text:
User:Egm4313.s12.team9/R1

= R1.1: Spring-dash pot system in parallel with a mass and applied force f(t) =

Initial Information
From [[media:Egm4313.s12.team9.1-4.jpg|Lecture slide 1-4]] and [[media:Egm4313.s12.team9.1-5.jpg|Lecture slide 1-5]]:

Variables:
 * {| style="width:100%" border="0" align="left"

k, c, m, f(t), y(t), y_k, y_c, f_i, f_k, f_c $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Methods
Kinematics:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle y = y_k = y_c$$
 * $$\displaystyle Derived (Eq.1)$$
 * }

Kinetics:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle f(t) = my'' + f_i$$
 * $$\displaystyle (Eq.2)$$
 * }

 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle f_i = f_k = f_c$$
 * $$\displaystyle (Eq.3)$$
 * }

 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle f(t) = my'' + f_k + f_c$$
 * $$\displaystyle Derived (Eq.2,Eq.3)$$
 * }

Spring Force:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle f_k = ky_k$$
 * $$\displaystyle (Eq.4)$$
 * }

Dash-Pot Force:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle f_c = cy'_c$$
 * $$\displaystyle (Eq.5)$$
 * }

Solution
Final Equation:
 * {| style="width:100%" border="0" align="left"

$$\displaystyle f(t) = my''_k + cy'_k + ky_k$$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle $$
 * }

=R1.2: Spring-mass-dashpot with applied force r(t) on the ball(Fig. 53, p.85, K2011)=

Initial Information
Variables:
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle k, c, m, r(t), y_k, f_k, f_c $$
 * $$\displaystyle (Eq.1)
 * $$\displaystyle (Eq.1)
 * }
 * }

Methods
Kinematics:

 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle y=y_k $$
 * $$\displaystyle (Eq.1)
 * $$\displaystyle (Eq.1)
 * }
 * }

Kinetics:

 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle my'' + f_k + f_c = r(t) $$
 * $$\displaystyle (Eq.2)
 * $$\displaystyle (Eq.2)
 * }
 * }

Spring Force:

 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle f_k = ky_k $$
 * $$\displaystyle (Eq.3)
 * $$\displaystyle (Eq.3)
 * }
 * }

Dashpot Force:

 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle f_c = cy'_c $$
 *
$$\displaystyle (Eq.4)
 *
$$\displaystyle (Eq.4)
 * }
 * }

Solution

 * {| style="width:0%" border="0" align="left"

$$\displaystyle my'' + ky_k + cy'_c = r(t) $$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

= R1.3 Spring-dashpot-mass system FBD and Derivation of Equation of Motion = This problem can be found on [[media:Egm4313.s12.team9.1-6.jpg|Lecture slide 1-6]].

Given
The spring-dashpot-mass system can be found on [[media:Egm4313.s12.team9.1-4.jpg|Lecture slide 1-4]]

Methods
Free body diagrams:

The general equilibrium expression in the y direction is:
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle \sum F_{y}=ma $$
 *
$$\displaystyle (Eq.1)
 *
$$\displaystyle (Eq.1)
 * }
 * }

From the FBD's it is apparent that:

 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle f_{I}=f_{k}=f_{c} $$
 *
$$\displaystyle (Eq.2)
 *
$$\displaystyle (Eq.2)
 * }
 * }

Where $$ f_{I} $$ is the internal force. Thus, substituting gives:

 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle ma=my''=f(t)-f_{I} $$
 *
$$\displaystyle (Eq.3)
 *
$$\displaystyle (Eq.3)
 * }
 * }

Solution
Rearranging to solve for $$f(t)$$ we get equation 2 from [[media:Egm4313.s12.team9.1-4.jpg|Lecture slide 1-4]]

 * {| style="width:0%" border="0" align="left"

$$\displaystyle my''+f_{I}=f(t) $$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

= R1.4: RLC Circuit Modeling =

Initial Information
From [[media:Egm4313.s12.team9.2-2.jpg|Lecture slide 2-2]], a general RLC circuit Kirchhoff's Voltage Law (KVL) equation, and two alternative formulations, are given:

 * {| style="width:100%" border="0" align="left"

V = LC \frac{d^{2}v_{c}}{dt^{2}} + RC \frac{dv_{c}}{dt}+ v_{c}$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq.2)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

{V}' = L{I}'' + R{I}' + \frac{1}{C}I$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq.3)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

V = L{Q}'' + R{Q}' + \frac{1}{C}Q$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq.4)
 * }
 * }

We are being asked to derive (3) and (4) from (2).

Methods
From [[media:Egm4313.s12.team9.2-2.jpg|Lecture slide 2-2]], capacitance is defined as,

 * {| style="width:100%" border="0" align="left"

Q = Cv_{c} \Rightarrow \int idt = Cv_{c} \Rightarrow i = C\frac{dv_{c}}{dt}$$ $$ Using this definition we can find equations (3) and (4).
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq.1)
 * }
 * }

Solution
Deriving (1), we get:

 * {| style="width:100%" border="0" align="left"

C\frac{d^{2}v_{c}}{dt^{2}} = \frac{di}{dt}$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq.{1}')
 * }
 * }

Also, by solving (1) for $$v_{c}$$, we obtain:

 * {| style="width:100%" border="0" align="left"

v_{c} = \frac{1}{C}\int idt$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq.{1}'')
 * }
 * }

Substituting equations (1), (1'), and (1") into (2)

 * {| style="width:100%" border="0" align="left"

V = L{I}' + RI + \frac{1}{C}\int I$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq.{2}')
 * }
 * }

Which is an "integro-differential equation." Therefore, to eliminate the integral we differentiate (2') with respect to t, to get:

 * {| style="width:0%" border="0" align="left"

$$\displaystyle {V}' = L{I}'' + R{I}' + \frac{1}{C}I$$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

Since $$Q = \int idt$$ from (1), substituting this into (2') yields:

 * {| style="width:0%" border="0" align="left"

$$\displaystyle V = L{Q}'' + R{Q}' + \frac{1}{C}Q $$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

= R1.5: General Solutions of 2nd Order Linear ODEs with Constant Coefficients =

Initial Information
These problems can be found on [[media:Lecture slide 2-5.jpg|Lecture slide 2-5]].

From pg. 59 problem 4,

 * {| style="width:100%" border="0" align="left"

{y}''+4{y}'+({{\pi }^{2}}+4)y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq.1)
 * }
 * }

And from pg. 59 problem 5,

 * {| style="width:100%" border="0" align="left"

{y}''+2\pi {y}'+{{\pi }^{2}}y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq.2)
 * }
 * }

Find a general solution for Equations (1) and (2) and check the answer by substitution.

Methods
Let $$\lambda =\frac{dy}{dx}$$ in Equations (1) and (2). The characteristic equation of Equation (1) is

 * {| style="width:100%" border="0" align="left"

{{\lambda }^{2}}+4\lambda +({{\pi }^{2}}+4)=0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

The case of the equation can be determined by $$\displaystyle {{a}^{2}}-4b$$, where $$\displaystyle a=4$$ and $$\displaystyle b={{\pi }^{2}}+4$$ , as shown below:

 * {| style="width:100%" border="0" align="left"

{{a}^{2}}-4b={{4}^{2}}-4({{\pi }^{2}}+4)=16-4{{\pi }^{2}}-16=-4{{\pi }^{2}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since this is below 0, the equation is a Case III equation with complex conjugate roots.

Using the same method, the characteristic equation of Equation (2) is

 * {| style="width:100%" border="0" align="left"

{{\lambda }^{2}}+2\pi \lambda +{{\pi }^{2}}=0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

The case of the equation can be determined by $$\displaystyle {{a}^{2}}-4b$$, where $$\displaystyle a=2\pi $$ and $$\displaystyle b={{\pi }^{2}}$$ , as shown below:
 * {| style="width:100%" border="0" align="left"

{{a}^{2}}-4b={{(2\pi )}^{2}}-4{{\pi }^{2}}=0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since this is equal to 0, the equation is a Case II equation with a real double root.

Solution: Problem 4
The roots of the characteristic equation is given by:

 * {| style="width:100%" border="0" align="left"

& \lambda =\frac{1}{2}\left( -a\pm \sqrt{{{a}^{2}}-4b} \right) \\ & \lambda =\frac{1}{2}\left( -4\pm \sqrt{-4{{\pi }^{2}}} \right) \\ & \lambda =-2\pm \pi i \\ \end{align}$$
 * $$\displaystyle\begin{align}
 * $$\displaystyle\begin{align}
 * }
 * }

Therefore, in this problem,
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \omega =\pi$$
 * }
 * }
 * }

From page 57, a real general solution in Case III is

 * {| style="width:100%" border="0" align="left"

y={{e}^{-ax/2}}(A\cos \omega x+B\sin \omega x)$$ where A and B are arbitrary constant coefficients. In this example, the general solution is
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * {| style="width:0%" border="0" align="left"

$$\displaystyle y={{e}^{-2x}}(A\cos \pi x+B\sin \pi x) $$
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style="width:100%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

To check by substitution, the first and second derivatives of the general solution are taken and substituted into Equation (1).

 * {| style="width:100%" border="0" align="left"

& {y}'=-2{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)+{{e}^{-2x}}\left( -A\pi \sin \pi x+B\pi \cos \pi x \right) \\ & {y}''=4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-2{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)-2{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+{{e}^{-2x}}(-A{{\pi }^{2}}\cos \pi x-B{{\pi }^{2}}\sin \pi x) \\ & {y}''=4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-4{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+{{e}^{-2x}}(-A{{\pi }^{2}}\cos \pi x-B{{\pi }^{2}}\sin \pi x) \\ \end{align}$$
 * $$\displaystyle\begin{align}
 * $$\displaystyle\begin{align}
 * }
 * }

Substituting into Equation (1),

 * {| style="width:100%" border="0" align="left"

4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-4{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+{{e}^{-2x}}(-A{{\pi }^{2}}\cos \pi x-B{{\pi }^{2}}\sin \pi x)+4\left( -2{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)+{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x) \right)+({{\pi }^{2}}+4)\left( {{e}^{-2x}}(A\cos \pi x+B\sin \pi x) \right)=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Multiplying and rearranging terms,
 * {| style="width:100%" border="0" align="left"

4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-8{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)+4{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)-4{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+4{{e}^{-2x}}(-A\pi \sin \pi x+B\pi \cos \pi x)+{{e}^{-2x}}(-A{{\pi }^{2}}\cos \pi x-B{{\pi }^{2}}\sin \pi x)+{{\pi }^{2}}{{e}^{-2x}}(A\cos \pi x+B\sin \pi x)=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * {| style="width:100%" border="0" align="left"

0=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since all terms cancel, the solution is
 * {| style="width:100%" border="0" align="left"

y={{e}^{-2x}}(A\cos \pi x+B\sin \pi x)$$.
 * $$\displaystyle
 * }
 * }

Solution: Problem 5
Equation (2) can be factored as shown below to determine the root.

