text:
User:Egm4313.s12.team17/Report 1

= Problem 1 - Equation of Motion of a Spring-Mass-Dashpot System in Parallel =

Given

 * k \, is the spring constant
 * c \, is the damping constant
 * y(t) is the displacement of the mass

Problem Statement
Derive the equation of motion for the system, with a mass m \, and applied force f(t).

Solution
Kinematics:
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$$  \displaystyle y_c = y_k = y $$ (Eq.1.1)
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 *
 * }

Kinetics:
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$$  \displaystyle my'' + f_I = f(t) $$     (Eq.1.2)
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 *
 * }

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$$  \displaystyle f_I = f_k +f_c $$     (Eq.1.3)
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 *
 * }

Equations:
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$$  \displaystyle f_k = ky_k = ky $$ (Eq.1.4)
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 * }

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$$  \displaystyle f_c = cy'_c = cy' $$     (Eq.1.5)
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 * }

By substitution of Eq. 1.4 and Eq. 1.5 into Eq. 1.3:
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$$  \displaystyle f_I = ky + cy' $$     (Eq.1.6)
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 *
 * }

Then substituting Eq. 1.6 into Eq. 1.2:
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$$  \displaystyle my'' + cy' + ky =f(t) $$     (Eq.1.7) Which is the equation of motion of a spring-dashpot system in parallel with a mass m \, and applied force f(t).
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 * }

= Problem 2 - Equation of Motion of a Spring-Mass-Dashpot System =

Given

 * k \, is the spring constant
 * c \, is the damping constant
 * m \, is the mass of the ball
 * y(t) is the displacement of the mass

Problem Statement
Derive the equation of motion of for the system, with an applied force r(t)  on the ball.

Solution
Kinematics:
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$$  \displaystyle y_c = y_k = y $$ (Eq.2.1)
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 *
 * }

Kinetics:
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$$  \displaystyle my'' + f_I = r(t) $$     (Eq.2.2)
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 *
 * }

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$$  \displaystyle f_I = f_k +f_c $$     (Eq.2.3)
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 *
 * }

Equations:
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$$  \displaystyle f_k = ky_k = ky $$ (Eq.2.4)
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 * }

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$$  \displaystyle f_c = cy'_c = cy' $$     (Eq.2.5)
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 *
 * }

By substitution of Eq. 2.4 and Eq. 2.5 into Eq. 2.3:
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$$  \displaystyle f_I = ky + cy' $$     (Eq.2.6)
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 *
 * }

Then substituting Eq. 2.6 into Eq. 2.2:
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$$  \displaystyle my'' + cy' + ky =r(t) $$     (Eq.2.7) Which is the equation of motion of a spring-mass-dashpot system with a mass m \, and applied force r(t).
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= Problem 3 - Equation of Motion of a Spring-Mass-Dashpot System =

Given

 * k \, is the spring constant
 * c \, is the damping constant
 * m \, is the mass of the ball
 * y(t) is the displacement of the mass
 * f(t) is the applied force
 * f_I  is the internal force

Equation of motion:
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$$  \displaystyle my'' + f_I = f(t) $$     (Eq.3.1)
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 *
 * }

Problem Statement
Draw the free-body diagrams and derive the Eq. 3.1 for this system.

Mass Free-body Diagram
 (Figure 3.1)

Dashpot Free-body Diagram

 (Figure 3.2)

Spring Free-Body Diagram

 (Figure 3.3)

Derivation
Cutting through the through the link that connects the mass to the dashpot exposes the internal force f_I

 (Figure 3.4)

Note that based on Newton's Third Law and referring to figures 3.1, 3.2, and 3.3, the following relation can be derived:
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\displaystyle f_I = f_c = f_k $$
 *
(Eq. 3.2)
 * }

Apply Newton's Second Law of Motion to the mass system:
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\sum F = m \frac{dV}{dt} $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq. 3.3)
 * }
 * }

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\sum F = f(t) - f_I = m \frac{dV}{dt} $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq. 3.4)
 * }
 * }

By definition:
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\frac{dV}{dt} = a(t) = y'' $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq. 3.5)
 * }
 * }

Substituting Eq. 3.5 into Eq. 3.4 results in:
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f(t) - f_I = ma(t) = my'' $$
 * $$\displaystyle
 * $$\displaystyle
 *
(Eq. 3.6)
 * }
 * }

Rearranging Eq. 3.6 results in:
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$$  \displaystyle my'' + f_I = f(t) $$     (Eq.3.1)
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 * }

= Problem 4 - Modeling: Electric Circuits =

Given
Charge Equation:
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$$  \displaystyle Q = \int I dt = C\nu_C $$     (Eq.4.1)
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 * }

Circuit Equations:
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$$  \displaystyle V = LC\frac{d^2\nu_C}{dt^2} + RC\frac{d\nu_C}{dt} + \nu_C $$     (Eq.4.2)
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 * }

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$$  \displaystyle V^\prime = LI^{\prime\prime} + RI^\prime + \frac{1}{C}I $$     (Eq.4.3)
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 * }

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$$  \displaystyle V = LQ^{\prime\prime} + RQ^\prime + \frac{1}{C}Q $$     (Eq.4.4)
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 * }

Problem Statement
Use Eq. 4.2 to derive the alternative forms of the circuit equation 4.3 and 4.4.

Solution
Solve for \nu_C \, in terms of I for Eq. 4.1:
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$$  \displaystyle \nu_C = \frac{1}{C}\int I dt = \frac{1}{C}Q $$     (Eq.4.5)
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 * }

Substitute Eq. 4.5 into Eq. 4.2:
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$$  \displaystyle V = LC\frac{d^2\nu_C}{dt^2} + RC\frac{d\nu_C}{dt} + \frac{1}{C}\int I dt $$ (Eq.4.6)
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 * }

Take the differential of Eq. 4.1 twice with respect to t \, in terms of Q and \nu_C \,:
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$$  \displaystyle Q^\prime = C\frac{d\nu_C}{dt} $$     (Eq.4.7)
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 * }

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$$  \displaystyle Q^{\prime\prime} = C\frac{d^2\nu_C}{dt^2} $$     (Eq.4.8)
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 * }

Substitute Eq. 4.5, 4.7, and 4.8 into Eq. 4.6:
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$$  \displaystyle V = LQ^{\prime\prime} + RQ^\prime + \frac{1}{C}Q $$     (Eq.4.4)
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 *

 * }

Take the differential of Eq. 4.4 with respect to t \,:
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$$  \displaystyle V^\prime = LQ^{\prime\prime\prime} + RQ^{\prime\prime} + \frac{1}{C}Q^\prime $$     (Eq.4.9)
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 * }

Take the differential of Eq. 4.1 three time with respect to t \, in terms of Q and I :
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$$  \displaystyle Q^\prime = I $$ (Eq.4.10)
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 * }

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$$  \displaystyle Q^{\prime\prime} = I^\prime $$     (Eq.4.11)
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 * }

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$$  \displaystyle Q^{\prime\prime\prime} = I^{\prime\prime} $$     (Eq.4.12)
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 *

 * }

Substitute Eq. 4.10, 4.11, and 4.12 into Eq. 4.9:
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$$  \displaystyle V^\prime = LI^{\prime\prime} + RI^\prime + \frac{1}{C}I $$     (Eq.4.3)
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 *

 * }

= Problem 5 - Homogeneous Linear Ordinary Differential Equations (ODEs) with Constant Coefficients =

Part 1 Problem Statement
Find a general solution to the following ODE:

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$$  \displaystyle y^{\prime\prime} + 4 y^\prime + (\pi^2 + 4)y = 0 $$     (Eq.5.1.1)
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 * }

Part 1 Solution
Characteristic Equation of Eq. 5.1.1:
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$$  \displaystyle x^2 + 4x + (\pi^2 + 4) = 0 $$     (Eq.5.1.2) Solve for the roots in Eq. 5.1.2 by using the quadratic equation:
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 * }
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$$  \displaystyle x = \frac{-4 \pm\sqrt{16-4(1)(\pi^2 + 4)}}{2} $$     (Eq.5.1.3)
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 * }

Simplifying Eq. 5.1.3 gives:
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$$  \displaystyle x = \frac{-4 \pm\sqrt{-4\pi^2}}{2} $$ $$  \displaystyle x = \frac{-4 \pm 2\pi\mathit{i}}{2} $$     (Eq.5.1.4)
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 * }

Eq. 5.1.4 gives the roots of:

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$$  \displaystyle x = -2\pm \pi\mathit{i} $$     (Eq.5.1.5)
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Additionally, it is known that when the roots of the characteristic equation are of the form:
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$$  \displaystyle \alpha\pm\beta\mathit{i} $$     (Eq.5.1.6)
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Which α and β are real.

