User:Egm4313.s12.team1.armanious/team1/R1

=Report 1=

Statement
Derive the equation of motion of a spring-dashpot system in parallel (see Figure 1 below), with a mass and applied force $$f(t).$$ Figure 1

Solution
Kinematics, Kinetics, and Relations equations as interpreted from picture

Kinematics:
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$$ \displaystyle y=y_{k}= y_{c} $$     (1.0)
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Kinetics:
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$$  \displaystyle f(t)=my'' + f_{k}+f_{c} $$     (1.1)
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Relations: -Force of spring relation:
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$$  \displaystyle f_{k}=ky_{k} $$     (1.2) -Force of dash-pot relation:
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$$  \displaystyle f_{c}=Cy'_{c} $$     (1.3)
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Putting kinematics, kinetics, and relations together:
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$$  \displaystyle y=y_{k}=y''_{c} $$     (1.4)
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Using kinetic relationship
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$$  \displaystyle f(t)=my''_{c}+f_{k}+f_{c} $$     (1.5)
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Using the kinematics and relations equations:
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$$  \displaystyle f_{k}=ky_{k}=ky_{c} $$     (1.6)
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$$  \displaystyle f(t)=my''_{c}+ky_{c}+f_{c} $$     (1.7)
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Using dash-pot relation:
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$$  \displaystyle f_{c}=Cy'_{c} $$     (1.8)
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$$  \displaystyle f(t)=my''_{c}+ky_{c}+Cy'_{c} $$     (1.9)
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Final Equation
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$$  \displaystyle f(t)=my''_{c}+Cy'_{c}+ky_{c} $$     (1.10)
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Author
Solved and typed by - Egm4313.s12.team1.silvestri 20:44, 25 January 2012 (UTC)

Reviewed By - --Egm4313.s12.team1.durrance 19:17, 30 January 2012 (UTC)

Edited by - --Egm4313.s12.team1.wyattling 19:49, 27 January 2012 (UTC)

Statement
Derive the equation of motion of the spring-mass-dashpot in Fig. 53, in Kreyszig 2011 p.85 (also see[[media:iea.s12.sec1.djvu| Sec 1 p. 1-4]]), with an applied force $$r(t)\!$$ on the ball.

Solution
Kinematics:
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$$  \displaystyle y=y_k=y_c $$     (2.0)
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Kinetics:


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$$  \displaystyle r(t)=my''+{{f}_{k}}+{{f}_{c}} $$     (2.1)
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$$  \displaystyle {{f}_{k}}=k{{y}_{k}} $$     (2.2)
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$$  \displaystyle {{f}_{c}}=c{{y}_{c}}' $$     (2.3)
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Rearranging equations:


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$$  \displaystyle r(t)=my''+k{{y}_{k}}+c{{y}_{c}}' $$     (2.4)
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$$  \displaystyle y=y_k=y_c $$     (2.5)
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Equation of motion:

$$  \displaystyle r(t)=my''+ky+cy' $$     (2.6)
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Author
Solved and typed by - --Egm4313.s12.team1.essenwein 23:06, 25 January 2012 (UTC)

Reviewed By - --Egm4313.s12.team1.stewart 01:56, 31 January 2012 (UTC)

Edited By - --Egm4313.s12.team1.wyattling 19:50, 27 January 2012 (UTC)

Statement
For the spring-dashpot system on p.1-4 [[media:iea.s12.sec1.djvu|(of sec.1 notes)]], draw the Free Body Diagram's and derive the equation of motion (2) p.1-4.

Solution
Free Body Diagrams:

By definition:


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$$  \displaystyle a=y'' $$     (3.0)
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Because only two forces are acting on the dashpot and they are in opposite directions, they are equal in magnitude.


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$$  \displaystyle ky = cv = f_{I} $$     (3.1) The resulting acceleration of the mass, m, must be the difference of the applied force $$f(t)\!$$ and the internal force $$f_{I}\!$$.
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$$  \displaystyle y'' \propto f(t)-f_{I} $$     (3.2)
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Using Newton's second law:
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$$  \displaystyle my'' = f(t) - f_{I} $$     (3.3)
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Rearranging, this yields:
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$$  \displaystyle my'' + f_{I} = f(t) $$     (3.4)
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Author
Solved and Typed By - Egm4313.s12.team1.armanious 20:09, 25 January 2012 (UTC)

Reviewed By - Egm4313.s12.team1.silvestri 19:59, 30 January 2012 (UTC)

Edited By - --Egm4313.s12.team1.wyattling 19:51, 27 January 2012 (UTC)

Statement
Using the Circuit equation (4.1) (See Sec.2 Pg.3 notes), derive equations (4.2) and (4.3) from it.