 * {| style="width:100%" border="0" align="left"

& {{(\lambda +\pi )}^{2}}=0 \\ & \lambda =-\pi \\ \end{align}$$
 * $$\displaystyle\begin{align}
 * $$\displaystyle\begin{align}
 * }
 * }

From page 56, a basis of solutions is:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {{e}^{-\pi x}}$$, $$\displaystyle x{{e}^{-\pi x}}$$
 * }
 * }
 * }

Therefore, the corresponding general solution for this equation is
 * {| style="width:0%" border="0" align="left"

$$\displaystyle y=({{c}_{1}}+{{c}_{2}}x){{e}^{-\pi x}}$$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

To check by substitution, the first and second derivatives of the general solution are taken and substituted into Equation (2).
 * {| style="width:100%" border="0" align="left"

{y}'=-\pi {{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)+{{c}_{2}}{{e}^{-\pi x}}$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

& {y}''={{\pi }^{2}}{{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)-\pi {{c}_{2}}{{e}^{-\pi x}}-\pi {{c}_{2}}{{e}^{-\pi x}} \\ & {y}''={{\pi }^{2}}{{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)-2\pi {{c}_{2}}{{e}^{-\pi x}} \\ \end{align}$$
 * $$\displaystyle\begin{align}
 * $$\displaystyle\begin{align}
 * }
 * }

Substituting into Equation (2),
 * {| style="width:100%" border="0" align="left"

{{\pi }^{2}}{{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)-2\pi {{c}_{2}}{{e}^{-\pi x}}+2\pi \left( -\pi {{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)+{{c}_{2}}{{e}^{-\pi x}} \right)+{{\pi }^{2}}\left( {{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x) \right)=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Multiplying and rearranging terms,
 * {| style="width:100%" border="0" align="left"

{{\pi }^{2}}{{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)+{{\pi }^{2}}{{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)-2{{\pi }^{2}}{{e}^{-\pi x}}({{c}_{1}}+{{c}_{2}}x)+2\pi {{c}_{2}}{{e}^{-\pi x}}-2\pi {{c}_{2}}{{e}^{-\pi x}}=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

0=0$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since all terms cancel, the solution is:
 * {| style="width:100%" border="0" align="left"

y=({{c}_{1}}+{{c}_{2}}x){{e}^{-\pi x}}$$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

= R1.6 Determination of ODE Order, Linearity, and Application of the Superposition Principle =

Initial Information
We are asked to determine the order, linearity and whether the principle of superposition can be applied to the following examples. The order of a differential equation is found by looking at the highest occurring derivative of the dependent variable. A differential equation is linear if the dependent variable and all of its derivatives occur linearly throughout the equation.

Background Theory

If we were to consider a function $$\bar{y}(x)$$, which is defined as the sum of a homogeneous solution $$\displaystyle{y_h(x)}$$ and a particular solution $$\displaystyle{y_p(x)} $$:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \bar{y}(x) = y_p(x) + y_h(x)$$
 * }
 * }
 * }

Homogeneous Solution:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle y_h''+p(x)y_h'+q(x)y_h=0$$
 * }
 * }
 * }

Particular Solution:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle y_p''+p(x)y_p'+q(x)y_p=r(x)$$
 * }
 * }
 * }

Summing up both the Homogeneous Solution and the Particular Solution will yield:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle (y_h+y_p)+p(x)(y_h'+y_p')+q(x)(y_h+y_p)=r(x)$$
 * }
 * }
 * }

Reducing the equation according to Linear Principle will yield:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle y_h+y_p=(y_h+y_p)= \bar{y}$$
 * }
 * }
 * }

 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle y_h'+y_p'=(y_h+y_p)'= \bar{y}'$$
 * }
 * }
 * }

Now, through substitution, we obtain:

 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \bar{y}''+p(x)\bar{y}'+q(x)\bar{y}=r(x)$$
 * }
 * }
 * }

which is see to be the same equation as:

 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle y''+p(x)y'+q(x)y=r(x)$$
 * }
 * }
 * }

Falling Stone

Governing Equation:

$$\displaystyle{y}''= g = constant$$

Order: Second Order ODE Linearity: Yes Superposition: Yes This can be proved using the previous method shown in background theory: Using:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \bar{y}(x) = y_p(x) + y_h(x)$$
 * }
 * }
 * }

The homogeneous equation is set as:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {y_h''}=0$$
 * }
 * }
 * }

And the particular solution is set as:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {y_p''}=g$$
 * }
 * }
 * }

Now, adding both equations, we get:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {y_h+y_p}=g$$
 * }
 * }
 * }

Which can be simplified to:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {(y_h+y_p)''}=g$$
 * }
 * }
 * }

Which, by superposition, equals
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \bar{y}''=g$$
 * }
 * }
 * }

Parachutist

Governing Equation:

$$\displaystyle{mv'}=mg-bv^2$$

Order: First Order ODE Linearity: No Superposition: No

This can be shown using the same method as the one shown in background theory. Start by rearranging the the given equation:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {mv'}=mg-bv^2$$
 * }
 * }
 * }

To look like:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {mv'+bv^2}=mg$$
 * }
 * }
 * }

Now, using the governing equation:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \bar{v}(x) = v_p(x) + v_h(x$$
 * }
 * }
 * }

The homogeneous solution is:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {mv_h'+mv_h^2}=0$$
 * }
 * }
 * }

And the particular solution is:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {mv_p'+mv_p^2}=mg$$
 * }
 * }
 * }

Now we add both of those equations together to get:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {mv_h'+mv_h^2+mv_p'+mv_p^2}=mg$$
 * }
 * }
 * }

We can simplify that equation down to:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {m(y_h+y_p)'+m(v_h^2+v_p^2)}=mg$$
 * }
 * }
 * }

We can see that:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle (v_h^2+v_p^2)\neq \bar{v}^2$$
 * }
 * }
 * }

The method of Superposition will not hold.

Outflowing water from a tank

Govering Equation:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {h'}=-k\sqrt{h}$$
 * }
 * }
 * }

Order: First Order ODE Linearity: No Superposition: No

This can be shown using the same method as the one shown in background theory. Rearrange the governing equation to look like:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {h'}=-k\sqrt{h}$$
 * }
 * }
 * }

Using:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \bar{v}(x) = v_p(x) + v_h(x)$$
 * }
 * }
 * }

The homogeneous solution is:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle mv_h'+mv_h^2 = 0$$
 * }
 * }
 * }

The particular solution is:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle mv_p'+mv_p^2 = mg$$
 * }
 * }
 * }

Now, adding both of those equations together, we get:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle mv_h'+mv_h^2+mv_p'+mv_p^2= mg$$
 * }
 * }
 * }

which will simplify to:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle m(y_h+y_p)'+m(v_h^2+v_p^2) = mg$$
 * }
 * }
 * }

You can see that:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle (v_h^2+v_p^2)\neq \bar{v}^2$$
 * }
 * }
 * }

So the method of Superposition does not hold.

Vibrating mass on a spring

Governing Equation:

$$\displaystyle{my''+ky}=0$$

Order: Second Order ODE Linearity: Linear Superposition: Yes

This can be shown using the same method as the one shown in background theory. Using: $$\bar{y}(x) = y_p(x)+y_h(x)$$

The homogeneous solution is:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle m{y}_h''+ky_h= 0$$
 * }
 * }
 * }

The particular solution is:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle m{y}_p''+ky_p= 0$$
 * }
 * }
 * }

Adding both equations will yield:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle m{y}_h+ky_h+m{y}_p+ky_p = 0$$
 * }
 * }
 * }

You can simplify the former equation using governing equation to get:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle m({y}_h+{y}_p)''+k(y_h+y_p) = 0$$
 * }
 * }
 * }

 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle m\bar{y}''+k\bar{y}= 0$$
 * }
 * }
 * }

Superposition can be applied.

Beats of a vibrating system

Governing Equation:

$${y}''+\omega _0^2y=\cos \omega t, \omega _0\approx \omega $$

Order: Second Order ODE Linearity: Linear Superposition: Yes

This can be shown using the same method as the one shown in background theory. Using:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \bar{y}(x) = y_p(x) + y_h(x)$$
 * }
 * }
 * }

The homogeneous solution is:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {y}_h''+\omega _0^2y_h= 0$$
 * }
 * }
 * }

The particular solution is:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {y}_p''+\omega _0^2y_p=\cos \omega t$$
 * }
 * }
 * }

Adding both equations will yield:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle {y}_h+\omega _0^2y_h+{y}_p+\omega _0^2y_p=\cos \omega t$$
 * }
 * }
 * }

You can simplify the former equation using the governing equation to get:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle ({y}_h+{y}_p)''+\omega _0^2(y_h+y_p)=\cos \omega t$$
 * }
 * }
 * }

 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \bar y''+\omega _0^2 \bar y=\cos \omega t$$
 * }
 * }
 * }

Superposition can be applied because the equations are the same.

Current I in an RLC Circuit

Governing Equation:

$$\displaystyle{V}' = L{I}'' + R{I}' + \frac{1}{C}I$$

Order: Second Order ODE Linearity: Yes Superposition: Yes

This can be shown using the same method as the one shown in background theory.

Using:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \bar{I}(x) = I_p(x) + I_h(x)$$
 * }
 * }
 * }

The homogeneous solution is:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle LI_h''+RI_h'+\frac{1}{C}I_h=0$$
 * }
 * }
 * }

The particular solution is:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle LI_p''+RI_p'+\frac{1}{C}I_p=E'$$
 * }
 * }
 * }

Adding both equations will yield:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle LI_h+RI_h'+\frac{1}{C}I_h+LI_p+RI_p'+\frac{1}{C}I_p= E'$$
 * }
 * }
 * }

You can simplify the former equation using the governing equation to get:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle L(I_h+I_p)''+R(I_h+I_p)'+\frac{1}{C}(I_h+I_p)= E'$$
 * }
 * }
 * }

 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle L\bar I''+R\bar I'+\frac{1}{C}\bar I= E'$$
 * }
 * }
 * }

Superposition can be applied because the equations are the same.

Beam Deformation

Governing Equation:

$$\displaystyle{EIy^{iv}}=f(x)$$

Order: Fourth Order ODE Linearity: Linear Superposition: Yes

This can be shown using the same method as the one shown in background theory.