Then the general solution of the ODE will be in the form:
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$$  \displaystyle y = C_1 e^{\alpha x} \cos(\beta x) + C_2 e^{\alpha x} \sin(\beta x) $$ (Eq.5.1.7) Therefore, the solution is:
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 * }
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$$ y = C_1 e^{-2x} \cos(\pi x) + C_2 e^{-2x} \sin(\pi x) \,$$ (Eq.5.1.8)
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 * }

Part 2 Problem Statement
Find a general solution to the following ODE:

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$$  \displaystyle y^{\prime\prime} + 2\pi y^\prime + \pi^2 y = 0 $$     (Eq.5.2.1)
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Part 2 Solution
Characteristic Equation of Eq. 5.2.1:
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$$  \displaystyle x^2 + 2\pi x + \pi^2 = 0 $$     (Eq.5.2.2)
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 * }

Solve for the roots in Eq. 5.2.2 by using the quadratic equation:
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$$  \displaystyle x = \frac{-2\pi \pm\sqrt{4\pi^2-4(1)(\pi^2)}}{2} $$     (Eq.5.2.3) Simplifying Eq. 5.2.3:
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 * }
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$$  \displaystyle x = \frac{-2\pi}{2} $$ $$  \displaystyle x = -\pi $$     (Eq.5.2.4) which is a real double root.
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 * }

Characteristic equations containing real double roots produces a general solution for the ODE in the form:
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$$  \displaystyle y = C_1 e^{-\alpha x} + C_2 x e^{-\alpha x} $$ (Eq.5.2.5) Therefore, the solution is:
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 * }
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$$ y = C_1 e^{-\pi x} + C_2 x e^{-\pi x} \,$$ (Eq.5.2.6)
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 * }

= Problem 6 - Order of an Ordinary Differential Equation (ODE) and Principle of Superposition = For problem reference.

Applying Superposition Principle
Consider the linear differential equation in standard form:
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$$  \displaystyle y^{\prime\prime} + p(x)y^\prime + q(x)y = r(x) $$     (Eq.6.1)
 *

 * }

Now consider a function $$ \bar y(x)$$, defined as the sum of the homogeneous solution y_h  and the particular solution y_p :
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$$  \displaystyle \bar y(x) = y_h(x)+ y_p(x) $$     (Eq.6.2)
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 * }

The homogeneous solution to Eq. 6.1 is:
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$$  \displaystyle y_h^{\prime\prime}+p(x)y_h^\prime+q(x)y_h=0 $$     (Eq.6.3)
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 * }

The particular solution to Eq. 6.1 is:
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$$  \displaystyle y_p^{\prime\prime}+p(x)y_p^\prime+q(x)y_p=r(x) $$     (Eq.6.4)
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 * }

Eq. 6.3 and Eq. 6.4 are now summed in order to get:
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$$  \displaystyle (y_h^{\prime\prime}+y_p^{\prime\prime})+p(x)(y_h^\prime+y_p^\prime)+q(x)(y_h+y_p)=r(x) $$     (Eq.6.5)
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 * }

If the equation is linear we can assume:
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$$  \displaystyle y_h^{\prime\prime}+y_p^{\prime\prime} \Rightarrow (y_h+y_p)^{\prime\prime} \Rightarrow \bar y^{\prime\prime} $$     (Eq.6.6) also,
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$$  \displaystyle y_h^\prime+y_p^\prime \Rightarrow (y_h+y_p)^\prime \Rightarrow \bar y^\prime $$     (Eq.6.7)
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Now we can substitute Eq. 6.6 and Eq. 6.7 into Eq. 6.5 which yields:
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$$  \displaystyle \bar y^{\prime\prime}+p(x)\bar y^\prime+q(x)\bar y=r(x) $$     (Eq.6.8) which is the same as:
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 * }

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$$  \displaystyle y^{\prime\prime} + p(x)y^\prime + q(x)y = r(x) $$     (Eq.6.1)
 *

 * }

Definition of Linearity
A differential equation is linear if it has the form:

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$$  \displaystyle a_n(x)\frac{d^ny}{dx^n}+a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}}+...+a_1(x)\frac{dy}{dx}+a_0(x)y=F(x) $$
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 * }

Requirements are the following:
 * 1) The coefficients $$(a_n(x), a_{n-1}(x), etc.)$$ and $$ F(x) $$ only depend on the independent variable x.
 * 2) No products of the function $$ y $$ and its derivatives.
 * 3) The function can only be raised to the first power, which goes the same for the derivatives.

Part 1 Problem Statement
Determine the following for the ODE shown below:
 * 1) The order
 * 2) Linearity (or lack of)
 * 3) Whether the principle of superposition can be applied

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$$  \displaystyle y^{\prime\prime} = g = constant $$     (Eq.6.1.1)
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Part 1 Solution

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Order: order ODE
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 * }
 * }

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Linear: Yes
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 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

Using the information from the "Applying Superposition Principle" section.

Recall the function:
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$$  \displaystyle \bar y(x) = y_h(x)+y_p(x) $$     (Eq.6.2)
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 *

 * }

Homogeneous Equation is:
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$$  \displaystyle y_h^{\prime\prime} = 0 $$     (Eq.6.1.2)
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 *

 * }

And Particular solution is:
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$$  \displaystyle y_p^{\prime\prime} = g $$ (Eq.6.1.3)
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Eq. 6.1.2 and Eq. 6.1.3 can be added to obtain:
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$$  \displaystyle y_h^{\prime\prime} +y_p^{\prime\prime} = g $$ (Eq.6.1.4)
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 * }

Now we can use Eq. 6.6 to simplify Eq. 6.1.4:
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$$  \displaystyle (y_h +y_p)^{\prime\prime} = g $$ (Eq.6.1.5)
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And the result is:
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$$  \displaystyle \bar y^{\prime\prime} = g $$ (Eq.6.1.6)
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 * }

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Superposition Principle can be applied since Eq. 6.1.6 is the same as Eq. 6.1.1.
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 * style="width:36%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

Part 2 Problem Statement
Determine the following for the ODE shown below:
 * 1) The order
 * 2) Linearity (or lack of)
 * 3) Whether the principle of superposition can be applied

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

$$  \displaystyle mv^\prime = mg-bv^2 $$     (Eq.6.2.1)
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 *

 * }

Part 2 Solution

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Order: order ODE
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

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

Linear: No, due to
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * style="width:11%; padding:10px; border:2px solid #8888aa" |

 * }
 * }

First rearrange the equation to:
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$$  \displaystyle mv^\prime+bv^2 = mg $$ (Eq.6.2.2)
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 * }

Then use the information from the "Applying Superposition Principle" section.

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$$  \displaystyle \bar v(x)=v_h(x)+v_p(x) $$     (Eq.6.2.3)
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 * }

Homogeneous equation is:
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$$  \displaystyle mv_h^\prime+bv_h^2 = 0 $$     (Eq.6.2.4)
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 * }

Particular Solution is:
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$$  \displaystyle mv_p^\prime+bv_p^2 =mg $$     (Eq.6.2.5)
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 * }

Adding Eq. 6.2.4 and Eq. 6.2.5:
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$$  \displaystyle mv_h^\prime+mv_p^\prime +bv_p^2+bv_h^2 =mg $$     (Eq.6.2.6)
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 *

 * }

Now Eq. 6.2.6 simplifies to:
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$$  \displaystyle m(v_h+v_p)^\prime+b(v_h^2+v_p^2) =mg $$     (Eq.6.2.7)
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 * }

Notice that:
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$$  \displaystyle (v_h^2+v_p^2)\neq (v_h+v_p)^2 \neq \bar v^2 $$     (Eq.6.2.8)
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Therefore, superposition principle is invalid.
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 * }
 * }

Part 3 Problem Statement
Determine the following for the ODE shown below:
 * 1) The order
 * 2) Linearity (or lack of)
 * 3) Whether the principle of superposition can be applied

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$$  \displaystyle h^\prime = -k\sqrt{h} $$     (Eq.6.3.1)
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 *

 * }

Part 3 Solution

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

Order: order ODE
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

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

Linear: No, due to the
 * style="width:13%; padding:10px; border:2px solid #8888aa" |
 * style="width:13%; padding:10px; border:2px solid #8888aa" |

 * }
 * }

First rearrange the equation to:
 * {| style="width:100%" border="0"

$$  \displaystyle h^\prime+k\sqrt{h}= 0 $$     (Eq.6.3.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using the information from the "Applying Superposition Principle" section:
 * {| style="width:100%" border="0"

$$  \displaystyle \bar h(x) = h_h(x)+h_p(x) $$     (Eq.6.3.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The Homogeneous equation is:
 * {| style="width:100%" border="0"

$$  \displaystyle h_h^\prime+k\sqrt{h_h}= 0 $$     (Eq.6.3.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The Particular solution is:
 * {| style="width:100%" border="0"

$$  \displaystyle h_p^\prime+k\sqrt{h_p}= 0 $$     (Eq.6.3.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Adding Eq. 6.3.4 and Eq. 6.4.5 together:
 * {| style="width:100%" border="0"

$$  \displaystyle h_h^\prime+h_p^\prime+k\sqrt{h_h}+k\sqrt{h_p}= 0 $$     (Eq.6.3.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Simplifying gives us:
 * {| style="width:100%" border="0"

$$  \displaystyle (h_h+h_p)^\prime+k(\sqrt{h_h}+\sqrt{h_p})= 0 $$     (Eq.6.3.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle (\sqrt{h_h}+\sqrt{h_p})\neq \sqrt{(h_h+h_p)} \neq \sqrt{\bar h} $$ (Eq.6.3.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

Therefore, superposition cannot be applied.
 * style="width:22%; padding:10px; border:2px solid #8888aa" |
 * style="width:22%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

Part 4 Problem Statement
Determine the following for the ODE shown below:
 * 1) The order
 * 2) Linearity (or lack of)
 * 3) Whether the principle of superposition can be applied

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

$$  \displaystyle my^{\prime\prime} + ky = 0 $$     (Eq.6.4.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 4 Solution

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

Order: order ODE
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

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

Linear: Yes
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

To determine if superposition is valid, we apply the information from the "Applying Superposition Principle" section.