Solution
Given:

Capacitance Equation:


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$$  \displaystyle Q=Cv_{c}\rightarrow \int idt=Cv_{c}\rightarrow i=C\frac{\mathrm{d} v_{c}}{\mathrm{d} t} $$ (4.0)
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Circuit Equation:


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$$  \displaystyle V=LC\frac{\mathrm{d^2}v_{c}}{\mathrm{d} t^2}+RC\frac{\mathrm{d} v_{c}}{\mathrm{d} t}+v_{c} $$     (4.1)
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First Alternate Circuit Equation:


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$$  \displaystyle LI''+RI'+\frac{1}{C}I=V' $$     (4.2)
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Second Alternate Circuit Equation:


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$$  \displaystyle LQ''+RQ'+\frac{1}{C}Q=V $$     (4.3)
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First we are asked to derive the alternate circuit equation (4.2) from the original circuit equation (4.1).


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$$  \displaystyle V=LC\frac{\mathrm{d^2}v_{c}}{\mathrm{d} t^2}+RC\frac{\mathrm{d} v_{c}}{\mathrm{d} t}+v_{c}\rightarrow LI''+RI'+\frac{1}{C}I=V' $$     (4.4)
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The first step is to take the derivative of the original equation so there is the $$V'\!$$ term on one side:


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$$  \displaystyle V'=LC\frac{\mathrm{d^3}v_{c}}{\mathrm{d} t^3}+RC\frac{\mathrm{d^2} v_{c}}{\mathrm{d} t^2}+\frac{\mathrm{d} v_{c}}{\mathrm{d} t} $$ (4.5)
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$$L\!$$ and $$C\!$$ are constants and therefore remain unchanged by the derivative.

Next use the capacitance equation(4.0) which states:
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$$  \displaystyle I=C\frac{\mathrm{d} v_{c}}{\mathrm{d} t} $$ (4.6)
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Take both the first and second derivative of this equation.

First Derivative:
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$$  \displaystyle I'=C\frac{\mathrm{d^2} v_{c}}{\mathrm{d} t^2} $$     (4.7)
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Second Derivative:
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$$  \displaystyle I''=C\frac{\mathrm{d^3} v_{c}}{\mathrm{d} t^3} $$     (4.8)
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Now substitute in the the $$I\!$$, $$I'\!$$, and $$I''\!$$ terms into the appropriate places. The equation below groups terms that need to be substituted.


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$$  \displaystyle V'=L(C\frac{\mathrm{d^3}v_{c}}{\mathrm{d} t^3})+R(C\frac{\mathrm{d^2} v_{c}}{\mathrm{d} t^2})+\frac{({C}\frac{\mathrm{d} v_{c}}{\mathrm{d} t})}{C} $$     (4.9)
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The final product after substitution is the alternate circuit equation(4.2):

$$  \displaystyle LI''+RI'+\frac{1}{C}I=V' $$     (4.2)
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We are also asked to derive the second alternate circuit equation(4.3)
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$$  \displaystyle V=LC\frac{\mathrm{d^2}v_{c}}{\mathrm{d} t^2}+RC\frac{\mathrm{d} v_{c}}{\mathrm{d} t}+v_{c}\rightarrow LQ''+RQ'+\frac{1}{C}Q=V $$     (4.3) from the original circuit equation(4.1).
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Once again use the capacitance equation(4.0) which also states:


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$$  \displaystyle Q=Cv_{c} $$     (4.12)
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Take the first and second derivative:

First Derivative:
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$$  \displaystyle Q'=C\frac{\mathrm{d} v_{c}}{\mathrm{d} t} $$ (4.13) Second Derivative:
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$$  \displaystyle Q''=C\frac{\mathrm{d^2} v_{c}}{\mathrm{d} t^2} $$     (4.14)
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The $$C\!$$ term is constant and does not change when the derivative is taken.

Group the original circuit equation like in the previous problem so that parentheses mark terms that should be substituted for $$Q\!$$ terms:
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$$  \displaystyle V=L(C\frac{\mathrm{d^2}v_{c}}{\mathrm{d} t^2})+R(C\frac{\mathrm{d} v_{c}}{\mathrm{d} t})+\frac{(Cv_{c})}{C} $$     (4.15)
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After substituting, the final product is the second alternate circuit equation(4.3): :{| style="width:100%" border="1" $$  \displaystyle LQ''+RQ'+\frac{1}{C}Q=V $$     (4.3)
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Author
Solved and typed by - Egm4313.s12.team1.stewart 20:20, 27 January 2012 (UTC)

Reviewed By - Egm4313.s12.team1.armanious 20:08, 30 January 2012 (UTC)

Edited By - --Egm4313.s12.team1.wyattling 19:52, 27 January 2012 (UTC)

Statement
Find a general solution for the following two problems found in Pp. R1.5 in Sec. 2 p. 2-5.