Using:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \bar{y}(x) = y_p(x) + y_h(x)$$
 * }
 * }
 * }

The homogeneous solution is:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle EIy_h^{iv} = 0$$
 * }
 * }
 * }

The particular solution is:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle EIy_p^{iv} = f(x)$$
 * }
 * }
 * }

Adding both equations will yield:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle EIy_h^{iv}+EIy_p^{iv} = f(x)$$
 * }
 * }
 * }

You can simplify the former equation using the governing equation to get:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle EI(y_h+y_p)^{iv} = f(x)$$
 * }
 * }
 * }

 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle EI(\bar y)^{iv} = f(x)$$
 * }
 * }
 * }

Superposition can be applied because the equations are the same.

Pendulum

Governing Equation:

$$\displaystyle{L\theta''+g*sin(\theta)}=0$$

Order: Second Order ODE Linearity: No Superposition: No

This can be shown using the same method as the one shown in background theory.

Using:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle \bar{\theta}(x) = \theta_p(x) + \theta_h(x)$$
 * }
 * }
 * }

The homogeneous solution is:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle L\theta_h'' +g\sin \theta_h =0$$
 * }
 * }
 * }

The particular solution is:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle L\theta_p'' +g\sin \theta_p =0$$
 * }
 * }
 * }

Adding both equations will yield:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle L\theta_h +g\sin \theta_h+L\theta_p +g\sin \theta_p =0$$
 * }
 * }
 * }

You can simplify the former equation using the governing equation to get:
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle L(\theta_h+\theta_p)''+g(\sin \theta_h +\sin \theta_p )=0$$
 * }
 * }
 * }

 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle (\sin \theta_h +\sin \theta_p )\neq \sin \bar \theta$$
 * }
 * }
 * }

Superposition cannot be applied because the equations are not the same.

= References =

Section 2 Lecture Notes

=Contributing Team Members=

title: User:Egm4313.s12.team9/R1


text:
User:Egm4313.s12.team9.ehlers.km

= Problem 2.4: 5 =

Initial Information
Problem Statement: Find a general solution and check by substitution.

$$ \; y'' + 2\pi y' + \pi^2 y = 0 $$

Methods
$$ \; y'' + 2 \pi y' + \pi^2 y = 0 \to \frac{d^2y}{dx^2}\ + 2 \pi \frac{dy}{dt}\ + \pi^2 y = 0 $$

$$ \; \lambda = \frac{d}{dx}\ $$

$$ \; y'' + 2 \pi y' + \pi^2 y = 0 \to \lambda^2 + 2 \pi \lambda + \pi^2 = 0 $$

Solve for $$ \; \lambda $$, through factoring.

$$ \; (\lambda + \pi) (\lambda + \pi) = 0 $$

Roots of $$ \; \lambda = -\pi, -\pi $$

Solution
General Solution:

$$ \; y = c_1 e^{ \lambda _1 x} + c_2e^{ \lambda _2 x} $$

So the general solution becomes:

$$ \; y = c_1 e^{-\pi x} + c_2 x e^{-\pi x} $$

$$ \; y = e^{-\pi x} (c_1 + c_2 x) $$

Check by Substitution:

To check by substitution, you need to find $$ \; y, y', y'' $$ and plug them into the original ODE.

$$ \; y= e ^ {-\pi x} (c_1 + c_2 x) $$

$$ \; y'= -\pi c_1e ^ {-\pi x} + c_2 (e ^ {-\pi x} -\pi x e ^ {-\pi x})= - \pi e ^{- \pi x} (c_1 + c_2 x) + c_2 e^ {- \pi x} $$

$$ \; y''= \pi ^{2} c_1e ^ {- \pi x} - \pi c_2 e ^ {-\pi x} -\pi c_2(e ^ {- \pi x} - \pi x e^ {- \pi x} )= \pi ^ {2} e^ {- \pi x} (c_1 + c_2 x) -2 \pi c_2 e^ {- \pi x} $$

Substitute the values of $$ \; y, y', y $$ into $$ \; {y}+2\pi {y}'+\pi ^{2}y=0 $$ :

$$ \; [ \pi ^ {2} e^ {- \pi x} (c_1 + c_2 x) - 2 \pi c_2e^ {- \pi x}] + 2 \pi [ - \pi e^ {- \pi x} (c_1 + c_2 x) + c_2e^ {- \pi x} ] + \pi ^ {2} [ e^ {- \pi x} (c_1 + c_2 x)]=0 $$

After substitution, it proves the general solution because the substitution equals zero.

= Problem 2.4: 6 =

Initial Information
Problem Statement: Find a general solution and check by substitution.
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle 10 y'' - 32 y' + 25.6 y = 0$$
 * $$\displaystyle $$
 * }

Methods
$$ \; 10 y'' - 32 y' + 25.6 y = 0 \to 10 \frac{d^2y}{dx^2} - 32 \frac{dy}{dx} + 25.6 y = 0 $$

$$ \; \lambda = \frac{d}{dx}\ $$

$$ \; 10 y'' - 32 y' + 25.6 y = 0 \to 10 \lambda^2 - 32 \lambda + 25.6 = 0 $$

Solve for $$ \; \lambda $$, by using the quadratic equation.

$$ \; \lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$

$$ \; \lambda = \frac{8}{5} $$

Solution
General solution:

$$ \; y = c_1 e^{ \lambda _1 x} + c_2e^{ \lambda _2 x} $$

The general solution then becomes:

$$ \; y = e^{\frac{8}{5}x}({c_1 + c_2 x}) $$

Check by Substitution:

To check by substitution, you need to find $$ \; y, y', y'' $$ and plug them into the original ODE.

$$ \; y = e^{\frac{8}{5}x}({c_1 + c_2 x}) $$

$$ \; y' = \frac{8}{5} e^{\frac{8}{5}x}(c_1 + c_2 x) + c_2 e^{\frac{8}{5}x} $$

$$ \; y'' = \frac{64}{25} e^{\frac{8}{5}x}(c_1 + c_2 x) + \frac{16}{5}c_2 e^{\frac{8}{5}x} $$

Substitute the values of $$ \; y, y', y $$ into $$ \; 10 y - 32 y' + 25.6 y = 0 $$ :

$$ \; \frac{64}{25} e^{\frac{8}{5}x}(c_1 + c_2 x) - \frac{64}{25} e^{\frac{8}{5}x}(c_1 + c_2 x) + 32 c_2 e^{\frac{8}{5}x} - 32 c_2 e^{\frac{8}{5}x} = 0 $$

After substitution, it proves the general solution because the substitution equals zero.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Egm4313.s12.team9.ehlers.km


text:
User:Egm4313.s12.team9.murray

Team Reports can be found on my team leaders page at:

http://en.wikiversity.org/wiki/User:Egm4313.s12.team9.lebo

title: User:Egm4313.s12.team9.murray


text:
User talk:Egm4313.s10.team9.delp

Hello my name is Denny... im on team domination!

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title: User talk:Egm4313.s10.team9.delp


text:
User:Egm4313.s10.team9.delp

report 1

Report 2

= R2.1: Non-Homogeneous L2-ODE-CC =

This problem is presented in three parts and will thus be solved in three parts:

Part 1 can be found on lecture slide 3-7

Part 2 can be found on lecture slide 3-7

Part 3 can be found on lecture slide 3-9

Given
Two roots:

$$\lambda_1=-2, \ \lambda_2=-2$$

Initial Conditions:

$$y(0)=1, \ y'(0)=0$$

Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $$r(x) \ $$.

Methods
From the characteristic equation:

$$(\lambda-\lambda_1)(\lambda-\lambda_2)=0 \ $$

We Have:

$$(\lambda+2)(\lambda-5)=0 \ $$

Expanding to the standard form:

$$\lambda^{2}-3\lambda-10 \ $$

Thus, the non-homogeneous L2-ODE-CC is:

$$y''-3y'-10y=r(x) \ $$

Its corresponding homogeneous solution is:

$$y(x)=c_1e^{5x}+c_2e^{-2x} \ $$

The overall solution $$y(x) \ $$is:

$$y(x)=c_1e^{5x}+c_2e^{-2x}+y_p(x) \ $$

and its derivative $$y'(x) \ $$ is:

$$y'(x)=5c_1e^{5x}-2c_2e^{-2x}+y'_p(x) \ $$

Satisfying the initial conditions:

$$y(0)=c_1+c_2+y_p(0)=1 \ $$            (eq1)

$$y'(0)=5c_1-2c_2+y'p(0) \ $$           (eq2)

Multiplying (eq1) by 2 and adding it to (eq2) yields:

$$2=7c_1+0c_2+2y_p(0)+y'_p(0) \ $$

Solving for $$c_1 \ $$:

$$7c_1=2-2y_p(0)-y'_p(0) \ $$

$$c_1=\frac{1}{7}(2-2y_p(0)-y'_p(0)) \ $$

Substituting $$c_1 \ $$ into (eq1) yields:

$$1=\frac{1}{7}(2-2y_p(0)-y'_p(0))+c_2+y_p(0) \ $$

Rearranging to solve for $$c_2 \ $$:

$$7=2-2y_p(0)-y'_p(0)+7c_2+7y_p(0) \ $$

$$7c_2=7-2+2y_p(0)-7y_p(0)+y'_p(0) \ $$

$$c_2=\frac{1}{7}(5-5y_p(0)+y'_p(0)) \ $$

Solution
Plugging $$c_1,c_2 \ $$ into $$y(x) \ $$:

$$y(x)=\frac{1}{7}(2-2y_p(0)-y'_p(0))e^{5x}+\frac{1}{7}(5-5y_p(0)+y'_p(0))e^{-2x}+y_p(x) \ $$

Given
Consider the particular case of the non-homogeneous L2-ODE-CC solved above with no excitation:

$$r(x)=0 \ $$

and plot the solution.

Methods
Since $$r(x)=0 \ $$, thus $$y_p(x)=0 \ $$ as well as its corresponding 1st and 2nd derivatives.

We can thus solve for the real values of $$c_1, c_2 \ $$:

$$c_1=\frac{2}{7}, \ c_2=\frac{5}{7}$$

Using MATLAB to plot $$y(x) \ $$ we use the following code:

Solution
Plugging in $$c_1, c_2 \ $$ to $$y(x) \ $$ we get:

$$y(x)=\frac{2}{7}e^{5x}+\frac{5}{7}e^{-2x}$$

The corresponding plot for $$y(x) \ $$ is:

Given
Part 3 asks us to generate 3 non-standard & non-homogeneous L2-ODE-CC that still admit the two initial values:

$$\lambda_1=-2, \ \lambda_2=5$$

found in figure 3a on page 3-7 as the two roots in the corresponding characteristic equation.

Methods
The standard form can be found from equation 1 on page 3-8 when $$M=1 \ $$. Non-standard forms can be found using any integer $$M\neq0,1 \ $$.

Thus, three non-standard corresponding equations can be found by using the values:

$$M=2,3,5 \ $$

and the characteristic equation:

$$M(\lambda-\lambda_1)(\lambda-\lambda_2)=0 \ $$.