Recall the function:
 * {| style="width:100%" border="0"

$$  \displaystyle \bar y(x) = y_h(x)+y_p(x) $$     (Eq.6.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Homogeneous Equation is:
 * {| style="width:100%" border="0"

$$  \displaystyle my_h^{\prime\prime} + ky_h = 0 $$     (Eq.6.4.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Particular Equation is:
 * {| style="width:100%" border="0"

$$  \displaystyle my_p^{\prime\prime} + ky_p = 0 $$     (Eq.6.4.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Adding Eq. 6.4.2 and Eq. 6.4.3 results in:
 * {| style="width:100%" border="0"

$$  \displaystyle my_h^{\prime\prime} + my_p^{\prime\prime}+ky_h+ky_p = 0 $$     (Eq.6.4.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Simplifying equation Eq. 6.4.4, we have:
 * {| style="width:100%" border="0"

$$  \displaystyle m(y_h+y_p)^{\prime\prime} + k(y_h+y_p) = 0 $$     (Eq.6.4.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

By using Eq. 6.2 from the "Applying Superposition Principle" section, Eq. 6.4.5 becomes:
 * {| style="width:100%" border="0"

$$  \displaystyle m\bar y^{\prime\prime} + k\bar y = 0 $$     (Eq.6.4.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

Since Eq. 6.4.6 and Eq. 6.4.1 are the same, the Superposition Principle can be applied.
 * style="width:38%; padding:10px; border:2px solid #8888aa" |
 * style="width:38%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

Part 5 Problem Statement
Determine the following for the ODE shown below:
 * 1) The order
 * 2) Linearity (or lack of)
 * 3) Whether the principle of superposition can be applied

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

$$  \displaystyle y^{\prime\prime} + {\omega_0}^2y = \cos \omega t, \omega_0\approx \omega $$     (Eq.6.5.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 5 Solution

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

Order: order ODE
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

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

Linear: Yes
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

Superposition can be applied by using the information in the "Applying Superposition Principle" section.

Recall the function:
 * {| style="width:100%" border="0"

$$  \displaystyle \bar y(x) = y_h(x)+y_p(x) $$     (Eq.6.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Homogeneous Equation is:
 * {| style="width:100%" border="0"

$$  \displaystyle y_h^{\prime\prime} + {\omega_0}^2y_h = 0 $$     (Eq.6.5.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Particular Equation is:
 * {| style="width:100%" border="0"

$$  \displaystyle y_p^{\prime\prime} + {\omega_0}^2y_p = \cos \omega t $$ (Eq.6.5.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Adding Eq. 6.5.2 and Eq. 6.5.3 we have:
 * {| style="width:100%" border="0"

$$  \displaystyle y_h^{\prime\prime} + y_p^{\prime\prime}+{\omega_0}^2y_h+{\omega_0}^2y_p = \cos \omega t $$ (Eq.6.5.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Simplifying Eq. 6.5.4we have:
 * {| style="width:100%" border="0"

$$  \displaystyle (y_h+y_p)^{\prime\prime}+{\omega_0}^2(y_h+y_p) = \cos \omega t $$ (Eq. 6.5.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using Eq. 6.2 from the "Applying Superposition Principle" we have:
 * {| style="width:100%" border="0"

$$  \displaystyle \bar y^{\prime\prime}+{\omega_0}^2\bar y = \cos \omega t $$ (Eq. 6.5.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

Since Eq. 6.5.1 and Eq. 6.5.6 are the same, the Superposition Principle is valid.
 * style="width:38%; padding:10px; border:2px solid #8888aa" |
 * style="width:38%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

Part 6 Problem Statement
Determine the following for the ODE shown below:
 * 1) The order
 * 2) Linearity (or lack of)
 * 3) Whether the principle of superposition can be applied

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

$$  \displaystyle LI^{\prime\prime} + RI^\prime + \frac{1}{C}I=E^\prime $$     (Eq.6.6.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 6 Solution

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

Order: order ODE
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

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

Linear: Yes
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

Using the information in the "Applying Superposition Principle" section, we can determine if superposition is valid.

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

$$  \displaystyle \bar I(x) = I_h(x)+I_p(x) $$     (Eq.6.6.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Homogeneous and Particular solutions are:
 * {| style="width:100%" border="0"

$$  \displaystyle LI_h^{\prime\prime} + RI_h^\prime + \frac{1}{C}I_h=0 $$     (Eq.6.6.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

and

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

$$  \displaystyle LI_p^{\prime\prime} + RI_p^\prime + \frac{1}{C}I_p=E^\prime $$     (Eq.6.6.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Adding Eq. 6.6.3 and Eq. 6.6.4:
 * {| style="width:100%" border="0"

$$  \displaystyle LI_h^{\prime\prime} +LI_p^{\prime\prime}+ RI_h^\prime +RI_p^\prime+\frac{1}{C}I_h+\frac{1}{C}I_p=E^\prime $$     (Eq.6.6.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Simplifying Eq. 6.6.5:
 * {| style="width:100%" border="0"

$$  \displaystyle L(I_h+I_p)^{\prime\prime}+ R(I_h+I_p)^\prime+\frac{1}{C}(I_h+I_p)=E^\prime $$     (Eq.6.6.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using Eq. 6.6.2 we have:
 * {| style="width:100%" border="0"

$$  \displaystyle L\bar I^{\prime\prime}+ R\bar I^\prime+\frac{1}{C}\bar I=E^\prime $$     (Eq.6.6.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

Eq. 6.6.7 and Eq. 6.6.1 are the same; therefore, the Superposition Principle can be applied.
 * style="width:40%; padding:10px; border:2px solid #8888aa" |
 * style="width:40%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

Part 7 Problem Statement
Determine the following for the ODE shown below:
 * 1) The order
 * 2) Linearity (or lack of)
 * 3) Whether the principle of superposition can be applied

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

$$  \displaystyle EIy^{iv} = f(x) $$     (Eq.6.7.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 7 Solution

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

Order: order ODE
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

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

Linear: Yes
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

To test if superposition is true for this ODE use the information in the "Applying Superposition Principle" section.

Recall the function:
 * {| style="width:100%" border="0"

$$  \displaystyle \bar y(x) = y_h(x)+y_p(x) $$     (Eq.6.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Homogeneous and Particular solutions are:
 * {| style="width:100%" border="0"

$$  \displaystyle EIy_h^{iv} = 0 $$   (Eq.6.7.2)
 * style="width:90%" |
 * style="width:90%" |
 *

 * }

and

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

$$  \displaystyle EIy_p^{iv} = f(x) $$   (Eq.6.7.3)
 * style="width:90%" |
 * style="width:90%" |
 *

 * }

Adding Eq. 6.7.2 and Eq. 6.7.3 we have:
 * {| style="width:100%" border="0"

$$  \displaystyle EIy_h^{iv}+EIy_p^{iv} = f(x) $$   (Eq.6.7.4)
 * style="width:90%" |
 * style="width:90%" |
 *

 * }

Simplifying Eq. 6.7.4:
 * {| style="width:100%" border="0"

$$  \displaystyle EI(y_h+y_p)^{iv} = f(x) $$   (Eq.6.7.5)
 * style="width:90%" |
 * style="width:90%" |
 *

 * }

Using Eq. 6.2 from the "Applying Superposition Principle" section we have:
 * {| style="width:100%" border="0"

$$  \displaystyle EI\bar y^{iv} = f(x) $$   (Eq.6.7.6)
 * style="width:90%" |
 * style="width:90%" |
 *

 * }

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

Since Eq. 6.7.6 is the same as Eq. 6.7.1, superposition can be applied.
 * style="width:32%; padding:10px; border:2px solid #8888aa" |
 * style="width:32%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

Part 8 Problem Statement
Determine the following for the ODE shown below:
 * 1) The order
 * 2) Linearity (or lack of)
 * 3) Whether the principle of superposition can be applied

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

$$  \displaystyle L{\theta}^{\prime\prime}+g\sin \theta = 0 $$     (Eq.6.8.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 8 Solution

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

Order: order ODE
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * style="width:11%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

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

Linear: No, due to
 * style="width:12%; padding:10px; border:2px solid #8888aa" |
 * style="width:12%; padding:10px; border:2px solid #8888aa" |

 * }
 * }

By using the information from the "Applying Superposition Principle" section, we can determine if superposition can be applied.

