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

$$  \displaystyle y_c = y_k = y $$ (Eq.1.1)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Kinetics:
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$$  \displaystyle my'' + f_I = f(t) $$     (Eq.1.2)
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 * style="width:95%" |
 * 
 * }


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

$$  \displaystyle f_I = f_k +f_c $$     (Eq.1.3)
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 * style="width:95%" |
 * 
 * }

Equations:
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$$  \displaystyle f_k = ky_k = ky $$ (Eq.1.4)
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 * style="width:95%" |
 * 
 * }


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$$  \displaystyle f_c = cy'_c = cy' $$     (Eq.1.5)
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 * style="width:95%" |
 * 
 * }

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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 * style="width:95%" |
 * 
 * }

Then substituting Eq. 1.6 into Eq. 1.2:
 * {| style="width:100%" border="0"

$$  \displaystyle my'' + cy' + ky =f(t) $$     (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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 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * 
 * }

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

$$  \displaystyle y_c = y_k = y $$ (Eq.2.1)
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 * style="width:95%" |
 * 
 * }

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

$$  \displaystyle my'' + f_I = r(t) $$     (Eq.2.2)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


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

$$  \displaystyle f_I = f_k +f_c $$     (Eq.2.3)
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 * style="width:95%" |
 * 
 * }

Equations:
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$$  \displaystyle f_k = ky_k = ky $$ (Eq.2.4)
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 * style="width:95%" |
 * 
 * }


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

$$  \displaystyle f_c = cy'_c = cy' $$     (Eq.2.5)
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 * style="width:95%" |
 * 
 * }

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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 * style="width:95%" |
 * 
 * }

Then substituting Eq. 2.6 into Eq. 2.2:
 * {| style="width:100%" border="0"

$$  \displaystyle my'' + cy' + ky =r(t) $$     (Eq.2.7) Which is the equation of motion of a spring-mass-dashpot system with a mass m \, and applied force r(t).
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * 
 * }

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

$$  \displaystyle my'' + f_I = f(t) $$     (Eq.3.1)
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 * style="width:95%" |
 * 
 * }

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
<p style="text-align:right"> (Figure 3.2)

Spring Free-Body Diagram
<p style="text-align:right"> (Figure 3.3)

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

<p style="text-align:right"> (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:
 * {| style="width:100%" border="0" align="left"

\displaystyle f_I = f_c = f_k $$
 * <p style="text-align:right;">(Eq. 3.2)
 * }

Apply Newton's Second Law of Motion to the mass system:
 * {| style="width:100%" border="0" align="left"

\sum F = m \frac{dV}{dt} $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(Eq. 3.3)
 * }
 * }


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

\sum F = f(t) - f_I = m \frac{dV}{dt} $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(Eq. 3.4)
 * }
 * }

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

\frac{dV}{dt} = a(t) = y'' $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(Eq. 3.5)
 * }
 * }

Substituting Eq. 3.5 into Eq. 3.4 results in:
 * {| style="width:100%" border="0" align="left"

f(t) - f_I = ma(t) = my'' $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">(Eq. 3.6)
 * }
 * }

Rearranging Eq. 3.6 results in:
 * {| style="width:100%" border="0"

$$  \displaystyle my'' + f_I = f(t) $$     (Eq.3.1)
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 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

= 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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 * style="width:95%" |
 * <p style="text-align:right">
 * }

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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 * style="width:95%" |
 * <p style="text-align:right">
 * }


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$$  \displaystyle V^\prime = LI^{\prime\prime} + RI^\prime + \frac{1}{C}I $$     (Eq.4.3)
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 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle V = LQ^{\prime\prime} + RQ^\prime + \frac{1}{C}Q $$     (Eq.4.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \nu_C = \frac{1}{C}\int I dt = \frac{1}{C}Q $$     (Eq.4.5)
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 * style="width:95%" |
 * <p style="text-align:right">
 * }

