
cos x
Taylor Series:
${\displaystyle f(x)=sum({\frac {1}{n!}})(f^{(}n)(x_{o}))(x-x_{o})^{n})}$
when ${\displaystyle x_{o}=0}$

${\displaystyle f(x)=\cos(x):}$

${\displaystyle f(0)=\cos(0)=1}$

${\displaystyle f'(0)=-\sin(0)=0}$

${\displaystyle f''(0)=-\cos(0)=-1}$

${\displaystyle f'''(0)=\sin(0)=0}$

${\displaystyle f(0)=sum({\frac {(-1)^{n}x^{2}n}{2n!}})}$

${\displaystyle f(0)=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+{\frac {x^{8}}{8!}}-{\frac {x^{1}0}{10!}}+....}$

n = 0, 1, 2, 3 ... N

2n = 0, 2, 4, 6, ... 2N

${\displaystyle cos(x)=sum({\frac {(-1)^{n}x^{2}n}{2n!}})}$

Part 2
sin x
Taylor Series:
${\displaystyle f(x)=sum({\frac {1}{n!}})(f^{(}n)(x_{o}))(x-x_{o})^{n})}$
when ${\displaystyle x_{o}=0}$

${\displaystyle f(x)=\sin(x):}$

${\displaystyle f(0)=\sin(0)=0}$

${\displaystyle f'(0)=\cos(0)=1}$

${\displaystyle f''(0)=-\sin(0)=0}$

${\displaystyle f'''(0)=-\cos(0)=-1}$

${\displaystyle f(0)=(-1)^{(}n+1)sum({\frac {x^{(}2n-1)}{(2n-1)!}})}$

${\displaystyle sin(x)={\frac {x}{1!}}-{\frac {x^{2}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+....}$

${\displaystyle sin(x)=(-1)^{(}n+1)sum({\frac {x^{(}2n-1)}{(2n-1)!}})}$

Part 3
exp x
Taylor Series:
${\displaystyle f(x)=sum({\frac {1}{n!}})(f^{(}n)(x_{o}))(x-x_{o})^{n})}$
when ${\displaystyle x_{o}=0}$

${\displaystyle f(x)=e^{x}:}$

${\displaystyle f(0)=e^{0}=1}$

${\displaystyle f'(0)=e^{0}=1}$

${\displaystyle f''(0)=e^{0}=1}$

${\displaystyle f'''(0)=e^{0}=1}$

${\displaystyle e^{x}={\frac {x}{1!}}+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+{\frac {x^{4}}{4!}}+....}$

${\displaystyle e^{x}=sum({\frac {x^{n}}{n!}})}$

#### Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

### Problem 6

#### Problem Statement

Part 1
Find and plot the solution for the L2-ODE-CC
${\displaystyle y''+4y'+13y=r(x)}$
Initial conditions: y(0) = 1, y'(0) = 0
No excitation: r(x) = 0

Part 2
In another Fig., superpose 3 Figs.: (a) this Fig.,
(b) the Fig. in R2.6 P. 5-6, (c) the Fig. in R2.1 P. 3-7.

#### Solution

Part 1
${\displaystyle y''+4y'+13y=0}$

The characteristic equation of the given ODE is:

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

Using the quadratic formula to solve for ${\displaystyle \lambda }$

${\displaystyle \lambda ={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$ where ${\displaystyle a=1,b=4,c=13}$

Solving to get ${\displaystyle \lambda ={\frac {-4\pm 6i}{2}}}$

${\displaystyle \lambda _{1}=-2+3i}$, ${\displaystyle \lambda _{2}=-2-3i}$

Therefore, the general solution of the given ODE is ${\displaystyle y=e^{-2x}[C_{1}\cos 3x+C_{2}\sin 3x]}$

Now we solve for ${\displaystyle C_{1}}$, ${\displaystyle C_{2}}$ using the given initial conditions

We have ${\displaystyle y(0)=1}$

Substituting ${\displaystyle x=0,y=1}$ into ${\displaystyle y=e^{-2x}[C_{1}\cos 3x+C_{2}\sin 3x]}$,

We get,

${\displaystyle 1=e^{-2(0)}[C_{1}\cos 3(0)+C_{2}\sin 3(0)]}$

${\displaystyle 1=C_{1}\cos 3(0)+C_{2}\sin 3(0)}$

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

We have ${\displaystyle y'(0)=0}$

Differentiating ${\displaystyle y=e^{-2x}[C_{1}\cos 3x+C_{2}\sin 3x]}$, we get:

${\displaystyle y'=-2e^{-2x}[C_{1}\cos 3x+C_{2}\sin 3x]+e^{-2x}[-3C_{1}\cos 3x+3C_{2}\sin 3x]}$
.