Plugging in these values of $$M \ $$ we get:

$$M=2 \rightarrow 2\lambda^{2}-6\lambda-20 \ $$

$$M=3 \rightarrow 3\lambda^{2}-9\lambda-30 \ $$

$$M=5 \rightarrow 5\lambda^{2}-15\lambda-50 \ $$

Solution
The corresponding non-standard and non-homogeneous L2-ODE-CC equations with matching roots are:

$$M=2 \rightarrow r(x)=2y''-6y'-20y \ $$

$$M=3 \rightarrow r(x)=3y''-9y'-30y \ $$

$$M=5 \rightarrow r(x)=5y''-15y'-50y \ $$

Author
Solved and typed by Egm4313.s10.team9.delp 04:24, 7 February 2012 (UTC)

Reviewed by

Edited by

title: User:Egm4313.s10.team9.delp


text:
User:Egm4313.s12.team12

IEA Spring 2012 Team 12

Members:

Lauren Anders

Patrick Greivell

Craig Pettigrew

Wagner Schulz

James Stadick

Daniel Stewart

Garrett Sutcliffe

Reports:

Report 1 (R1)

Report 2 (R2)

Report 3 (R3)

Report 4 (R4)

Report 5 (R5)

Report 6 (R6)

Report 7 (R7)

title: User:Egm4313.s12.team12


text:
User:Egm4313.s12.team12/R1

Statement
Given: A spring-dashpot mass system in parallel.

Find: A second-order ordinary differential equation to describe the system's motion.

Diagram of the Entire System:

Free Body Diagram of the forces acting on the Spring:

Free Body Diagram of the forces acting on the Dashpot:

Free Body Diagram of the forces acting on the Mass:

Assumptions
1) Let the position direction (y) be positive as the mass system moves to the right.

2) The mass of the spring and the dashpot are negligible in terms of the force they apply on the mass.

3) Let Fc be the force caused by the dashpot on the mass.

4) Let Fs be the force caused by the spring on the mass.

5) f(t) will denote the variable pulling force that acts on the mass over time.

6) During motion, the spring and dashpot will not be stretched beyond their elastic limits, and will remain connected to the mass.

Solution
In order to develop an equation of motion, Newton's Second law will be used. The sum of all the forces in the direction of motion, y, will be balanced by the body's mass and acceleration. As shown in the above FBD, the forces acting on the body are the variable force, the force of the spring, and the force of the dashpot. The weight of the body also acts, but not in this given direction of motion, so it is not considered. A reference zero for the initial displacement, y0, will be considered from the location of the mass when the entire system is in equilibrium, i.e., when the body is not moving.

The force of the spring Fs is equation to the magnitude of the product of its stretched displacement, y, and a given spring constant, k. Therefore, we can write:

 * $$F_s = ky_k$$

The force of the dashpot is equal to its equivalent displacement, y, and a constant c. Therefore:

 * $$F_c = cy'_{c}$$

If the sum of all the forces acting on the body is equal to the body's mass and acceleration, which is equal to the second derivative of the body's displacement y, the following equation can be written:

 * $$\sum F_y = my'' = f(t) - F_c - F_k$$

If their equivalent values are substituted into this equation, the following is formed:

 * $$\sum F_y = my'' = f(t) - cy'_{c} - ky_k$$

However, in order to properly write a solvable second-order ODE (ordinary differential equation), the equation given must be in terms of one variable. In this instance, it can be shown that the equation can be written in terms of y. If the body is pulled a certain distance y due to the variable force f(t), then the dashpot and spring will be forced to extend by the same displacement amount y in order to remain in contact with the mass. Otherwise, the spring and the dashpot will be disconnected. Therefore, we can write:

Subsequently, the first and second derivatives of the variable displacement y can be taken to yield the following:

 * $$y' = y'_{k} = y'_{c}$$

 * $$y = y_{k} = y''_{c}$$

Therefore, it is shown that the displacement of both the dashpot and the spring is equal to the displacement of the mass. Therefore, we can write the following second-order ODE in terms of one variable y :

Then, if the equation is manipulated, the equation can be written to solve for the variable force:

Finally, the equation can be written in standard from by dividing by the mass m to yield the function:

Author & Resources
Author: Daniel Stewart 128.227.38.74 19:09, 1 February 2012 (UTC)

Resource: Kreyszig, Erwin, Herbert Kreyszig, and E. J. Norminton. Advanced Engineering Mathematics. Hoboken, NJ: Wiley, 2011. Print.

Statement
Derive the equation of motion of the spring-mass-dashpot in Fig. 53, in Kreyszig 2011 p.85, with an applied force $$r(t)\!$$ on the ball.

Source: Kreyszig, Erwin, Herbert Kreyszig, and E. J. Norminton. Advanced Engineering Mathematics. Hoboken, NJ: Wiley, 2011.

Solution
Kinematic Equations:

$$ \displaystyle y=y_k=y_c $$

Kinetics:

$$ \displaystyle my''+{{f}_{k}}+{{f}_{c}}= r(t)$$

Constitutive relations:

$$ \displaystyle {{f}_{k}}=k{{y}_{k}}$$

$$ \displaystyle {{f}_{c}}=c{{y}_{c}}'$$

Rearranging the equations:

$$ \displaystyle my''+k{{y}_{k}}+c{{y}_{c}}' = r(t)$$

Equation of motion:

 * {| style="width:100%" border="1"

$$ \displaystyle my''+ky+cy' = r(t)$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Author & Resources
Author: Garrett Sutcliffe

Statement
For the spring dashpoint mass system on p1-4, draw the FBDs and derive the equation of motion (2). p1-4

Solution

Start with Equilibrium equation

$$ \sum F = ma $$

$$ a = \frac{d^2y}{dx^2} =y'' $$

$$ ma = m'' $$

also the forces of the spring internal force and dashpoint are equal

$$ f_I = f_k = f_c $$

so plugging into equilibrium equation

$$ f(t) - f_I = my'' $$

Which gives you equation 2 by solving for f(t)

$$ my'' + f_I = f(t) $$

Author & Resources
Author: Patrick Greivell

Resource(s): Kreyszig, Erwin, Herbert Kreyszig, and E. J. Norminton. Advanced Engineering Mathematics. Hoboken, NJ: Wiley, 2011.

Statement
Derive

$$ V = LQ'' + RQ' + \frac{Q}{C} $$           (1)

$$ V' = LI'' + RI' + \frac{I}{C} $$          (2)

From

$$ V= LC\frac{dV^2_c}{dt^2} + RC\frac{dV_c}{dt} + V_c $$

Solution
Solving for (1)

If $$ Q = CV_c $$  (3)

We can differentiate twice to find:

$$ Q' = C\frac{dV_c}{dt} $$ (4)

$$ Q'' = C\frac{d^2V_c}{dt^2} $$ (5)

and we can also rearrange (3) to find

$$ V_c = \frac{Q}{C} $$ (6)

Equations (4), (5), and (6) can then be substituted into the initial equation to find:

$$ V = LQ'' + RQ' + \frac{Q}{C} $$

Solving for (2)

We first differentiate the initial equation to find:

$$ V' = LC\frac{d^3V_c}{dt^3} + RC\frac{d^2V_c}{dt^2} + \frac{dV_c}{dt} $$

We then differentiate

$$ I = C\frac{dV_c}{dt} $$ (7)

to get

$$ I' = C\frac{d^2V_c}{dt^2} $$ (8)

$$ I'' = C\frac{d^3V_c}{dt^3} $$ (9)

Equations (7), (8), and (9) can then be substituted into the initial equation to find:

$$ V' = LI'' + RI' + \frac{I}{C} $$

Author & Resources
Cpettigrew 01:40, 1 February 2012 (UTC)

Statement
K 2011 p.59 pbs. 4-5 Find a general solution. Check your answer by substitution.

Solution
Problem 4: Consider an ODE of the form:  $$ \displaystyle {y}'' + 4y' + (\pi ^{2}+4)y = 0$$  First we notice that the ODE is homogenous, linear, and has constant coefficients.  Therefore, we find the auxiliary equation, which is of the form:  $$ \displaystyle \lambda^2 + a\lambda + b = 0 $$  Looking at the coefficients we obtain:  $$ \displaystyle \lambda^{2} + 4\lambda + (\pi^{2} + 4) = 0 $$  Next we obtain the roots of this equation with the quadratic formula: $$ \displaystyle \ \frac{-4 \pm \sqrt{(-4)^2 - 4(1)(\pi^2 + 4)}}{2(1)} $$  This can be simplified to:  $$ \displaystyle -2\pm\frac{\sqrt{16 - 4\pi^2 - 16}}{2} $$  $$ \displaystyle -2\pm\frac{\sqrt{-4\pi^2}}{2} $$<br\> We notice that by looking inside the radical, we will be obtaining complex roots. Continuing with simplification we get: <br\><br\> $$ \displaystyle -2\pm{i\pi} $$<br\><br\> The general solution to the case of complex roots is: <br\><br\> $$ \displaystyle y = e^{\frac{-ax}{2}}(Acos\omega{x} + Bsin\omega{x}) $$ <br\> <br\> Where complex roots are of the form: <br\><br\> $$ \displaystyle -\frac{1}{2}a\pm{i\omega} $$<br\><br\> Knowing $$ \displaystyle a = 4 $$ and $$ \displaystyle \omega = \pi $$, we get the final general solution: <br\><br\> $$ \displaystyle y = e^{-2x}(Acos\pi{x} + Bsin\pi{x}) $$<br\><br\>

Problem 5: Consider an ODE of the form: <br\><br\> $$ \displaystyle {y}'' + 2\pi y' + \pi ^{2}y = 0$$ <br\> This ODE is linear, and has constant coefficients. Therefore, its general solution is of the form: <br\><br\> $$ \displaystyle {y}=e^{\lambda x} $$ <br\> Therefore: <br\><br\> $$ \displaystyle {y}'=\lambda e^{\lambda x} $$ <br\> And: <br\><br\> $$ \displaystyle {y}''={\lambda}^2 e^{\lambda x} $$ <br\> The ODE can be rearranged so that: <br\><br\> $$ \displaystyle e^{\lambda x}(\lambda^2 + 2\pi \lambda + \pi^2)=0 $$ <br\> And since: <br\><br\> $$ \displaystyle e^{\lambda x} \neq 0 $$ <br\> Then: <br\><br\> $$ \displaystyle \lambda^2 + 2\pi \lambda + \pi^2=0 $$ <br\> $$ \displaystyle (\lambda + \pi)=0 $$ <br\> $$ \displaystyle \lambda =- \pi $$ <br\> Since: <br\><br\> $$ \displaystyle \lambda_{1} = \lambda_{2} $$ <br\> The general solution then becomes: <br\><br\> $$ \displaystyle y = c_{1}e^{\lambda x} + c_{2}xe^{\lambda x} $$ <br\> $$ \displaystyle y = e^{\lambda x} (c_{1} + c_{2}x) $$ <br\> Therefore: <br\><br\> $$ \displaystyle y = e^{-\pi x} (c_{1} + c_{2}x) $$ <br\>