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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 * style="width:95%" |
 * <p style="text-align:right">
 * }

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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 * style="width:95%" |
 * <p style="text-align:right">
 * }


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$$  \displaystyle Q^{\prime\prime} = C\frac{d^2\nu_C}{dt^2} $$     (Eq.4.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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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 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

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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 * style="width:95%" |
 * <p style="text-align:right">
 * }

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)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle Q^{\prime\prime} = I^\prime $$     (Eq.4.11)
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 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle Q^{\prime\prime\prime} = I^{\prime\prime} $$     (Eq.4.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Substitute Eq. 4.10, 4.11, and 4.12 into Eq. 4.9:
 * {| style="width:100%" border="0"

$$  \displaystyle V^\prime = LI^{\prime\prime} + RI^\prime + \frac{1}{C}I $$     (Eq.4.3)
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 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

= 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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 * style="width:95%" |
 * <p style="text-align:right">
 * }

Part 1 Solution
Characteristic Equation of Eq. 5.1.1:
 * {| style="width:100%" border="0"

$$  \displaystyle x^2 + 4x + (\pi^2 + 4) = 0 $$     (Eq.5.1.2) Solve for the roots in Eq. 5.1.2 by using the quadratic equation:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle x = \frac{-4 \pm\sqrt{16-4(1)(\pi^2 + 4)}}{2} $$     (Eq.5.1.3)
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 * <p style="text-align:right">
 * }

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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 * style="width:95%" |
 * <p style="text-align:right">
 * }

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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 * style="width:95%" |
 * <p style="text-align:right">
 * }

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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 * style="width:95%" |
 * <p style="text-align:right">
 * }

Which &alpha; and &beta; are real.

Then the general solution of the ODE will be in the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y = C_1 e^{\alpha x} \cos(\beta x) + C_2 e^{\alpha x} \sin(\beta x) $$ (Eq.5.1.7) Therefore, the solution is:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ y = C_1 e^{-2x} \cos(\pi x) + C_2 e^{-2x} \sin(\pi x) \,$$ (Eq.5.1.8)
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 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

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


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

$$  \displaystyle y^{\prime\prime} + 2\pi y^\prime + \pi^2 y = 0 $$     (Eq.5.2.1)
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 * style="width:95%" |
 * <p style="text-align:right">
 * }

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)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solve for the roots in Eq. 5.2.2 by using the quadratic equation:
 * {| style="width:100%" border="0"

$$  \displaystyle x = \frac{-2\pi \pm\sqrt{4\pi^2-4(1)(\pi^2)}}{2} $$     (Eq.5.2.3) Simplifying Eq. 5.2.3:
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 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle x = \frac{-2\pi}{2} $$ $$  \displaystyle x = -\pi $$     (Eq.5.2.4) which is a real double root.
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 * style="width:95%" |
 * <p style="text-align:right">
 * }

Characteristic equations containing real double roots produces a general solution for the ODE in the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y = C_1 e^{-\alpha x} + C_2 x e^{-\alpha x} $$ (Eq.5.2.5) Therefore, the solution is:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ y = C_1 e^{-\pi x} + C_2 x e^{-\pi x} \,$$ (Eq.5.2.6)
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 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

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


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$$  \displaystyle y^{\prime\prime} + p(x)y^\prime + q(x)y = r(x) $$     (Eq.6.1)
 * <p style="text-align:right">
 * }

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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 * style="width:95%" |
 * <p style="text-align:right">
 * }