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

$$  \displaystyle \bar \theta (x) = \theta_h(x)+\theta_p(x) $$     (Eq.6.8.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Homogeneous and Particular equations are:
 * {| style="width:100%" border="0"

$$  \displaystyle L{\theta_h}^{\prime\prime}+g\sin \theta_h = 0 $$   (Eq.6.8.3)
 * style="width:90%" |
 * style="width:90%" |
 *

 * }

and

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

$$  \displaystyle L{\theta_p}^{\prime\prime}+g\sin \theta_p = 0 $$   (Eq.6.8.4)
 * style="width:90%" |
 * style="width:90%" |
 *

 * }

Adding Eq. 6.8.3 and Eq. 6.8.4 we have:
 * {| style="width:100%" border="0"

$$  \displaystyle L{\theta_h}^{\prime\prime}+ L{\theta_p}^{\prime\prime}+g\sin \theta_h+g\sin \theta_p = 0 $$   (Eq.6.8.5)
 * style="width:90%" |
 * style="width:90%" |
 *

 * }

Simplifying Eq. 6.8.5:
 * {| style="width:100%" border="0"

$$  \displaystyle L({\theta_h}+ {\theta_p})^{\prime\prime}+g(\sin \theta_h+\sin \theta_p) = 0 $$   (Eq.6.8.6)
 * style="width:90%" |
 * style="width:90%" |
 *

 * }

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

$$  \displaystyle (\sin \theta_h+\sin \theta_p) \neq \sin ({\theta_h+\theta_p}) \neq \sin {\bar \theta} $$   (Eq.6.8.7)
 * style="width:90%" |
 * style="width:90%" |
 *

 * }

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

Therefore, the Superposition Principle cannot be applied.
 * style="width:28%; padding:10px; border:2px solid #8888aa" |
 * style="width:28%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

= References =

= Contributing Team Members =
 * Allan Axelrod
 * 1) Solved problem 2
 * 2) Reviewed all problems
 * Michael Deaver
 * 1) Solved problem 4
 * 2) Reviewed all problems
 * Max Hintz
 * 1) Solved problem 5
 * 2) Reviewed all problems
 * Kelvin Li
 * 1) Solved problem 6
 * 2) Reviewed all problems
 * Thomas Wheeler
 * 1) Solved problem 3
 * 2) Reviewed all problems
 * Chen Ying
 * 1) Solved problem 1
 * 2) Reviewed all problems

= Links = Team 17 Wiki

title: User:Egm4313.s12.team17/Report 1


text:
File:PipeTaborConceptMap.jpg

Summary
A concept map defining the knowledge domain for learning the pipe and tabor. This concept map is a work in progress...

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

title: File:PipeTaborConceptMap.jpg


text:
Bromley by Bow Centre/Study Circles on Wikiversity/Getting Started

How can I get involved?

 * As a member you identify your contributions and get a set of personal pages where you can engage in discussions.
 * Consider what would be the most suitable username for you to use.
 * [ You can start an account here].

How can I edit pages?

 * Some editing tips
 * Edit Interface
 * How to start a new page

 * You don't need "sign" your submissions to content pages — that's recorded automatically and shown on the History of the page. But you are definitely encouraged to sign your contributions to the Discussion page and any Talk page. Do this with four tildes (~). Here's more about signatures.

 * Wikipedia's guidance on contributing generally apply to Wikiversity as well. If you don't want your writing to be edited or redistributed by others, do not submit it.

 * Be sure to visit the Colloquium for general discussions. For real-time chat with other Wikiversity users, check out our Chat page.


text:
User:EGM4313.s12.team6.davis

Team 6

= Reports =

R1

R2

title: User:EGM4313.s12.team6.davis


text:
Draft:Original research/Literature

Literature is the art of written work, and can, in some circumstances, refer exclusively to published sources.

The two major classifications of literature are poetry and prose (which can be further sub-divided into fiction and non-fiction).

Literature may consist of texts based on factual information (journalistic or non-fiction), as well as on original imagination, such as polemical works as well as autobiography, and reflective essays as well as belles-lettres. Literature can be classified according to historical periods, genres, and political influences. The concept of genre, which earlier was limited, has broadened over the centuries. A genre consists of artistic works which fall within a certain central theme, and examples of genre include romance, mystery, crime, fantasy, erotica, and adventure, among others.

Two others are science fiction and horror.

"Literature" is a highly ambiguous term: at its broadest, it can mean any sequence of words that has been preserved for transmission in some form or other (including oral transmission); more narrowly, it is often used to designate imaginative works such as narrative stories, poems, and plays; more narrowly still, it is used as an honorific and applied only to those works which are considered to have particular merit.

Theory of literature
Def. "the body of [all] written work" is called literature.

The theory of literature involves methods of studying and investigating literature, its "nature and function"; "literary theory, criticism, and history; and general, comparative, and national literature."

Def. “[t]he theory or the philosophy of the interpretation of literature and literary criticism” is called literary theory.

Literary criticism
Def. “[t]he study, discussion, evaluation, and interpretation of literature” is called literary criticism.

"But my point is that one constant within this struggle remains: that an oppositional culture of non-dominant groups has to define itself against the practices and ideology of the dominant group (or groups), and this inevitably has consequences for form. Indeed, only a very unsophisticated literary criticism could conceive of form and content as distinct entities."

Compositions
A newspaper, or online, feature article is composed of the following:
 * 1) a lede,
 * 2) topic sentence,
 * 3) a body, and an
 * 4) ending.

The ratio of each of these may depend on the audience. In an inverted pyramid style the ratios are about 5:3:2 for lead (including topic sentence), body, and ending.

There is also what's called a "news-peg" or "hook", something that will interest a reader, usually the first sentence or the title.

The following elements should be present: What, When, Where, Why, Who, and How. Nearly all of these elements must appear somewhere in the story.

Debates
Def. “a type of literary composition, taking the form of a discussion or disputation” is called debate.

Debate occurs in public meetings, academic institutions, and legislative assemblies.

Debating is also carried out for educational and recreational purposes, usually associated with educational establishments and debating societies.

Debating topics covered a broad spectrum of topics while the debating societies allowed participants from both genders and all social backgrounds, making them an excellent example of the enlarged public sphere of the Age of Enlightenment. Debating societies were a phenomenon associated with the simultaneous rise of the public sphere, a sphere of discussion separate from traditional authorities and accessible to all people that acted as a platform for criticism and the development of new ideas and philosophy.

John Henley, a clergyman, founded an Oratory in 1726 with the principal purpose of "reforming the manner in which such public presentations should be performed." He made extensive use of the print industry to advertise the events of his Oratory, making it an omnipresent part of the London public sphere. Henley was also instrumental in constructing the space of the debating club: he added two platforms to his room in the Newport district of London to allow for the staging of debates, and structured the entrances to allow for the collection of admission. These changes were further implemented when Henley moved his enterprise to Lincoln's Inn Fields. The public was now willing to pay to be entertained, and Henley exploited this increasing commercialization of British society. By the 1770s, debating societies were firmly established in London society.

Dramas
Def. “[a] composition, normally in prose, telling a story and intended to be represented by actors impersonating the characters and speaking the dialogue” is called a drama.

Essays
Def. “[a] written composition of moderate length exploring a particular issue or subject” is called an essay.

An essay is a piece of writing which is often written from an author's personal point of view.

Fictions
Def. a "[l]iterary type using invented or imaginative writing, instead of real facts, usually written as prose” is called fiction.

Romances
The "popularity of old-fashioned romance novels featuring conventional and traditional gender roles seems to defy the stances of the modern-day women's liberation movement."

A romance novel might be characterized as a "hyper-romantic, contrived and extremely unrealistic tales of handsome, manly heroes falling in love with virginal women, enduring a series of adventures, then inexorably ending in a happy resolution."

“Romance novels offer an escape from daily life with the belief that true love really exists.”

"Romance novels [portrayed by the partial cover image on the right] are at once the most scorned and popular form of literature in the world, accounting for as much as 40% of total book sales in much of the world. The average romance reader (and writer) is female, ambitious, leads a very full and busy life, and has an above-average education and intelligence. The livelihood of some of the world’s most critically-acclaimed (mostly male) authors depends on the revenue base generated from the sale of the remarkably diverse genre called ‘romance’, written by and bought overwhelmingly by women."

Mysteries
Def. a "suspenseful, sensational genre of story, book, play or film", such as a "detective story, mystery novel, whodunit, crime fiction", is called a thriller.

"The scripts for the series [I Love a Mystery] were usually themed towards the dark and supernatural, with perhaps the most famous, or infamous (depending on your point of view) scenario being “Temple of the Vampires,” which aroused a great deal of censorial comment when first broadcast as a twenty-episode serial from January 22 through February 16, 1940."

Crimes
"Accounts of true crime have always been enormously popular among readers. The subgenre would seem to appeal to the highly educated as well as the barely educated, to women and men equally. The most famous chronicler of true crime trials in English history is the amateur criminologist William Roughead, a Scots lawyer who between 1889 and 1949 attended every murder trial of significance held in the High Court of Justiciary in Edinburgh, and wrote of them in essays published first in such journals as The Juridical Review and subsequently collected in best-selling books with such titles as Malice Domestic, The Evil That Men Do, What Is Your Verdict?, In Queer Street, Rogues Walk Here, Knave's Looking Glass, Mainly Murder, Murder and More Murder, Nothing But Murder, and many more…. Roughead's influence was enormous, and since his time "true crime" has become a crowded, flourishing field, though few writers of distinction have been drawn to it."