Author & Resources
Problems 4 and 5 were solved by [James Stadick] and [Wagner Schulz]

Statement
For each ODE determine the order, linearity (or lack of), and show whether the principle of superposition can be applied:

1) $$   \displaystyle  y''=g=constant $$

2) $$   \displaystyle  mv' = mg - bv^2 $$

3) $$   \displaystyle  h' = -k\sqrt{h} $$

4) $$   \displaystyle  my'' + ky = 0 $$

5) $$   \displaystyle y'' + \omega_{0}^2y = \cos \omega t, \omega_{0}\approx\omega $$

6) $$   \displaystyle  LI'' + RI' + \frac{C}I = E' $$

7) $$   \displaystyle  EIy = f(x) $$

8) $$   \displaystyle  L\theta'' + g\sin\theta = 0 $$

Solution
1) $$   \displaystyle  y'' = g = constant $$<br\><br\> a) Order of ODE<br\> - The order of an ODE is just the highest derivative seen in the equation<br\>
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This ODE is of the 2nd order
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b) Linearity of ODE<br\> - An ODE is said to be linear if the variable in the equation is always seen to the first power<br\>
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This ODE is linear
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c) Principle of Superposition<br\> - The Principle of Superposition states that for a linear homogeneous Ordinary Differential Equation where $$ \displaystyle y_1(t) $$ and $$ \displaystyle y_2(t) $$ are solutions then so is $$ \displaystyle y(t) = c_1y_1(t) + c_2y_2(t) $$. Therefore in order for this principle to be applied the equation has to be a linear homogeneous ODE.<br\>
 * {| style="width:100%" border="1"

Yes the Principle of Superposition can be applied
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 * }
 * }

2) $$ \displaystyle mv' = mg - bv^2 $$<br\><br\> a) Order of ODE<br\>
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This ODE is of the 1st order
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b) Linearity of ODE<br\>
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This ODE is non-linear
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c) Superposition Principle<br]>
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No the Principle of Superposition cannot be applied
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 * }

3) $$ \displaystyle h' = -k\sqrt{h} $$<br\><br\> a) Order of ODE<br\>
 * {| style="width:100%" border="1"

This ODE is of the 1st order
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b) Linearity of ODE<br\>
 * {| style="width:100%" border="1"

This ODE is non-linear
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 * }
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c) Principle of Superposition<br\>
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No the Principle of Superposition cannot be applied
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 * }
 * }

4) $$ \displaystyle my'' + ky = 0 $$<br\><br\> a) Order of ODE<br\>
 * {| style="width:100%" border="1"

This ODE is of the 2nd order
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

b) Linearity of ODE<br\>
 * {| style="width:100%" border="1"

This ODE is linear
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 * }
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c) Principle of Superposition<br\>
 * {| style="width:100%" border="1"

Yes the Principle of Superposition can be applied
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 * }
 * }

5) $$ \displaystyle y'' + \omega_{0}^2y = \cos \omega t, \omega_{0} \approx \omega $$<br\><br\> a) Order of ODE<br\>
 * {| style="width:100%" border="1"

This ODE is of the 2nd order
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 * }
 * }

b) Linearity of ODE<br\>
 * {| style="width:100%" border="1"

This ODE is Linear
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 * }
 * }

c) Principle of Superposition<br\>
 * {| style="width:100%" border="1"

Yes the Principle of Superposition can be applied
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

6) $$ \displaystyle LI'' + RI' + \frac{1}{c}I = E' $$<br\><br\> a) Order of ODE<br\>
 * {| style="width:100%" border="1"

This ODE is of the 2nd order
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

b) Linearity of ODE<br\>
 * {| style="width:100%" border="1"

This ODE is linear
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

c) Principle of Superposition<br\>
 * {| style="width:100%" border="1"

Yes the Principle of Superposition can be applied
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 * style="width:95%" |
 * }
 * }

7) $$ \displaystyle EIy' = f(x) $$<br\><br\> a) Order of ODE'''<br\>
 * {| style="width:100%" border="1"

This ODE is of the 4th order
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 * }
 * }

b) Linearity of ODE<br\>
 * {| style="width:100%" border="1"

This ODE is linear
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 * style="width:95%" |
 * }
 * }

c) Principle of Superposition<br\>
 * {| style="width:100%" border="1"

Yes the Principle of Superposition can be applied
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

8) $$ \displaystyle L\theta'' = g\sin\theta = 0 $$<br\><br\> a) Order of ODE<br\>
 * {| style="width:100%" border="1"

This ODE is of the 2nd order
 * style="width:95%" |
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 * }
 * }

b) Linearity of ODE<br\>
 * {| style="width:100%" border="1"

This ODE is non-linear
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 * }
 * }

c) Principle of Superposition<br\>
 * {| style="width:100%" border="1"

No the Principle of Superposition cannot be applied
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 * }
 * }

Author & Resources
Author: Lauren Anders

Resource: Kreyszig, Erwin, Herbert Kreyszig, and E. J. Norminton. Advanced Engineering Mathematics. Hoboken, NJ: Wiley, 2011. Print.

title: User:Egm4313.s12.team12/R1


text:
Dominant group/Relative synonyms

The difference between exact synonyms and relative synonyms is similar to the difference between identical twins and fraternal or sororal twins.

This resource is experimental in the sense that a number of relative synonyms for dominant group are not relative synonyms when either word1 (dominant synonym) or word2 (synonym group) is actually used in context other than synonymy for either word. Finding these cross-over terms may help to better restrict the overall contextual meanings for the various relative synonyms and dominant group.

Two different words having very close meanings are relative synonyms, while two different words having exactly the same meanings are absolute synonyms.

Synonymous definitions
A synonymous definition is a definition “defining a single word [or symbol] by giving another single word [or symbol] which has the same meaning.” But, synonymous definitions have limitations:
 * 1) “some words have no exact synonyms” ,
 * 2) a synonymous definition “cannot be used in the construction of precising or theoretical definitions.”, and
 * 3) no synonym should appear in the definiens of a genus–differentia definition.

Def. a word having a very close meaning to another word is called a relative synonym.

Dominant group
Examples from primary sources are to be used to prove or disprove each hypothesis. These can be collected per subject or in general.

 * 1) Accident hypothesis: dominant group is an accident of whatever processes are operating.
 * 2) Artifact hypothesis: dominant group may be an artifact of human endeavor or may have preceded humanity.
 * 3) Association hypothesis: dominant group is associated in some way with the original research.
 * 4) Bad group hypothesis: dominant group is the group that engages in discrimination, abuse, punishment, and additional criminal activity against other groups. It often has an unfair advantage and uses it to express monopolistic practices.
 * 5) Control group hypothesis: there is a control group that can be used to study dominant group.
 * 6) Entity hypothesis: dominant group is an entity within each field where a primary author of original research uses the term.
 * 7) Evolution hypothesis: dominant group is a product of evolutionary processes, such groups are the evolutionary process, produce evolutionary processes, or are independent of evolutionary processes.
 * 8) Identifier hypothesis: dominant group is an identifier used by primary source authors of original research to identify an observation in the process of analysis.
 * 9) Importance hypothesis: dominant group signifies original research results that usually need to be explained by theory and interpretation of experiments.
 * 10) Indicator hypothesis: dominant group may be an indicator of something as yet not understood by the primary author of original research.
 * 11) Influence hypothesis: dominant group is included in a primary source article containing original research to indicate influence or an influential phenomenon.
 * 12) Interest hypothesis: dominant group is a theoretical entity used by scholarly authors of primary sources for phenomena of interest.
 * 13) Metadefinition hypothesis: all uses of dominant group by all primary source authors of original research are included in the metadefinition for dominant group.
 * 14) Null hypothesis: there is no significant or special meaning of dominant group in any sentence or figure caption in any refereed journal article.
 * 15) Object hypothesis: dominant group is an object within each field where a primary author of original research uses the term.
 * 16) Obvious hypothesis: the only meaning of dominant group is the one found in Mosby's Medical Dictionary.
 * 17) Original research hypothesis: dominant group is included in a primary source article by the author to indicate that the article contains original research.
 * 18) Primordial hypothesis: dominant group is a primordial concept inherent to humans such that every language or other form of communication no matter how old or whether extinct, on the verge of extinction, or not, has at least a synonym for dominant group.
 * 19) Purpose hypothesis: dominant group is written into articles by authors for a purpose.
 * 20) Regional hypothesis: dominant group, when it occurs, is only a manifestation of the limitations within a region. Variation of those limitations may result in the loss of a dominant group with the eventual appearance of a new one or none at all.
 * 21) Source hypothesis: dominant group is a source within each field where a primary author of original research uses the term.
 * 22) Term hypothesis: dominant group is a significant term that may require a 'rigorous definition' or application and verification of an empirical definition.

Relative synonymy
"If a given group includes only absolute synonyms, it is taken as such for further steps. If the group includes at least one relative synonym, all of them are taken for the operation referred in mathematics as transitive closure."

"Homonyms are listed one per line. Absolute synonyms, relative synonyms and antonyms are hyphen-separated. If a relative synonym of the target term has some absolute synonyms these are listed after it comma-separated. So are the absolute synonyms of the antonyms."

Distinguishing between absolute and relative synonyms "is an important distinction for a domain-specific terminology."

"[O]ne might exchange terms almost intuitively, as though one concept of a word is fundamentally the same as its relative "synonym.""

"The text phrase "see..." indicates that "..." is considered a relative synonym."

Synonyms
With a two-word term, genus classes (or categories ) may consist of three varieties:
 * 1) category1 + category2,
 * 2) "dominant" + category2, or
 * 3) category1 + "group".

Here "category1" refers to any category of relative synonyms for "dominant" (adjective or noun), and "category2" is the same for "group" (noun or verb).

The added possibilities of "group" as a verb suggests usages like "dominantly grouped", with the adverb "dominantly", or "dominant grouping".

These variations in part due to analyzing a two-word term allow additional possibilities to consider in forming genus differentia definitions.

The term dominant group appears to be used to identify entities of importance. The genera differentia for possible definitions of dominant group fall into the following set of orderable pairs:

'Orderable' means that any synonym from within the first category can be ordered with any synonym from the second category to form an alternate term for dominant group; for example, "superior class", "influential sect", "master assembly", "most important group", and "dominant painting". "Dominant" falls into category 171. "Group" is in category 61. Further, any word which has its most or much more common usage within these categories may also form an alternate term, such as "ruling group", where "ruling" has its most common usage in category 739, or "dominant party", where "party" is in category 74. "Taxon" or "taxa" are like "species" in category 61. "Society" is in category 786 so there is a "dominant society".