Substituting ${\displaystyle x=0,y'=0}$ into ${\displaystyle y'=-2e^{-2x}[C_{1}\cos 3x+C_{2}\sin 3x]+e^{-2x}[-3C_{1}\cos 3x+3C_{2}\sin 3x]}$, we get:

${\displaystyle 0=-2e^{-2(0)}[C_{1}\cos 3(0)+C_{2}\sin 3(0)]+e^{-2(0)}[-3C_{1}\cos 3(0)+3C_{2}\sin 3(0)]}$

${\displaystyle 0=-2C_{1}+3C_{2}}$

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

${\displaystyle 2(1)=3C_{2}}$

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

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

Hence the solution of the given ODE is ${\displaystyle y=e^{-2x}[\cos 3x+2/3\sin 3x]}$

Fig. 1:

Part 2
${\displaystyle x=0:0.2:20}$
${\displaystyle y=e^{-2x}[\cos 3x+2/3\sin 3x]}$
${\displaystyle y_{2}=e^{5x}-5xe^{5x}}$
${\displaystyle y_{3}={\frac {5}{7}}e^{-2x}+{\frac {2}{7}}e^{5x}}$

Fig. 2:

Fig. 3:

Fig. 4:

#### Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

### Problem 7

#### Problem Statement

Consider the same system as in the Example p.7-3, i.e., the same L2-ODE-CC (4) p.5-5 and initial condi- (2) p.3-4, but with the following excitation: ${\displaystyle r(x)=7e^{5x}-2x^{2}}$

#### Solution

${\displaystyle y''-10y'+25y=7e6{5x}-2x^{2}}$

${\displaystyle y_{p}=C_{1}x^{2}e^{5x}-C_{2}x^{3}}$

Replacing ${\displaystyle y}$ with ${\displaystyle y_{p}}$ and after simplifying we get,

${\displaystyle 2C_{1}e^{5x}-C_{2}x(25x^{2}-30x+6)=7e^{5x}-2x}$

The root here is ${\displaystyle \lambda =5}$. So we can solve for our constants,

${\displaystyle C_{1}={\frac {7}{2}}}$
${\displaystyle C_{2}={\frac {2}{481}}}$

${\displaystyle y(x)=y_{h}+y_{p}=c_{1}e^{5x}+c_{2}xe^{5x}+{\frac {7}{2}}x^{2}e^{5x}-{\frac {2}{481}}x^{3}}$

Using the initial conditions y(0)=4 and y'(0)=-5 we can solve for our constants,

${\displaystyle c_{1}=4}$
${\displaystyle c_{2}=-1}$

So the solution is,

${\displaystyle y=4e^{5x}+xe^{5x}+{\frac {7}{2}}x^{2}e^{5x}-{\frac {2}{481}}x^{3}}$

#### Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

### Problem 8

#### Problem Statement

Plot the error between the exact derivative and the approximate derivative, i.e.

${\displaystyle xe^{\lambda x}-{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}}$ from

${\displaystyle -15\leq x\leq 15}$

For ε = .0001, .0003, .0006, and .001 and λ =.3

#### Solution

Since the above equation is the error between the exact derivative and the approximate derivative, it must be plotted with the correct values of \epsilon and \lambda, from x = -15 to 15

ε=.0001

ε=.0003

ε=.0006

ε=.001

#### Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.

### Problem 9

#### Problem Statement

Find the complete solution for ${\displaystyle y''-3y'+2y=4x^{2}}$, with the initial conditions
${\displaystyle y(0)=1}$, ${\displaystyle y'(0)=0}$
plot the solution y(x)

#### Solution

Particular Solution
${\displaystyle y_{p}(x)=Ax^{2}+Bx+c}$
${\displaystyle y_{p}'=2Ax+B}$
${\displaystyle y_{p}''=2A}$
${\displaystyle 2A-6Ax-3B+2Ax^{2}+2Bx+2c=4x^{2}}$
${\displaystyle 2A=4}$

${\displaystyle -6A+2B=0}$

${\displaystyle 2A-3B+2C=0}$

${\displaystyle A=2}$
${\displaystyle B=6}$
${\displaystyle C=7}$

${\displaystyle y_{p}(x)=2x^{2}+6x+7}$

Homogeneous Solution
${\displaystyle y_{h}(x)=c_{1}e^{x}+c_{2}e^{(}2x)}$
${\displaystyle y_{h}'=c_{1}e^{x}+2c_{2}e^{(}2x)}$

Initial conditions
${\displaystyle y(0)=1}$
${\displaystyle 1=c_{1}+c_{2}}$

${\displaystyle y'(0)=0}$
${\displaystyle 0=c_{1}+2c_{2}}$

${\displaystyle c_{1}=2,c_{2}=-1}$

General Solution
${\displaystyle y_{g}=2e^{x}-e^{(}2x)+2x^{x}+6x+7}$

#### Honor Pledge

On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.