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

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 ${\displaystyle r=-1/4}$
${\displaystyle y_{h}=y_{1,h}+y_{2,h}=y_{1,h}+xy_{1,h}}$
${\displaystyle y_{h}(x)=cbar_{1}e^{-x/4}+cbar_{2}xe^{-x/4}}$

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

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

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 Statement

Solution

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

Problem Statement

Solution

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.....

Honor Pledge

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

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 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 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 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

Honor Pledge

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

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.