Dominance groups
"Dominance group" is a cross-over phrase. The primary meaning of "dominance" falls under "authority". The secondary meaning is under "influence" like "dominant".

Dominant class
The earliest sociological use of dominant group occurs in the article "Art in a Democracy" from 1923.
 * 1) "It is but natural that when aristocratic ideals should impose themselves upon any polity the art of that polity should reflect the taste, the culture, the ways of life, and the very being of the dominant classes.", bold added, and “At any rate, here is confirmation of the thesis that art voices the will of the dominant group in society.”
 * 2) "The prevalence of the Russian power is not the prevalence of the rude barbarians, that constitute the bulk of the nation, but of the dominant class of proprietors, which is equally civilized with the same class in any other part of Europe." from 1822.
 * 3) "[B]ecause political democracies generally arise from a compromise between contending organized elites that are unstable to impose their will unilaterally or the unilateral action of one dominant group, usually the armed forces, this does not bode well for democratization in situations in which the armed forces are inextricably tied to the interests of a dominant (and antidemocratic) agrarian class."

Dominant family

 * 1) "Dominant" occurs in "dominant families" and "dominant influence".
 * 2) "The poet here seems to assume that the province of Connaught took its name from the race of Conn of the Hundred Battles, who were the dominant family of that province in MacCoise's time." from 1857.

Dominant firm

 * 1) In this article dominant group does not occur but "dominant" does: "dominant player", "dominant firm", and "[f]or policy selection, the policymaker is dominant".
 * 2) "If such a fringe existed and became a threat to the dominant firm then we would expect the dominant group, by predatory action, to do something about it."

Dominant genera

 * 1) In a freshwater ecosystem such as the large shallow Chinese lake, Lake Taihu, depending upon the time of year (summer or winter), location (Meiliang Bay or lake center), wind speed and direction, "temperature, underwater light climate, nutrients and grazing by zooplankton and" fish, the lake has a dominant species, dominant genera, and/or a dominant group. Bold added.

Dominant party
An early occurrence of a dominant group relative synonym is from 1820.
 * 1) "There was a violent war party, who wished all the resources of the State to be placed at the disposal of the national government; there was a peace party, composed of members of both parties, determined to put every obstacle in the way of the administration; while the Federalists generally regarded the declaration of war as an act of tyranny, and the measures of the dominant party as an infringement of State rights." [Bold added]

Dominant religions
The phrase “die Dominanten Religionen von ganz Europa” occurs in 1726.

Dominant schools

 * 1) "It is arguably now the dominant school of economics, being taught in most major universities.", from Wikiversity Overview of economic schools of thought. When the context use of "school" is meant as an assemblage or association, the two-word phrase "dominant school" is a relative synonym for the technical term "dominant group". However, when school is meant as "school of thought", then "dominant school" is not a relative synonym. For example, "The metropolitan thesis, also known as the Laurentian thesis, is one of the dominant schools in Canadian historical thought.", from Wikipedia Metropolitan thesis. Likewise, "dominant school of belief" is also not a relative synonym. For example, "In Bhutan the Drukpa Lineage is the dominant school and state religion.", from Wikipedia Drukpa Lineage.
 * 2) "English School of Painting, a dominant school of painting in England, active in 1750–1850.", per Wikipedia The English School. This appears to be a relative synonym because "grouping" has a secondary meaning with respect to composition of art and "school of painting" is differentiated from "school of philosophy" or "school of belief".
 * 3) "During this period it had become the dominant school among Congregationalists, had led to division among Presbyterians, resulting in the creation of a new religious denomination, the New School Presbyterian (1838-69), had founded all the seminaries of the Congregationalists and several of the Presbyterians, had furnished the vital forces from which had spring the American Board of Commissioners for Foreign Missions, had established a series of colleges from Amherst in the East to Pacific University in the West, and led in a great variety of practical efforts to extend the kingdom of Christ on earth.", from Wikipedia New England theology. Bold added. Here the relative synonymy is via "religion".
 * 4) "Zhe School and the school which was supported by the royal court (Yuanti School) were the dominant schools during the early Ming period.", per Wikipedia Ming Dynasty painting.
 * 5) "Lebanon has become the dominant school in Western Pennsylvania winning the district championship five of the last six years, and in 2003, 2004, and 2005 sending record numbers of students to National Forensic League national tournament.", per Wikipedia Mt. Lebanon School District.

Dominants group

 * 1) "[A]n increase in dominance interactions ... [led] to a new group of dominants that eventually laid reproductive eggs." Bold added.

Dominant sizees
"The granite is relatively fine grained (1–2 mm dominant grain size) and isotropic without any signs of ductile or brittle deformation."

Dominant society

 * 1) "A related, but separate, definition relies on a linguistic identity that differs from that of the dominant society [5]."
 * 2) "They were extended toward a privileged representative of the dominant society who was usually (not always) visibly accompanied by a person the group had long accepted into their midst."

Dominant species

 * 1) "At the low density given by the spherically symmetric wind model (see Table 1), the dominant species in the gas are atomic ions while as the gas number density increases, the recombination of ions takes place and the gas composition is governed by neutral-phase chemistry, that is, the dominant species are neutral atoms and molecules although electrons and some ions are still present in relatively large amounts (for example, C+, O+ and He+)."
 * 2) "From the geometry and scaling laws the contribution of a clump of radius R to the number of particles N of each species is ... where ηC is the carbon abundance and we are implicitly assuming that all the carbon is in the form of the dominant species in each zone."
 * 3) "[I]n any limited country, the species which is most common, that is abound most in individuals, and the species which are most widely diffused within their country (and this is a different consideration from wide range, and to a certain extent from commonness), often give rise to varieties sufficiently well-marked to have been recorded in botanical works. Hence it is the most flourishing, or as they may be called, the dominant species -- those which range widely over the world, are the most diffused in their own country, and are the most numerous in individuals, -- which oftenest produce well-marked varieties, or, as I consider them, incipient species."

Dominant teams
When the word "team" refers to more than a duality, "dominant team" becomes a relative synonym for dominant group.
 * 1) "So far Mt. Lebanon has taken 2 PHUL championships out of the 7 Finals appearances, and has been a dominant team in the Pittsburgh region for many years.", from Wikipedia Mt. Lebanon School District.

Dominant types

 * 1) "new dominant type ... supplanted the previous dominant type"
 * 2) "All visibly different colony types were isolated and of dominant colony types, several isolates were taken."

Dominating groups
"Dominating" appears to have its most common meaning under "government".

Hegemonic class

 * 1) "Thus Dennis Cosgrove, one of the dominant British members of this landscape school, argued that 'in class societies, where surplus production is appropriated by the dominant group, symbolic production is likewise seized as hegemonic class culture to be imposed on all classes' (Cosgrove, 1983, p. 5)" Bold added.

Higher types
"Lovejoy (1936) discusses how the great Chain of Being was "temporalized" during the 18th century; by this, he has in mind the tendency to think that the natural ordering of living things from those of higher type down to those of lower type also represented an historical progression."

Highest classes
"Though we generally use the words "classical" and "Classics" in a high-cultural sense, this use is derivative and figuratively loaded. These words derive from the Latin adjective classicus,-a,-um, originally referring to someone belonging to the highest of the five classes of Roman citizens in the division of the Roman people according to property and wealth made by Servius Tullius (Livy 1.43.5; Aulus Gellius 6[7].13.1, 3)."

The most popular meaning of "highest" is "210. TOP". The second most popular meaning is "36. SUPERIORITY". Any highest class may not be a dominant group, unless that which puts them as highest is also a criterion of dominance.

"Servius Tullius was the legendary sixth king of ancient Rome, and the second of its Etruscan dynasty. He reigned 578-535 BC. Roman and Greek sources describe his servile origins and later marriage to a daughter of Lucius Tarquinius Priscus, Rome's first Etruscan king, who was assassinated in 579 BC. Servius was variously said to have been the first Roman king to accede without election by the Senate, having gained the throne by popular support, at the contrivance of his mother-in-law; and the first to be elected by the Senate without reference to the people." The classes including the highest class of the centuriate classes were apparently created during rule of Servius Tullius.

"The oldest surviving source for the overall political developments of the Roman kingdom and Republic is Cicero's De republica ("On the State"), written in 44 BC. The main literary sources for Servius' life and achievements are the Roman historian Livy (59 BC – AD 17), whose Ab urbe condita was generally accepted by the Romans as the standard, most authoritative account; Livy's near contemporary Dionysius of Halicarnassus, and Plutarch (c.46 – 120 AD); their own sources included works by Quintus Fabius Pictor, Diocles of Peparethus, Quintus Ennius and Cato the Elder. Livy's sources probably included at least some official state records, he excluded what seemed implausible or contradictory traditions, and arranged his material within an overarching chronology. Dionysus and Plutarch offer various alternatives not found in Livy, Livy's own pupil, the etruscologist, historian and emperor Claudius offered yet another, based on Etruscan tradition. "

This would put an early origin of "highest class" at 2544 b2k, with reference to the reign of Servius Tullius from 2578 b2k to 2535 b2k, where the most recent date of origin for "highest class" is 2535 b2k. As a relative synonym for dominant group, highest class (classicus) puts the origin of the concept behind dominant group at 2535 b2k, well before the exact usage in 1826, or 174 b2k in application to insect distributions by William Kirby.

"In each case it is found that the lower social classes have mortality rates significantly higher than those of the upper social classes with the mortality rate of the lowest class being approximately twice that of the highest class on a six-category scale."

Influential groups

 * 1) "Finally, and perhaps most importantly, London was the institutional home of the most influential group of astronomers of the period: the Royal Astronomical Society."
 * 2) "During the first five years (the 'formative era') the Branch flourished under the guidance of what was collectively the most influential group of amateur astronomers in the country."

Influential sets

 * 1) "Together with Leonard Searle, he wrote an influential set of papers which established that stellar disks are truncated at about four exponential scale-lengths, and that the vertical scale-height of disks is constant with radius."
 * 2) "Until now Themo has been best known for an influential set of questions on Aristotle's Meteorologica, which is closely related to similar sets by Nicole Oresme and, putatively, Simon Tunsted."

Major groups

 * 1) "Siah-Posh (black-Robed) Kafirs was the former designation of the major and dominant group of the Hindu Kush Kafirs inhabiting the Bashgul (Kam) valley of the Kafiristan, now called  Nuristan."

Majority groups
"Majority group" is a phrase which may be a cross-over. When "majority" means "superior", "majority group" is a relative synonym for dominant group. But, when "majority" only refers to a "plurality" in context, "majority group" is not a relative synonym for "dominant group".
 * 1) "For majority group members, the experience of being in a minority enclave will thus tend to have an aggravating effect on attitudes, resulting in defensive intolerance toward the numerically dominant minority group."