Fantases
Def. a "literary genre generally dealing with themes of magic and [fictive] medieval technology" is called fantasy.

Eroticas
Def. literature "relating to or tending to arouse sexual desire or excitement" is called erotic literature, or erotica.

On the left is a cover image from the new erotica e-book by Elizabeth Black called "PURR a Puss 'n Boots Twisted Tale".

Adventures
Adventure fiction is a genre of fiction in which an adventure, an exciting undertaking involving risk and physical danger, forms the main storyline.

"An adventure is an event or series of events that happens outside the course of the protagonist's ordinary life, usually accompanied by danger, often by physical action. Adventure stories almost always move quickly, and the pace of the plot is at least as important as characterization, setting and other elements of a creative work."

Science fictions
Def.' "[f]iction in which advanced technology and/or science is a key element" is called science fiction.

Horrors
Def. a "genre of fiction, meant to evoke a feeling of fear and suspense" is called horror.

The Mammoth Book of Best New Horror specifically and exclusively focuses on and publishes current horror fictional literature.

Laws
Def. “[a] written or understood rule that concerns behaviors and the appropriate consequences thereof” is called law.

Narratives
Def. "the systematic recitation of an event or series of events" is called a narrative.

A narrative is a constructive format (as a work of speech, writing, song, film, television, video games, photography or theatre) that describes a sequence of non-fictional or fictional events.

Further, the word "story" may be used as a synonym of "narrative", but can also be used to refer to the sequence of events described in a narrative. A narrative can also be told by a character within a larger narrative. An important part of narration is the narrative mode, the set of methods used to communicate the narrative through a process narration.

Philosophies
Def. a written work “that seeks truth through reasoning rather than empiricism” is called philosophy.

"The Tengrism tangraïsme or sometimes (in Mongolian: Тэнгэр шүтлэг, Tenger shütleg, worship (or religion) of heaven) was the major belief of Xiongnu and Xianbei which consisted of the Turkish population, Mongolian, Hungarian and Bulgarian in antiquity. It focuses around the divinity of the eternal sky Tengri (also transliterated Tangri, Tanrı, Tangra, etc.), and incorporates elements of shamanism, animism, totemism and ancestor worship." Translated using Google Translate.

The Daychin Tengry - Turkic pagan god is shown in the image on the right.

That a text of this philosophy, or religion, probably existed is suggested by the stamp on the left used by Güyük Khan, in a letter of 1246.

Plays
Def. "[a] literary composition, intended to be represented by actors impersonating the characters and speaking the dialogue" or "a theatrical performance featuring actors" is called a play.

Poems
Def. a literary piece written in verse” is called a poem.

Geography
At the right is an allegorical painting in the Pinacoteca do Estado de São Paulo, Brazil, 1915.

It is one among many connected to the literature of the time preceded by a long and baroque heritage expressed even in the final years of the nineteenth century in various regions and in art and culture.

On the left is a form of geographic literature consisting of a geographic map showing the locations of Turkic peoples and their languages.

History
Def. “[a] [written] record or narrative description of past events” is called history.

On the right is an illustration of the "Hof van Beselare" in Flanders, Belgium, from the history book, Flandria Illustrata by Antoon Sanders.

Natural sciences
Def. a written work “studying phenomena or laws of the physical world” is called natural science.

An example of a natural science, or sciences, journal is shown on the right with its cover image.

Abstractions
Def.
 * 1) a "separation from worldly objects",
 * 2) "the withdrawal from one's senses",
 * 3) the "act of focusing on one characteristic of an object rather than the object as a whole group of characteristics; the act of separating said qualities from the object or ideas",
 * 4) the "act of comparing commonality between distinct objects and organizing using those similarities; the act of generalizing characteristics; the product of said generalization",
 * 5) an "idea of an unrealistic or visionary nature", or
 * 6) any "generalization technique that ignores or hides details to capture some kind of commonality between different instances for the purpose of controlling the intellectual complexity of engineered systems, particularly software systems"

is called an abstraction.

The image on the right is an example of the literature of abstraction.

Intellects
On the right, as an example of intellect literature, is an image of the Issue 2 cover of Intellect Magazine.

Def. "the faculty of thinking, judging, abstract reasoning, and conceptual understanding; the cognitive faculty" is called an intellect.

Humanities
On the right is a cover for the book A New Companion to Digital Humanities, a companion to a book on the humanities.

Sciences
On the right is a cover image for performing health sciences literature reviews.

Engineering
Def. the "application of [mathematics and the physical sciences] to the needs of humanity and the development of technology" is called engineering

On the right, as an example of engineering literature, is a cover image of the List of Periodical Engineering Literature.

Humor
Def. the "quality of being amusing, comical, funny" is called humor, or humour.

Literary history
"To justify the existence and the methods of literary history is entirely superfluous nowadays, and it is no less superfluous to dwell upon the differences and likenesses between it and literary criticism. Our common sense tells us, if we do away with prejudices and futile scholarly discussions, that literary history, working in its own field, is trying neither to replace nor to oppose literary criticism. Literary history thinks that it can help literary criticism; can clear a path for it; can lighten its task of understanding, judging, and classifying literary works and the great movements of human thought. It offers its services as a devoted auxiliary, modest and self-effacing. It has no imperialistic designs: it covers enough territory already to have no need to encroach on that of a neighbor. It prepares the material for the critic but puts no restrictions on the way [she] should use it. If he has faith in impressionistic criticism, if [she] believes that the literary critic should surrender himself to the emotion produced by the book [she] is studying and then should express this emotion with precision and delicacy, he is free to do so. Literary history asks [her] only to base his personal reaction on facts that have been historically verified, to define [her] position clearly, and, when communicating a purely personal reaction to the public, not to believe or to make others believe that he is giving any added information about the work or its writer. "Impressionism", says Lanson, "is the only method that puts us in touch with beauty. Let us, then, use it for this purpose, frankly, but let us limit it to this, rigorously. To distinguish knowing from felling, what we may know from what we should feel; to avoid feeling when we can know, and thinking that we know when we feel: to this, it seems to me, the scientific method of literary history can be reduced."1"

Sweden
The Rök runestone is considered the first piece of written Swedish literature and thus it marks the beginning of the history of Swedish literature.

It was proposed that the inscription has nothing at all to do with the recording of heroic sagas and that it contains riddles which refer only to the making of the stone itself.

The Rök runestone inscription is not connected to heroic deeds in war. Instead it deals with the conflict between light and darkness, warmth and cold, life and death.

"The Rök runestone from central middle Sweden, [is] dated to around 800 CE [...] Combining perspectives and findings from semiotics, philology, archaeology, and history of religion, the study presents a completely new interpretation which follows a unified theme, showing how the monument can be understood in the socio-cultural and religious context of early Viking Age Scandinavia. The inscription consists, according to the proposed interpretation, of nine enigmatic questions. Five of the questions concern the sun, and four of them, it is argued, ask about issues related to the god Odin. A central finding is that there are relevant parallels to the inscription in early Scandinavian poetry, especially in the Eddic poem Vafþrúðnismál."

Hypotheses

 * 1) Literature may also exist from other species or ancestors of hominins.

title: Draft:Original research/Literature


text:
User:EGM4313.s12.team6.davis/R1

Report Problem 5
This problem consists of finding a general solution to two linear, second order differential equations with constant coefficients. The problems are taken from the tenth edition of Erwin Kreyszig's Advanced Engineering Mathematics and can be found as problems 4 and 5 of problem set 2.2 on page 59.

Given:
$$y''+6y'+(\pi^2+4)y=0$$

Find:
The homogeneous solution to the differential equation.

Solution:
Use the method of undetermined coefficients, also known as the method of trial solution, to solve the differential equation.

Let   $$y=e^{\lambda x}$$

Therefore:

 * $$y'=\lambda e^{\lambda x}$$
 * and
 * $$y''=\lambda^2 e^{\lambda x}$$

So,

 * $$y''+6y'+(\pi^2+4)y=0$$

can be written as

 * $$e^{\lambda x}(\lambda^2+a\lambda+b)=0$$  where   $$a=6$$   and   $$b=\pi^2+4$$

Since the exponential $$e^{\lambda x}$$ can never equal zero,

 * $$\lambda^2+a\lambda+b=0$$

This equation can be solved for lambda using the quadratic equation :

 * $$\lambda =\frac{-a\pm \sqrt[]{a^2-4b}}{2}$$

Substituting values:

 * $$\lambda =\frac{-(6)\pm \sqrt[]{(6)^2-4(\pi^2+4)}}{2}$$

Which evaluates to:

 * $$\lambda =-3\pm \sqrt[]{-19.478}$$

 * or

 * $$\lambda =-3\pm 2.207i$$

Since the discriminant is less than zero we let $$\omega=2.207$$ and the homogeneous solution will be of the form:

 * $$y=e^{\frac{-ax}{2}}(Acos(\omega x)+Bsin(\omega x))$$

Substituting values we have the homogeneous solution:

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

$$y=e^(Acos(2.21 x)+Bsin(2.21 x))$$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 *
 * }
 * }

Given:
$$y''+2\pi y'+\pi^2y=0$$

Find:
The homogeneous solution to the differential equation.