Given

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$$  \displaystyle y'' + 4y' + (\pi^2 + 4)y = 0 $$     (5.0)
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Solution
The characteristic equation of this linear ODE with constant coefficients (L2-ODE-CC) is:


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$$  \displaystyle \lambda^2+4\lambda+(\pi^2+4)=0 $$     (5.1)
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Note: The discriminant of the characteristic equation is negative which gives roots of the form:
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$$  \displaystyle \lambda=-\frac{1}{2}a\pm bi $$ (5.2)
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The general equation for this case of L2-ODE-CC is:


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$$  \displaystyle y(x)=c_1e^{ax}\cos bx+c_2e^{ax}\sin bx $$ (5.3)
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By applying the quadratic equation to the characteristic equation, we see that the roots are:
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$$  \displaystyle \lambda =-2\pm (4\pi)i $$     (5.4)
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This yields a general solution of:
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$$  \displaystyle y(x)=c_1e^{-2x}\cos (4\pi*x)+c_2e^{-2x}\sin (4\pi*x) $$     (5.5)
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Given

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$$  \displaystyle y''=2\pi*y'+\pi^2y $$     (5.6)
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Solution
This ODE is a linear ODE with constant coefficients (L2-ODE-CC).

First, equation (5.6) is rearranged into general form:
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$$  \displaystyle y''-(2\pi)y'-(\pi^2)y=0 $$     (5.7)
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The characteristic equation is:
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$$  \displaystyle \lambda^2-2\pi\lambda-\pi^2=0 $$     (5.8)
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Substituting the characteristic equation coefficients into the quadratic formula, we find:
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$$  \displaystyle \lambda_1=\lambda_2=\lambda=\pi $$     (5.9)
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Therefore, the general solution is:
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$$  \displaystyle y(x)=c_1e^{\pi*x}+c_2xe^{\pi*x} $$     (5.10)
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Author
Solved and Typed By - Egm4313.s12.team1.durrance 20:30, 25 January 2012 (UTC)

Reviewed By - Egm4313.s12.team1.silvestri 20:09, 30 January 2012 (UTC)

Edited By - --Egm4313.s12.team1.wyattling 19:53, 27 January 2012 (UTC)

Statement
For each ODE in Fig.2 in Kreyszig 2011 p.3 (except the last one involving a system of 2 ODEs), determine the order, linearity (or lack of), and show whether the principle of superposition can be applied.

Given

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$$  \displaystyle y''=g=const. $$     (6.0)
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$$  \displaystyle mv'=mg-bv^2 $$     (6.1)
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$$  \displaystyle h'=-k\sqrt{h} $$     (6.2)
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$$  \displaystyle my''+ky=0 $$     (6.3)
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$$  \displaystyle y''+w_{0}^2y=\cos wt, w_{0}/approx w $$ (6.4)
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$$  \displaystyle LI''+RI'+\frac{C}I=E' $$     (6.5)
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$$  \displaystyle EIy=f(x) $$     (6.6)
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$$  \displaystyle L\Theta''+g\sin\Theta=0 $$     (6.7)
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Solution
For equations 6.0-6.7, we need to determine the order, whether or not its linear, and whether or not the superposition principal can be applied. The order of an equation is determined by the highest derivative present. The equation is said to be linear if the variable from which the derivatives are taken, is independently linear. Finally, superposition may be applied if the equation is in fact linear.


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$$  \displaystyle y''=g=const. $$     (6.0)
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 * } Order: 2nd Linearity: Yes, all $$\displaystyle y $$ values are to a power of one. Superposition: Yes


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$$  \displaystyle mv'=mg-bv^2 $$     (6.1)
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 * } Order: 1st Linearity: No, the second $$\displaystyle v $$ in the equation is to the power of 2 Superposition: No, the equation is non-linear


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$$  \displaystyle h'=-k\sqrt{h} $$     (6.2)
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 * } Order: 1st Linearity: No, the second $$\displaystyle h $$ in the equation is to $$\displaystyle \frac{2} $$ power Superposition: No, the equation is non-linear


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$$  \displaystyle my''+ky=0 $$     (6.3)
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 * } Order: 2nd Linearity: Yes, each $$\displaystyle y $$ in the equation is to to a power of 1 Superposition: Yes, the equation is linear and homogeneous


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$$  \displaystyle y''+w_{0}^2y=\cos wt, w_{0}=w $$     (6.4)
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 * } Order: 2nd Linearity: Yes, each $$\displaystyle y $$ in the equation is to to a power of 1 Superposition: Yes


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$$  \displaystyle LI''+RI'+\frac{C}I=E' $$     (6.5)
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 * } Order: 2nd Linearity: Yes, each $$\displaystyle y $$ in the equation is to to a power of 1 Superposition: Yes


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$$  \displaystyle EIy=f(x) $$     (6.6)
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 * } Order: 4th Linearity: Yes, each $$\displaystyle y $$ in the equation is to to a power of 1 Superposition: Yes


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$$  \displaystyle L\Theta''+g\sin\Theta=0 $$     (6.7)
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 * } Order: 2nd Linearity: No, the term $$\displaystyle \sin\Theta $$ is non-linear Superposition: No, the equation is non-linear.

Author
Solved and typed by - --Egm4313.s12.team1.rosenberg 15:13, 30 January 2012 (UTC)

Reviewed By - Egm4313.s12.team1.armanious 02:37, 1 February 2012 (UTC)

Edited By - --Egm4313.s12.team1.wyattling 19:55, 27 January 2012 (UTC)