Master classes

 * 1) "Once created, device class objects are registered with an instance of the master class."
 * 2) "For ATIC, a possible set of defined classes would be a master class event, and sub-classes header, silicon, scintillator, bgo and track."
 * 3) "As they advanced in the knowledge of the dictionary, and in reading, they were promoted to the second company, and afterward to the first of the master's class. from 1811.

Most important interests
"Even the actual promoters of the most important interests of mankind have seldom anticipated, in idea, the progressive consequences of their own plans."

Patrician classes
"According to Livy, the first 100 men appointed as senators by Romulus were referred to as "fathers" (patres), and the descendants of those men became the Patrician class. The patricians were distinct from the plebeians because they had wider political influence, at least in the times of the Republic."

"Venetian Humanism in an Age of Patrician Dominance" is a book title verifying dominance by the Patrician class.

"Until the year 445 B.C. [-2445 b2k] a regular marriage (iustae nuptiae) could be contracted only between patricians - members of the ruling class." Bold added.

"'Consequently, the ruling class accumulated all the wealth for themselves, and the slave-population filled the country, while the real Italian population decreased terribly, worn out by poverty, taxation, and military service. And when there was a respite from these things they found themselves unemployed, because the land was owned by rich men who used slaves instead of freemen on their farms.' Whatever the origin of this passage may be, it shows the necessary result of the military expansion of Rome,"

Predominant groups

 * 1) "Sub-dominant group, in entom., in Kirby's system, those groups of insects which either never enter the tropics or those tropical insects whose range does not exceed 50° in the Old World of 43° in the New : in contradistinction to his predominant groups and dominant groups, in the first of which he includes groups extending from the arctic region, where vegetation ceases, to the equator, and in the second those which reach to the tropics but fall short of the polar circles." Bold added.
 * 2) "The bacterial members within the dominant group of our clone library belonged to unclassified taxa, while the second predominant group consisted of members of the phylum Proteobacteria." Bold added.

Predominant types

 * 1) "To illustrate the problems that [have] arisen from COSP and representativeness issues, consider a situation where we utilize the raw land use information as 1-km resolution to get the predominant land use type for the 12-km grid cell." Bold added.

Ruling classes
"The caste-like nature of the Venetian nobility always looms large in the writings of Renaissance observers of the Republic's constitution.1 It was natural that they, and later writers too, should find this quality impressive because from the last years of marco Polo's life until the terrible war with the Turks in the middle of the seventeenth century, the Venetian nobility was remarkable for its policy of exclusiveness. The patricians were organized as a distinct social class with jealously guarded privileges. In order to avoid contamination by inferiors and to keep power in their own hands, they refused to accept new members; they adopted strict rules regarding marriages; and they kept careful birth and marriage records in the famous Golden Books. But the ruling class had not always had this exclusive character."

Ruling elites
Def. "[t]he nobility, or the hereditary ruling class" is called an aristocracy.

Def. a governing group from the aristocracy is called ruling elites.

"Also using the state apparatus for agencies that serve the ruling elites, such as the security forces, presidential staff, central bank, diplomatic service, customs and collection agencies. " Bold added.

"Use of aggressive nationalistic rhetoric by ruling elites, especially destructive forms of communal irredentism or communal solidarity (e.g., "ethnic cleansing", "defending the faith"). " Bold added.

Ruling groups

 * 1) An example of this contains "Syria's Ruling, Military Group" and "Causes for Its Dominance", in the title. "[T]he ruling element consists at its core of a close kinship group which draws strength simultaneously, but in decreasing intensity, from a tribe, a sect-class, and an ecologic-cultural division of the people." "[T]he core of the ruling element of Iraq also consists of a kinship group (closely related members of the Begat section of the Albui Nasir tribe); rests essentially on members of a minority sect (Sunni Arabs) and on country rather than city people (on middle and lower middle class families from the country towns of the Arab north-western part of Iraq". The term dominant group does not occur in the article.

Superior classes
"Then we reach F, the lowest of those yet superior classes of intellect, with which this volume is chiefly concerned."

Superior sizes

 * 1) "The superior size and albedo of Venus completely turn the scale, with the result that Venus at her brightest is about 12 times brighter than Mercury at his brightest."
 * 2) "There is no reason to question but that they are simply ordinary meteors, which from their superior size and unusually slow speed have survived to reach the earth's surface."

The dominants
“While Kanter sees the differentiation between dominants and tokens as originating with the dominant group, it seems logical that such differentiation might also originate with the token group in certain situations. Specifically, when tokens are of a higher social status than dominants, they may be inclined to differentiate themselves from the dominants in order to improve their overall status or to avoid any social stigma which may be associated with the dominants in the larger society.”

“This means that the dominant group were close together in a confined circle while the subdominants were more widely separated around the outside of that circle. We shall expect then that the dominants would fight among themselves more than the subdominants.”

As a group of dominants is likely synonymous with a dominant group, so “the dominants” is a likely synonym for a dominant group.

In German, the word “Dominanten” may mean a dominant or dominants. The phrase “den Dominanten” occurs in 1774 usually referring to “die Dominante” in music. The plural is “die Dominanten” (the dominants). But, “the dominant” translates to “die dominierende”.

A teaching textbook for piano from 1826 contains “Sind die Grundtöne der Dur- und Moll-Tonarten, jedoch ohne die mindeste Beziehung auf ihre Dominanten, durch den Zirkel angesetzt, so werden die Dominanten dann hinterher beigefügt.”

Texts for musical composition contain "the dominants", such as “Doch leide« dieses Berkoth eine Ausnahme, wenn die Dominanten häufig von kommen.” from 1781.

The earliest occurrence so far for "the dominants" comes from 1757, “so sind solche entweder tonische Noten, Dominanten, oder Unterdominanren, und die Dominanten können simple oder tonische Dominanten seyn.”

Upper classes

 * 1) "[S]ince their rise to power, an upper class has been differentiating itself from the rest of the community and, within the ranks of this class, even a group of millionaires, waxing rich from fat commissions on state contracts, has reared its head." Bold added.

Upper departments
The most popular usage of the word "department" is category "998. JURISDICTION". The second most popular usage of "jurisdiction" is category 739. If an author's intended usage is closer to category 739, then "upper department" may be a relative synonym for dominant group.

"For the impression of the commercial arts is often conspicuous in the upper departments of life, before it reaches those of inferior condition; but the circle gradually widens." Neither word "dominant" nor "group" appears in the text of Dunbar's book from 1780, digitized by Google.

Hypotheses

 * 1) Dominant group may have been first used in 1826, but at least one of its synonyms may have occurred in the oldest human language.

title: Dominant group/Relative synonyms


text:
User:Egm6322.s12.TEAM1.KIM/HW1

Problem 5: Proof problems
 Solved without assistance

Given
$$\displaystyle f = \sum_{i}g_i$$

Find
1. Show that if g is odd, then f is odd. 2. Show that if g is even, then f is even.

Solution
If g is odd, g(x) is following that $$\displaystyle g_i(-x) = -g_i(x)$$ $$\displaystyle f(-x) = \sum g_i(-x) = -\sum g_i(x) = -f(x) $$ This shows that if g is odd, f is also odd. If g is even, g(x) is following that $$\displaystyle g_i(-x) = g_i(x)$$ $$\displaystyle f(-x) = \sum g_i(-x) = \sum g_i(x) = f(x)$$ This result shows that if g is even, f is also even.

Given
$${{p}_{n\left( x \right)}}=\sum\limits_{i=0}^{\left[ {}^{n}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 2n-2i \right)!{{x}^{n-2i}}}{{{2}^{n}}i!\left( n-i \right)!\left( n-2i \right)!}}$$

Find
Show $$\displaystyle P_2k(x) $$ is even $$\displaystyle P_{2k+1}(x)$$ is odd When k = 0,1,2,3,....,

Solution
$${{p}_{n\left( x \right)}}=\sum\limits_{i=0}^{\left[ {}^{n}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 2n-2i \right)!{{x}^{n-2i}}}{{{2}^{n}}i!\left( n-i \right)!\left( n-2i \right)!}}$$ When n= 2k,the above equation is following that $${{p}_{2k\left( x \right)}}=\sum\limits_{i=0}^{\left[ {}^{2k}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 4k-2i \right)!{{x}^{2k-2i}}}{{{2}^{2k}}i!\left( 2k-i \right)!\left( 2k-2i \right)!}} = \sum\limits_{i=0}^{\left[ {}^{2k}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 4k-2i) \right)!{{x}^{2(k-i)}}}{{{2}^{2k}}i!\left( 2k-i \right)!\left( 2k-2i \right)!}}= \sum\limits_{i=0}^{\left[ {}^{2k}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 4k-2i) \right)!{{(-x)}^2{(k-i)}}}{{{2}^{2k}}i!\left( 2k-i \right)!\left( 2k-2i \right)!}}$$ The above results show that if n is even, f(x) = f(-x). $$\displaystyle P_{2k}(x) = P_{2k}(-x)$$ When n = 2k+1, the equation is following that $${{p}_{2k+1\left( x \right)}}=\sum\limits_{i=0}^{\left[ {}^{2k+1}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 4k-2i+2 \right)!{{x}^{2k-2i+1}}}{{{2}^{2k+1}}i!\left( 2k-i+1 \right)!\left( 2k-2i+1 \right)!}}=\sum\limits_{i=0}^{\left[ {}^{2k+1}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 4k-2i+2 \right)!{{(x)x}^{2(k-i)}}}{{{2}^{2k+1}}i!\left( 2k-i+1 \right)!\left( 2k-2i+1 \right)!}}$$ When x= -x, $${{p}_{2k+1\left( x \right)}}= \sum\limits_{i=0}^{\left[ {}^{2k+1}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 4k-2i+2 \right)!{(-x){(-x)}^{2(k-i)}}}{{{2}^{2k+1}}i!\left( 2k-i+1 \right)!\left( 2k-2i+1 \right)!}}$$ This shows that $$\displaystyle P_{2k+1}(x) = -P_{2k+1}(-x)$$

title: User:Egm6322.s12.TEAM1.KIM/HW1


title: File:Pea2.s12.sec43.c.djvu


text:
User:Egm4313.s12.team6.jagolinzer

Wiki Page for Scott Jagolinzer of Team 6.

Problems
R1.4

R2

Back to Team page

title: User:Egm4313.s12.team6.jagolinzer


text:
Talk:University of Florida/Egm4313/s12.team6

Discussion Page

Report Drafts
Here we can post the parts that we have worked for us to review.