Solution:
Use the method of undetermined coefficients, also known as the method of trial solution, to solve the differential equation.

Let   $$y=e^{\lambda x}$$

Therefore:

 * $$y'=\lambda e^{\lambda x}$$
 * and
 * $$y''=\lambda^2 e^{\lambda x}$$

So,

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

can be written as

 * $$e^{\lambda x}(\lambda^2+a\lambda+b)=0$$  where   $$a=2\pi$$ and $$b=\pi^2$$

Since the exponential  $$e^{\lambda x}$$   can never equal zero,

 * $$\lambda^2+a\lambda+b=0$$

This equation can be solved for lambda using the quadratic equation :

 * $$\lambda =\frac{-a\pm \sqrt[]{a^2-4b}}{2}$$

Substituting values we can see that the discriminant is zero:

 * $$\lambda =\frac{-2\pi\pm \sqrt[]{4\pi^2-4\pi^2}}{2}$$

Therefore, only one solution can be found from this equation:

 * $$y_{1}=e^{-\pi x}$$

For the second solution reduction of order must be used:

Let:

 * $$y_{2}=uy_{1}$$

Then,

 * $$y_{2}'=u'y_{1}+y_{1}'u$$

and
 * $$y_{2}=uy_{1}+2u'y_{1}'+y_{1}''u$$

Substituting these into the original equation gives:

 * $$(uy_{1}+2u'y_{1}'+y_{1}u)+a(u'y_{1}+y_{1}'u)+ buy_{1}=0$$

Which simplifies to:
 * $$uy_{1}'+u'(2y_{1}'+ay_{1})+u(y_{1}+ay_{1}'+by_{1})=0$$

Since,

 * $$2y_{1}'=-ae^{\frac{-a}{2}x}=-ay_{1}$$

and
 * $$y_{1}''+ay_{1}'+by_{1}=0$$

What remains is:

 * $$u''y_{1}=0$$

or
 * $$u''=0$$

Therefore,

 * $$u=C_{1}x+C_{2}$$

Now we let  $$C_{1}=1$$   and   $$C_{2}=0$$   so that:

 * $$u=x$$

Now,

 * $$y_{2}=xy_{1}=xe^{-\pi x}$$

Using  $$y_{1}$$   and   $$y_{2}$$   we have the general solution:

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

$$y=(C_{1}+C_{2}x)e^{-\pi x}$$
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 *
 * }
 * }

title: User:EGM4313.s12.team6.davis/R1


text:
User:Egm4313.s12.team14.pickett

[| Report 1]

[| Report 2]

title: User:Egm4313.s12.team14.pickett


title: File:Testrender2.JPG


title: File:R1.3.jpg


text:
Definitions/Rigorous

What is a rigorous definition?

By collecting definitions from primary sources where it is claimed and hopefully defended that a particular definition is rigorous, it should be possible to provide relatively simple tests that demonstrate rigor (or rigour).

Hopefully, the rigorous definition will be a particular type of definition.

As the figure at the upper right suggests, a rigorous definition may be the limit of a series of definitions that are closer and closer approximations to the concept to be defined. This figure is a series of more accurate approximations to the area under the curve.

Definitions
Def. a "statement of the meaning of a word or word group or a sign or symbol" is called a definition.

Rigorous
Def. "allowing no abatement or mitigation [unyielding or inflexible] " is called rigorous.

Theory of rigorous definitions
“[D]efinitions are always of symbols, for only symbols have meanings for definitions to explain.” “A definition can be expressed in either of two ways,”
 * 1) writing “about the symbol to be defined, or”
 * 2) writing “about its referent.”

In the theory of definition, “the symbol being defined is called the definiendum, and the symbol or set of symbols used to explain the meaning of the definiendum is called the definiens.” “The definiens is not the meaning of the definiendum, but another symbol or group of symbols which, according to the definition, has the same meaning as the definiendum.”

“Anyone who introduces a new symbol has complete freedom to stipulate what meaning is to be given it.”

One reason for the introduction of a new symbol is that “[t]he emotive suggestions of familiar words are often disturbing to one interested only in their literal or informative meanings.”

Def. "[m]anifesting, exercising, or favoring rigour; allowing no abatement or mitigation; scrupulously accurate; exact; strict; severe; relentless" is called rigorous.

"[A]s, ... a rigorous definition or demonstration".

Axiomatic definitions
It has been stated that "the rigorous definition of distance" fulfills "the three axioms that define an Euclidean metric" so that a "generalized metric can be defined using as distance an appropriate function ... that fulfills the three axioms of an Euclidean metric". Having met these three axioms as a criteria of an Euclidean metric, the definition of the generalized metric is said to be a "rigorous definition of distance".

An axiomatic definition is a rigorous definition: "the definition must clearly state the rules that are considered as binding, and on the other hand give the implementor enough freedom to achieve efficiency by leaving certain less important aspects undefined." This rigorous definition is for "an axiomatic definition of the programming language PASCAL".

Constipation
Def. “a bowel movement only every three days or less frequently” is called constipation. This is a rigorous definition of constipation.

Formal definition of limit
Def. (Formal definition of a limit)

Let $$f(x)$$ be a function defined on an open interval $$D$$ that contains $$c$$, except possibly at $$x=c$$.

Let $$L$$ be a number.

If, for every $$\varepsilon>0$$, there exists a $$\delta>0$$ such that for all $$x\in D$$ with
 * $$0 < \left| x - c \right| < \delta,$$

we have
 * $$\left| f(x) - L \right| < \varepsilon$$, then

 * $$ \lim_{x \to c} f(x) = L $$ is called a limit.

This precise definition of limit is a rigorous definition, the formal definition, from Wikibooks Calculus/Formal Definition of the Limit.

Infinitesimals
The Greek mathematician Archimedes (c.287 BC–c.212 BC), in The Method of Mechanical Theorems, was the first to propose a logically rigorous definition of infinitesimals. His Archimedean property defines a number x as infinite if it satisfies the conditions |x|>1, |x|>1+1, |x|>1+1+1, ..., and infinitesimal if x≠0 and a similar set of conditions holds for 1/x and the reciprocals of the positive integers. A number system is said to be Archimedean if it contains no infinite or infinitesimal members.

Religion
Def. a "conjunction of ... a dogma, a moral law, and a cult or form of worship" is called a religion.

René Guénon proposed a rigorous definition of the term "religion".

Riemann integral
The Riemann integral [is] the first rigorous definition of the integral of a function on an interval.

"For Riemann's definition of his integral, see section 4, "Über der Begriff eines bestimmten Integrals und den Umfang seiner Gültigkeit", pages 101-103.

Let $$f$$ be a non-negative real-valued function of the interval $$[a,b]$$.

Let $$S = \{ (x, y) | 0 < y < f(x) \}$$ be the region of the plane under the graph of the function $$f$$ and above the interval $$[a,b]$$ [on the x-axis]. We are interested in measuring the area of $$S.$$

Def. a finite sequence of numbers of the form

 * $$\, a = x_0 < x_1 < x_2 < \cdots < x_n = b\,$$

is called a partition of an interval $$[a,b]$$.

Def. each xi, xi+1 is called a subinterval of the partition.

Notation: let $$\Delta x$$ represent $$(x_{i+1}-x_i).$$

Notation: let the symbol $$s$$ represent the area of the region $$S$$.

Def. the length of the longest subinterval,

 * $$\max (x_{i+1}-x_i),$$

where 0 ≤ i ≤ n-1" is called "[t]he mesh or norm of a partition.

Def. a partition together with a finite sequence of numbers t0, tn−1 subject to the conditions that for each i, xi ≤ ti ≤ xi+1" is called a tagged partition $$P(x,t)$$ of an interval [a,b].

Let the two partitions $$P(x,t)$$ and $$Q(y,s)$$ be partitions of the interval [a,b].

Def. for each integer i with 0 ≤ i ≤ n there exists an integer r(i) such that xi = yr(i) and such that ti = sj for some j with r(i) ≤ j < r(i+1)" $$Q(y,s)$$ is called "a refinement of $$P(x,t)$$.

Def. one tagged partition is greater or equal to another when the former is a refinement of the latter is called a partial order on the set of all tagged partitions.

The Riemann integral of ƒ equals s if the following condition holds:

 * For all $$\epsilon > 0$$, there exists $$\delta > 0$$ "such that for any tagged partition $$x_0,\ldots,x_n$$ and $$t_0,\ldots,t_{n-1}$$ whose mesh is less than $$\delta$$, we have

 * $$\left|\sum_{i=0}^{n-1} f(t_i) (x_{i+1}-x_i) - s\right| < \varepsilon.$$

 * $$ \lim_{\Delta x \to 0} \sum_{i=0}^{n-1} f(t_i) (x_{i+1}-x_i) = s $$ is denoted by:

 * $$\int_{a}^{b}f(x)\,dx = s.$$

Stem cells
A rigorous definition of the term “stem cell” is as follows.

A cell is a stem cell if and only if it has the properties:
 * 1) unlimited self-renewal and
 * 2) within-tissue multipotentiality.