The homogeneous solution to Eq. 6.1 is:
 * {| style="width:100%" border="0"

$$  \displaystyle y_h^{\prime\prime}+p(x)y_h^\prime+q(x)y_h=0 $$     (Eq.6.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The particular solution to Eq. 6.1 is:
 * {| style="width:100%" border="0"

$$  \displaystyle y_p^{\prime\prime}+p(x)y_p^\prime+q(x)y_p=r(x) $$     (Eq.6.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Eq. 6.3 and Eq. 6.4 are now summed in order to get:
 * {| style="width:100%" border="0"

$$  \displaystyle (y_h^{\prime\prime}+y_p^{\prime\prime})+p(x)(y_h^\prime+y_p^\prime)+q(x)(y_h+y_p)=r(x) $$     (Eq.6.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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,
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y_h^\prime+y_p^\prime \Rightarrow (y_h+y_p)^\prime \Rightarrow \bar y^\prime $$     (Eq.6.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now we can substitute Eq. 6.6 and Eq. 6.7 into Eq. 6.5 which yields:
 * {| style="width:100%" border="0"

$$  \displaystyle \bar y^{\prime\prime}+p(x)\bar y^\prime+q(x)\bar y=r(x) $$     (Eq.6.8) which is the same as:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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


 * style="width:95%" |
 * style="width:95%" |

$$  \displaystyle y^{\prime\prime} + p(x)y^\prime + q(x)y = r(x) $$     (Eq.6.1)
 * <p style="text-align:right">
 * }

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


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$$  \displaystyle a_n(x)\frac{d^ny}{dx^n}+a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}}+...+a_1(x)\frac{dy}{dx}+a_0(x)y=F(x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$  \displaystyle y^{\prime\prime} = g = constant $$     (Eq.6.1.1)
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 * <p style="text-align:right">
 * }

Part 1 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 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)
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 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle y_h^{\prime\prime} = 0 $$     (Eq.6.1.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle y_p^{\prime\prime} = g $$ (Eq.6.1.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Eq. 6.1.2 and Eq. 6.1.3 can be added to obtain:
 * {| style="width:100%" border="0"

$$  \displaystyle y_h^{\prime\prime} +y_p^{\prime\prime} = g $$ (Eq.6.1.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now we can use Eq. 6.6 to simplify Eq. 6.1.4:
 * {| style="width:100%" border="0"

$$  \displaystyle (y_h +y_p)^{\prime\prime} = g $$ (Eq.6.1.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

And the result is:
 * {| style="width:100%" border="0"

$$  \displaystyle \bar y^{\prime\prime} = g $$ (Eq.6.1.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

Superposition Principle can be applied since Eq. 6.1.6 is the same as Eq. 6.1.1.
 * style="width:36%; padding:10px; border:2px solid #8888aa" |
 * 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)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Part 2 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: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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 * <p style="text-align:right">
 * }

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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 * style="width:95%" |
 * <p style="text-align:right">
 * }

Homogeneous equation is:
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$$  \displaystyle mv_h^\prime+bv_h^2 = 0 $$     (Eq.6.2.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle mv_p^\prime+bv_p^2 =mg $$     (Eq.6.2.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Adding Eq. 6.2.4 and Eq. 6.2.5:
 * {| style="width:100%" border="0"

$$  \displaystyle mv_h^\prime+mv_p^\prime +bv_p^2+bv_h^2 =mg $$     (Eq.6.2.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now Eq. 6.2.6 simplifies to:
 * {| style="width:100%" border="0"

$$  \displaystyle m(v_h+v_p)^\prime+b(v_h^2+v_p^2) =mg $$     (Eq.6.2.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle (v_h^2+v_p^2)\neq (v_h+v_p)^2 \neq \bar v^2 $$     (Eq.6.2.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

Therefore, superposition principle is invalid.
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

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


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

$$  \displaystyle h^\prime = -k\sqrt{h} $$     (Eq.6.3.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }


 * {| 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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }


 * {| 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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }


 * {| 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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }


 * {| 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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

and


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

$$  \displaystyle EIy_p^{iv} = f(x) $$   (Eq.6.7.3)
 * style="width:90%" |
 * style="width:90%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }


 * {| 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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }

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%" |
 * <p style="text-align:right">
 * }


 * {| 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