Report 1
R1.4 by Scott Jago

R1.5 by Scott Davis

Problems
We can post who will do what problems here?


text:
Learning the abc's/D

D is the fourth letter in the English alphabet. Dog
 * Words that start with D

Doll

Dingo
 * Words with D in the middle

Add

Addition

Wedding

Needle

Saddle Had
 * Words that end with D

Add

Round

Mad

Sad

Bed

title: Learning the abc's/D


text:
Talk:Big bang

Comment
Again, the layout of this page appears to be good for the style of presentation, but the grammar ought to be improved on - there are even section titles that do not begin with a capital letter; e.g. "discovery of accelerated expansion". Also, there are a good deal of inline citations provided before a punctuation mark, when they ought to be placed after. Wilhelmina Will 02:03, 27 January 2012 (UTC)

title: Talk:Big bang


text:
University of Florida/Egm4313/s12.team6.reports

The status of a report can be found here.

Here we can collect and compile our reports.


text:
User:Egm6322.s12.team2.shields

Matt Shields' HW problems

Report 8
R8.1:

'''Construct the Gram matrix for the Fourier basis functions (3)p.40-1 using the scalar product (2)p.42-8. What is the property of the Gram matrix?'''

Solution (Note: solved independently):

Fourier basis functions:
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$$  \displaystyle \textbf{B} = \left\{1,cos(n\omega\theta),sin(n\omega\theta)\right\}, n = 1,2,.... $$     (8.1.1)
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Gram matrix:
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$$  \displaystyle \mathbf{\Gamma} = [\mathbf{\Gamma_{ij}}] = [(\mathbf b_i \cdot \mathbf b_j)] \in \mathbb{R} $$     (8.1.2)
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Definition of inner product for functions:
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$$  \displaystyle \langle g,h \rangle = \int_a^b g(x)h(x)dx $$     (8.1.3)
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To form the Gram matrix for the Fourier basis functions given in Eq. 8.1.1, it is necessary to compute the inner product for all combinations of basis functions; ie, $$g(\theta)=1,h(\theta)=cos(n\omega\theta); g(\theta)=1,h(\theta)=sin(n\omega\theta); g(\theta)=cos(n\omega\theta),h(\theta)=cos(n\omega\theta);$$ etc. Using Eq. 3 for all permutations, the components of the Gram matrix are computed below. The diagonal terms are given by:

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$$  \displaystyle \begin{align} \langle 1,1 \rangle &= 1 \\ &=\Gamma_{11} \end{align} $$     (8.1.4)
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$$  \displaystyle \begin{align} \langle \text{cos}(n\omega\theta),\text{cos}(n\omega\theta) \rangle& = \int_0^T \text{cos}^2(n\omega\theta)d\theta \\ &= \int_0^T \frac{1}{2} + \frac{1}{2}\text{cos}(2n\omega\theta) d\theta \\ &=\left[\frac{\theta}{2} + \frac{\text{sin}(2n\omega\theta)}{4n\omega}\right]_0^T \\ &= \frac{T}{2} + \frac{\text{sin}(2n\omega T)}{4n\omega} \\ & =\Gamma_{22} \end{align} $$     (8.1.5)
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$$  \displaystyle \begin{align} \langle \text{sin}(n\omega\theta),\text{sin}(n\omega\theta) \rangle& = \int_0^T \text{sin}^2(n\omega\theta)d\theta \\ &= \int_0^T \frac{1}{2} - \frac{1}{2}\text{cos}(2n\omega\theta) d\theta \\ &=\left[\frac{\theta}{2} - \frac{\text{sin}(2n\omega\theta)}{4n\omega}\right]_0^T \\ &= \frac{T}{2} - \frac{\text{sin}(2n\omega T)}{4n\omega} \\ & =\Gamma_{33} \end{align} $$     (8.1.6)
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And the off-diagonal terms, noting that due to symmetry of the integral, $$\Gamma_{ij} = \Gamma_{ji}$$ for $$i\neq j$$:

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$$  \displaystyle \begin{align} \langle 1,\text{cos}(n\omega\theta) \rangle& = \int_0^T (1)(\text{cos}(n\omega\theta))d\theta \\ &= \frac{1}{n\omega}[\text{sin}(n\omega\theta)]_0^T \\ &= \frac{1}{n\omega}\text{sin}(n\omega T)\\ & =\Gamma_{12} = \Gamma_{21} \end{align} $$     (8.1.7)
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$$  \displaystyle \begin{align} \langle 1,\text{sin}(n\omega\theta) \rangle& = \int_0^T (1)(\text{sin}(n\omega\theta))d\theta \\ &= -\frac{1}{n\omega}[\text{cos}(n\omega\theta)]_0^T \\ &= -\frac{1}{n\omega}\left(\text{cos}(n\omega T)-1\right) \\ & =\Gamma_{13} = \Gamma_{31} \end{align} $$     (8.1.8)
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$$  \displaystyle \begin{align} \langle \text{cos}(n\omega\theta),\text{sin}(n\omega\theta) \rangle& = \int_0^T (\text{cos}(n\omega\theta))(\text{sin}(n\omega\theta))d\theta \\ &= \int_0^T \frac{1}{2}\text{sin}(2n\omega\theta) d\theta \\ &= -\frac{1}{4n\omega}[\text{cos}(2n\omega\theta)]_0^T \\ &= -\frac{1}{4n\omega}\left(\text{cos}(2n\omega T)-1\right) \\ & =\Gamma_{13} = \Gamma_{31} \end{align} $$     (8.1.9)
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Substituting the results of Eq. 8.1.4-8.1.9 into Eq. 8.1.2 yields the Gram matrix for the Fourier basis functions:
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$$  \displaystyle \mathbf{\Gamma} = \begin{bmatrix} 1 & \frac{1}{n\omega}\text{sin}(n\omega T) & -\frac{1}{n\omega}\left(\text{cos}(n\omega T)-1\right) \\ \frac{1}{n\omega}\text{sin}(n\omega T) & \frac{T}{2} + \frac{\text{sin}(2n\omega T)}{4n\omega} & -\frac{1}{4n\omega}\left(\text{cos}(2n\omega T)-1\right)\\ -\frac{1}{n\omega}\left(\text{cos}(n\omega T)-1\right) & -\frac{1}{4n\omega}\left(\text{cos}(2n\omega T)-1\right) & \frac{T}{2} - \frac{\text{sin}(2n\omega T)}{4n\omega} \end{bmatrix} $$     (8.1.10)
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By inspection, it is clear that the columns of $$\mathbf{\Gamma}$$ are linearly independent as one cannot form one vector from a linear combination of the other two; thus, the Gram matrix for the Fourier basis functions is invertible for all values of n.

R8.1(cont)

Show that $$\left\{cos\theta,sin\theta\right\}$$ are linearly independent using the Wronskian; see (2)p.35-1

Solution (Note: Solved independently):

Wronskian:
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$$  \displaystyle W = \begin{vmatrix} u_1 & u_2\\ \dot{u_1} & \dot{u_2} \end{vmatrix} $$     (8.1.11) in which $$\dot{u}$$ represents the derivative of u. Thus, for the Fourier basis functions in Eq. 8.1.1,
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$$  \displaystyle \begin{align} W &= \begin{vmatrix} \text{cos}\theta & \text{sin}\theta\\ \dot{\text{cos}\theta } & \dot{\text{sin}\theta} \end{vmatrix}\\ &= \begin{vmatrix} \text{cos}\theta & \text{sin}\theta\\ -\text{sin}\theta & \text{cos}\theta \end{vmatrix} \\ &= \text{cos}^2\theta + \text{sin}^2\theta = 1 \neq 0 \end{align} $$     (8.1.12)
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As the determinant is nonzero the basis functions are linearly independent.

R8.7

Consider the following polynomial:

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$$  \displaystyle \begin{align} q &\in \mathcal{P}_n \Rightarrow q(x) = \sum_{i=0}^n c_ix^i \\ n &= 5; c_0 = -4,c_1=-7,c_2=3,c_3=2,c_4=-5,c_5=1 \end{align} $$     (8.1.13) Find $$d_i$$ such that
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$$  \displaystyle q(x) = \sum_{i=0}^n d_iP_i(x) \in \mathcal{P}_n $$     (8.1.14) '''and plot Eq. 8.1.12 and 8.1.13 on separate figures'''
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Solution (Note: Solved independently):

The goal of this problem is essentially to complete change of basis such that the polynomial q(x) given in Eq. 8.1.13 is mapped onto the orthogonal basis of Legendre polynomials $$P_i$$. This requires the coefficients to the polynomials which can be obtained using Eq. (3)p.42-16 in the class notes, adapted here in Eq. 8.1.15:
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$$  \displaystyle d_i = \frac{2i+1}{2} \int_{x=-1}^{x=1} P_i(x) q(x)dx $$     (8.1.15)
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The Legendre polynomials are well known and can be found online; see for example Wolfram's website (http://mathworld.wolfram.com/LegendrePolynomial.html). Thus, it is only necessary to multiply the polynomials and integrate from -1 to 1; this yields the required coefficients $$d_0$$ through $$d_5$$. For brevity, the algebra for the first coefficient will be included and only the values of the remaining coefficients will be presented. All calculations were verified using Mathematica.

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$$  \displaystyle \begin{align} d_0& = \frac{2(0)+1}{2} \int_{x=-1}^{x=1} (1) (x^5-5x^4+2x^3+3x^2-7x-4)dx \\ &= \frac{1}{2}\left[\frac{1}{6}x^6 - x^5 + \frac{1}{2} x^4 +x^3 -\frac{7}{2}x^2 -4x\right]_{-1}^1 \\ &= -4 \end{align} $$     (8.1.16)
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Similarly;
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$$  \displaystyle d_1=-5.370; d_2= -0.857; d_3=1.244;d_4=-1.143;d_5=0.127 $$     (8.1.17)
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To verify that the mapping of q(x) onto the Legendre polynomial basis functions, the original polynomial and the summation of $$\sum_{i=0}^{n=5}d_i P_i$$ are plotted separately from $$-10<x<10$$. As expected, the figures are identical.

title: User:Egm6322.s12.team2.shields


text:
User talk:Egm4313.sp12.team4.diroma

Hi there. I moved your page to User:Egm4313.sp12.team4.diroma/report1. I think this is a better location for it since it is not related to the main page of Wikiversity. Let me know if you have questions. --HappyCamper 05:52, 27 January 2012 (UTC)

title: User talk:Egm4313.sp12.team4.diroma


text:
User talk:Ashaker13

Your editing experiments
You can continue to participate at Wikiversity and keep other community members from reverting or removing your edits as vandalism by conducting your editing experiments in the sandbox, and in your own user space when you login or create an account. You can contact me or the Wikiversity community with any questions you may have. Thank you. --Mathonius 06:49, 27 January 2012 (UTC)

title: User talk:Ashaker13


text:
User:Tiloz

Swedish hardcore gamer. Ex game developer. Studying computer technology and electricity at ITG Skövde.

title: User:Tiloz