Def. a cell that has only "the properties [of] unlimited self-renewal and within-tissue multipotentiality" is called a stem cell.

This definition has limited flexibility in that it “does not necessarily exclude cross-tissue plasticity.”

Increasing rigor
The beam width is the single most important characteristic of a laser beam profile.

For astigmatic beams a more rigorous definition of the beam width is

 * $$ d_{\sigma x} = 2 \sqrt{2} \left( \langle x^2 \rangle + \langle y^2 \rangle + \gamma \left( \left( \langle x^2 \rangle - \langle y^2 \rangle \right)^2 + 4 \langle xy \rangle^2 \right)^{1/2} \right)^{1/2} $$

and
 * $$ d_{\sigma y} = 2 \sqrt{2} \left( \langle x^2 \rangle + \langle y^2 \rangle - \gamma \left( \left( \langle x^2 \rangle - \langle y^2 \rangle \right)^2 + 4 \langle xy \rangle^2 \right)^{1/2} \right)^{1/2}. $$

This definition also incorporates information about x-y-correlation $$ \langle xy \rangle $$, but for circular symmetric beams, both definitions are the same.

σ is the standard deviation of the horizontal or vertical marginal distribution.

 * "$$ \langle x^2 \rangle = \frac{1}{P} \int{I(x,y) (x - \langle x \rangle )^2 dx dy}, $$
 * $$ \langle xy \rangle = \frac{1}{P} \int{I(x,y) (x - \langle x \rangle ) (y - \langle y \rangle ) dx dy},$$
 * $$ \langle y^2 \rangle = \frac{1}{P} \int{I(x,y) (y - \langle y \rangle )^2 dx dy} $$
 * $$ \langle x \rangle = \frac{1}{P} \int{I(x,y) x dx dy}, $$
 * $$ \langle y \rangle = \frac{1}{P} \int{I(x,y) y dx dy}$$

The beam power is
 * $$ P = \int{I(x,y) dx dy} $$

and
 * $$ \gamma = \sgn \left( \langle x^2 \rangle - \langle y^2 \rangle \right) = \frac{\langle x^2 \rangle - \langle y^2 \rangle}{|\langle x^2 \rangle - \langle y^2 \rangle|}. $$

The beam profile is the intensity (I) of the beam as measured at specific rectilinear coordinates (x,y): $$I(x,y).$$

Hypotheses

 * 1) A rigorous definition should be possible for each word or symbol.

title: Definitions/Rigorous


text:
User talk:Algamicagrat

Hello
Testing 123 123 123 I dont care.Algamicagrat 02:40, 26 January 2012 (UTC)

Cutting and pasting
Hi there. Thanks for your contributions to Wikiversity, but please avoid cutting-and-pasting text from Wikipedia to Wikiversity. Also, I recommend integrating your pages with an existing K-12 type educational resource at Wikiverisity. Please let me know if you have questions. --HappyCamper 03:48, 28 January 2012 (UTC)

User:Life421
Hello Algamicagrat, you've created the userpage User:Life421, but that user currently doesn't exist on this project. Mathonius (talk) 04:05, 16 March 2012 (UTC)

title: User talk:Algamicagrat


text:
User:Egm6322.s12.TEAM1.REPORT/REPORT 1

Problem 1: Constructing the Gram Matrix for Fourier Basis Functions
 Solved without assistance

Given Equations
Fourier Series with basis functions:

Scalar (inner) product:

Wronskian:

Find: Gram Matrix and Properties
A. Construct the Gram Matrix for the given Fourier basis functions using the scalar product. B. Determine the property of this Gram Matrix. C. Show that $$

\left \{ \cos \theta\ \, \sin \theta\ \ \right \} $$ are linearly independent using the Wronskian.

Solution
A. From Wolfram Mathworld the Gram matrix is defined as a set $$ \mathbf{V} $$ of $$ \mathbf{m} $$ vectors (points in $$ \mathbb R ^n $$), the Gram Matrix $$ \mathbf{G} $$ is the matrix of all possible inner products of $$ \mathbf{V} $$, i.e. where $$ \mathbf{A}^T $$ denotes the transpose. The Gram matrix determines the vectors $$ \mathbf{v}_i $$ up to isometry. For this problem the value of $$ \mathbf{v}_i = g(x)= \cos n \omega\ \theta\ \ $$ and $$ \mathbf{v}_j = h(x)= \sin n \omega\ \theta\ \ $$ Then,

Problem 2: Pose Question to be solved
 Solved without assistance

Problem 3: Pose Question to be solved
 Solved without assistance

Problem 4: Pose Question to be solved
 Solved without assistance

Problem 5: Pose Question to be solved
 Solved without assistance

Problem 6: Pose Question to be solved
 Solved without assistance

Problem 7: Pose Question to be solved
 Solved without assistance

Solution: ????????????????????
=Contributing Members & Referenced Lecture=

=References=

title: User:Egm6322.s12.TEAM1.REPORT/REPORT 1


text:
User:Egm4313.s12.team11.sheider

=Team Page= Team 11

=Report 1 Problems= /R1.5: K 2011 p.59 pb. 4/

/R1.5: K 2011 p.59 pb. 5/

=Report 2 Problems= /R2.4: K 2011 p.59 pb. 5/

/R2.4: K 2011 p.59 pb. 6/

=Report 3 Problems=

/R3.3: Lecture 7 p.7-11/

=Report 4 Problems=

/R4.2a: Lecture 7 p.7-26 pb. 1-2/

=Report 5 Problems= /R5.1: Lecture 7 p.7-32 pb. 1-5/

/R5.8: Lecture 8 p.8-16/

=Report 6 Problems= /R.6.6: Lecture 10 p.10-9/

=Report 7 Problems= /R.7.4: K 2011 p.482 pb. 12/

/R.7.4: K 2011 p.482 pb. 13/

title: User:Egm4313.s12.team11.sheider


text:
User:Egm4313.s12.team11.sheider/Egm4313.s12.team11.sheider

=Report 1 Problems= ../R1.5: K 2011 p.59 pb. 4/

../R1.5: K 2011 p.59 pb. 5/

=Report 2 Problems= ../R2.4: K 2011 p.59 pb. 5/

../R2.4: K 2011 p.59 pb. 6/

=Report 3 Problems=

../R3.3: Lecture 7 p.7-11/

=Report 4 Problems=

../R4.2a: Lecture 7 p.7-26 pb. 1-2/

=Report 5 Problems= ../R5.1: Lecture 7 p.7-32 pb. 1-5/

../R5.8: Lecture 8 p.8-16/

=Report 6 Problems= ../R.6.6: Lecture 10 p.10-9/ =Report 7 Problems= ../R.7.4: K 2011 p.482 pb. 12/

../R.7.4: K 2011 p.482 pb. 13/


text:
User talk:Ran1983

 Hello Ran1983, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --HappyCamper 06:01, 26 January 2012 (UTC)

title: User talk:Ran1983


text:
User:Samsujata

Hi! My name is Salman Gurung. For more info, see Wikipedia English

title: User:Samsujata


title: File:Egm4313.s12.team8.tclamb.r1.1.png


text:
User:Inlandmamba

Hi, My name is neither "Inland" nor "Mamba". And neither "Inlandmamba". So you must be thinking why Inlandmamba then? When I first registered on Wikipedia I intended to use my real name, but I had a look at the names of other senior editors here and found that none of them were actually using their real names. I read the policy here of making usernames and Therefore I decided to make a name that would be different. The first thing that struck my mind was Snakes, as I am a big fan of snakes. Yeah some people are against my so called hobby of collecting every article or picture on this topic, but I have been into this hobby since my childhood.

I am a Chartered Accountant by profession, and some of my editing has been in accountancy articles. However, this is a fairly small proportion of my total work on Wikipedia. I have a passion of technology and I am very curious about the latest technologies that are invented. I had written some articles in the start to see how the Wikipedia practically works. Even though due to a misunderstanding I got blocked earlier, but now I am back again and will try my level best to become a part of this community by helping out new comers and cleaning up some articles with what ever knowledge I have and I can find via reliable sources.

title: User:Inlandmamba


text:
Category:Egm4313.s12.team8

Resources for Team 8

title: Category:Egm4313.s12.team8


title: File:CORDIC.Background.4.A.20120126.pdf


text:
User:Egm6322.s12.TEAM1.KIM

http://en.wikiversity.org/wiki/User:Egm6322.s12.TEAM1.HW http://en.wikiversity.org/wiki/User:Egm6322.s12.TEAM1.KIM/HW1 http://en.wikiversity.org/wiki/User:Egm6322.s12.TEAM1.KIM/HW2 http://en.wikiversity.org/wiki/User:Egm6322.s12.TEAM1.KIM/HW3

title: User:Egm6322.s12.TEAM1.KIM


text:
User talk:Egm6322.s12.TEAM1.KIM

GG

title: User talk:Egm6322.s12.TEAM1.KIM


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User:Egm6322.s12.team2.Xia/RP7.3

=R*8.4 -Legendre Function =

Given
the following pairs: $$\displaystyle \{P_{0},\, Q_{0}\},\,\{P_{1},\! Q_{1}\},\cdots\,,\{P_{4},\, Q_{4}\}$$ (8.4.1) Where

Find
1.Plot the 8.4.1 in separate figures. 2.Observe even-ness and odd-ness of 8.4.1 and guess the value of the scalar products:

$$\displaystyle =\int_{\mu=-1}^{\mu=+1}P_{i}(\mu)Q_{i}(\mu)du=?$$

Solution
$$\displaystyle $$

title: User:Egm6322.s12.team2.Xia/RP7.3


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User:Egm6322.s12.team2.Xia/RP8.4

=R*8.4 -Legendre Function =

Given
The following pairs: $$\displaystyle \{P_{0},\, Q_{0}\},\,\{P_{1},\! Q_{1}\},\cdots\,,\{P_{4},\, Q_{4}\}$$ (8.4.1)

where $$\displaystyle \{P_{i},\, Q_{i}\} $$ are Legendre polynomials.

Find
1.Plot the 8.4.1 in separate figures. 2.Observe even-ness and odd-ness of 8.4.1 and guess the value of the scalar products:

$$\displaystyle =\int_{\mu=-1}^{\mu=+1}P_{i}(\mu)Q_{i}(\mu)du=?$$

Solution
1.The Legendre polynomial is as follows: $$\displaystyle {P}_{n}(x)=\sum^{m}_{r=0}(-1)^r \frac{(2n-2r)!x^{n-2r}}{2^{n}r!}(n-r)!(n-2r)! $$ where $$\displaystyle m = \frac{n}{2}$$. $$\displaystyle {P}_{0}(x) = 1 $$

$$\displaystyle {P}_{1}(x) = x $$

$$\displaystyle {P}_{2}(x) = \frac{1}{2}(3x^2-1) $$

$$\displaystyle {P}_{3}(x) = \frac{1}{2}(5x^3-3x) $$

$$\displaystyle {P}_{4}(x) = \frac{35\,{x}^{4}}{8}-\frac{15\,{x}^{2}}{4}+\frac{3}{8} $$ and $$\displaystyle {Q}_{0}(x) = \frac{1}{2}\mathrm{log}\left(\frac{1+x}{1-x}\right) $$

$$\displaystyle {Q}_{1}(x) = \frac{1}{2}x \mathrm{log}\left(\frac{1+x}{1-x}\right)-1 $$

$$\displaystyle {Q}_{2}(x) = \frac{1}{4}(3x^2-1)\mathrm{log}\left(\frac{1+x}{1-x}\right)-\frac{3}{2}x $$

$$\displaystyle {Q}_{3}(x) = \frac{1}{4}(5x^3-3x)\mathrm{log}\left(\frac{1+x}{1-x}\right)-\frac{5}{2}x^2+\frac{2}{3} $$

$$\displaystyle {Q}_{4}(x) = \frac{35\,{x}^{4}\,\mathrm{log}\left( -\frac{x}{x-1}-\frac{1}{x-1}\right) }{16}-\frac{15\,{x}^{2}\,\mathrm{log}\left( -\frac{x}{x-1}-\frac{1}{x-1}\right) }{8}+\frac{3\,\mathrm{log}\left( -\frac{x}{x-1}-\frac{1}{x-1}\right) }{16}-\frac{35\,{x}^{3}}{8}+\frac{55\,x}{24} $$ Using Maxima with GNUplot we get the following figures: 2. It is easy to find that $$\displaystyle {P}_{0}(x) = 1={P}_{0}(-x) $$

$$\displaystyle {P}_{1}(x) = x= -{P}_{1}(-x) $$

$$\displaystyle {P}_{2}(x) = \frac{1}{2}(3x^2-1)= {P}_{2}(-x)$$

$$\displaystyle {P}_{3}(x) = \frac{1}{2}(5x^3-3x)= -{P}_{2}(-x) $$ Thus, we are able to guess that $$\displaystyle {P}_{2n}(x)={P}_{2n}(-x) $$ are even-ness while $$\displaystyle {P}_{2n+1}(x)=-{P}_{2n+1}(-x) $$ are odd-ness. Using the same method, we will find that $$\displaystyle {Q}_{2n}(x)=-{Q}_{2n}(-x) $$ are odd-ness while $$\displaystyle {Q}_{2n+1}(x)={Q}_{2n+1}(-x) $$ are even-ness. From the above, we can easily find that $$\displaystyle {P}_{i}(x){Q}_{i}(x) $$ are odd-ness, Since the integral of smooth odd functions from -a to a are zero. That's to say $$\displaystyle =\int_{\mu=-1}^{\mu=+1}P_{i}(\mu)Q_{i}(\mu)du= 0$$

title: User:Egm6322.s12.team2.Xia/RP8.4


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User:Egm6322.s12.team2.Xia/RP8.5

=R*8.5 -Even or Odd Function =

Given
Consider the functions $$\displaystyle g_{i}:\,\mathbb{R}\rightarrow\mathbb{R}$$ and the function $$\displaystyle f=\sum_{i}g_{i}\,:\,\mathbb{R}\rightarrow\mathbb{R}$$

(8.5.1)

Find
Show that: 1. If $$\displaystyle g_{i}$$ is odd, then $$\displaystyle f $$ is odd. 2. If $$\displaystyle g_{i}$$ is even, then $$\displaystyle f$$  is even.

Solution
1. If $$\displaystyle g_{i} $$ is odd, we have $$\displaystyle g_{i}(-x)=-g_{i}(x) $$ then $$\displaystyle {\color{blue}f(-x)}=\sum_{i}g_{i}(-x)=g_{1}(-x)+g_{2}(-x)+\cdots+g_{i}(-x)$$ $$\displaystyle =-g_{1}(x)-g_{2}(x)-\cdots-g_{i}(x)=-\sum_{i}g_{i}(x)={\color{blue}-f(x)}$$ thus $$\displaystyle f$$ is odd. 2. If $$\displaystyle g_{i}$$ is even, we have $$\displaystyle g_{i}(-x)=g_{i}(x)$$ then $$\displaystyle {\color{blue}f(-x)}=\sum_{i}g_{i}(-x)=g_{1}(-x)+g_{2}(-x)+\cdots+g_{i}(-x)$$ $$\displaystyle =g_{1}(x)+g_{2}(x)+\cdots+g_{i}(x)=\sum_{i}g_{i}(x)={\color{blue}f(x)}$$ thus $$\displaystyle f$$ is even.

title: User:Egm6322.s12.team2.Xia/RP8.5


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User:Egm4313.s12.team13.ryser

Problem Statement
Find and plot the solution for $$ y''-10y'+25=0 \!$$.

Initial Conditions: $$ y(0)=1, y'(0)=0 \!$$

No excitation: $$ r(x)=0 \!$$

Background Information
Finding Homogeneous Solutions in cases where $$ \Delta=a^2-4b=0 \!$$

The two roots are a double root:

$$ \lambda_1=\lambda_2=\lambda=-a/2\!$$ (1)

The first homogeneous solution is:

$$ y_{h,1}(x)= e^{yx} \!$$ (2)

To find the 2nd homogeneous solution we use a trial solution:

$$ y_{h,2}=u(x)y_{h,1}(x) \!$$where u(x) is an unknown function of x (3)

$$ y'_{h,2}=uy'_{h,1}+u'y_{h,1} \!$$ (4)

$$ y_{h,2}=uy_{h,1}+2u'y'_{h,1}+u''y_{h,1}\!$$ (5)

Adding up equations (3), (4), and (5) we get:

$$ 0=u[y_{h,1}+ay'_{h,1}+by_{h,1}]+u'[2y'_{h,1}+ay_{h,1}]+uy_{h,1} \!$$ (6) We can simplify equation (6) by applying known equations:

$$ 0=u''y_{h,1} \!$$ (7)

Hence:

$$ u''=0 \!$$ (8)

Integrating with respect to x twice yields:

$$ u(x)=xc_3 \!$$ (9)

So $$ y_{h,2}=(x+c_3)y_{h,1}\!$$ (10)

The final homogeneous solution is:

$$ y_h=c_1y_{h,1}+c_2xy_{h,1} \!$$ (11)

Solution
$$ y''-10y'+25y=r(x) \!$$ (12)

The characteristic equation is:

$$ \lambda^2-10\lambda+25=0\!$$ (13)

The first homogeneous root is:

$$ \lambda=5 \!$$ (14)

Hence the homogeneous solution is:

$$ y_h(x)=c_1e^{5x}+c_2xe^{5x} \!$$ (15)

Apply the initial conditions y(0)=1 and y'(0)=0:

$$ y_h(0)= c_1 =1 \!$$ (16)

$$ y_h'(x)= 5c_1e^{5x}+5c_2xe^{5x}+c_2 \!$$
 (17)

$$ y_h'(0)=5+c_2 \!$$
 (18)

Hence:

$$c_1=1 \!$$
 (19)

$$ c_2=-5 \!$$
 (20)

The homogeneous equation is:

$$ y_h(x)=e^{5x}-5xe^{5x} \!$$
 (21)

Since r(x)=0 the particular equation is zero. Hence the solution to the problem is:

$$ y(x)=e^{5x}-5xe^{5x} \!$$
 (22)

The plot of the function is:

References:

Section 5 Notes

title: User:Egm4313.s12.team13.ryser