User:Egm4313.s12.team3/Report1

Report 1 =Problem 1: Spring-dashpot system in parallel=

Problem Statement
For a spring-dashpot system in parallel with mass and an applied force f(t), derive the equation of motion. See [[media:iea.s12.sec1.djvu| Sec. 1 notes]] for a similar example. Figure of a spring-dashpot system in parallel:

Problem Solution
Force Body Diagram for the mass The kinematics for this problem:
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$$\displaystyle y=y_k=y_c $$      (1.1)
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The kinetics for this problem:
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$$ \displaystyle ma+{{f}_{c}}+{{f}_{k}}=f(t) $$     (1.2)
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For a spring:
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$$\displaystyle f_k=ky_k $$     (1.3) For a dashpot:
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$$\displaystyle f_c=c{y_c}' $$     (1.4)
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Taking equation 1.3 and 1.4 and plugging it back into equation 1.2 yields:
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$$\displaystyle f(t)=my''+c{y_c}'+ky_k $$     (1.5)
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Take the time derivative of Equation 1.1 to get:
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$$\displaystyle {y}'={y_k}'={y_c}' $$     (1.6)
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Substituting Equation 1.1 and 1.6 into Equation 1.5 you get:
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$$ \displaystyle my''+cy'+ky=f(t) $$
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=Problem 2: Spring-mass-dashpot system=

Problem Statement
Derive the equation of motion for the spring-mass-dashpot system with an applied force referred to on pages 1-4 of [[media:iea.s12.sec1.djvu| section 1 notes]].

Problem Solution
Force Body Diagram Kinematics:
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$$\displaystyle y=y_c=y_k $$     (2.1) Kinetics:
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$$\displaystyle ma+{{f}_{c}}+{{f}_{k}}=f(t) $$     (2.2)
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Hookes Law:
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$$  \displaystyle {{f}_{k}}=k{{y}_{k}} $$     (2.3) Dashpot:
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$$  \displaystyle {{f}_{c}}=c{{y}_{c}}' $$     (2.4)
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$$\displaystyle ma=my'' $$     (2.5) Combining equations 2.2-2.5:
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$$\displaystyle my''+c{{y}_{c}}'+k{{y}_{k}}=f(t) $$     (2.6)
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Take the time derivative of Equation 2.1 to get:
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$$\displaystyle {y}'={y_k}'={y_c}' $$     (2.7) Final equation of motion:
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$$ \displaystyle my''+cy'+ky=f(t) $$
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=Problem 3: Spring-Dashpot-Mass System in Series=

Problem Statement
For the following system (as can be seen in the figure below) comprised of a spring, a dashpot, and a mass in series solve for the equation of motion and draw the free body diagrams.

Problem Solution
For information regarding the order and linearity of ordinary differential equations, see [[media:iea.s12.sec1.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 1]]. Also, for a similar problem look into [[media:iea.s12.sec1.djvu| Sec 1 of Loc Vu-Quoc's class EGM4313 p.1-4]]. Force Body Diagrams Kinematics


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$$\displaystyle {y}=y_k+y_c $$     (3.1)
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Where:
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$$\displaystyle {y_k}=Spring\,Displacement $$
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&
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$$\displaystyle {y_c}=Dashpot\,Displacement $$
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Kinetics
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$$\displaystyle {a}=d^2y/dt^2={y}'' $$
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From the previous equation for acceleration and the mass (represented by the variable m) the sum of the forces can represented as follows:
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$$\displaystyle m{y}''=f(t)-f_I $$     (3.2) Where:
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$$\displaystyle f_I=The\,Internal\,Forces\,on\,the\,Mass $$     (3.3)
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Taking the second derivative of Equation 3.1:
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$$\displaystyle {(y=y_k+y_c)}'' $$
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Equals
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$$\displaystyle {y}={y_k}+{y_c}'' $$     (3.4)
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From the free body diagrams above through geometry :
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$$\displaystyle f_I=f_k $$     (3.5)
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And:
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$$\displaystyle f_k=f_c $$     (3.6)
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Which finally means that:
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$$\displaystyle f_I=f_c=f_k $$     (3.7)
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With:
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$$\displaystyle f_k=k{y_k} $$     (3.8)
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&
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$$\displaystyle f_c=C{y_c}' $$     (3.9)
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From Equation 3.7 we can solve for $$\displaystyle {y_c}'$$
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$$\displaystyle k{y_k}=C{y_c}' $$     (3.10)
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$$\displaystyle {y_c}'=(k/C){y_k} $$     (3.11) Then solving for $$\displaystyle {y_c}''$$ equals:
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$$\displaystyle {(y_c')}'=((k/C)y_k)'\,\,==>{y_c}''=(k/C){y_k}' $$     (3.12)
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Then plugging Equation 3.12 into Equation 3.4:
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$$\displaystyle {y}=y_k+(k/C){y_k}' $$     (3.13)
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Substituting Equation 3.13 for $$\displaystyle {y}''$$
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$$\displaystyle m({y_k''+(k/C){y_k}'})=f(t)-f_I $$     (3.14)
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And solving for $$\displaystyle f(t)$$ with $$\displaystyle f_I=$$ Equation 3.5
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$$\displaystyle m({y_k''+(k/C){y_k}'})+f_k=f(t) $$     (3.15)
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Finally with Equation 3.8 substituted into $$\displaystyle f_k$$ in Equation 3.15 gives the final answer of:
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$$ \displaystyle m({y_k''+(k/C){y_k}'})+k{y_k}=f(t) $$
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=Problem 4: Deriving circuit equations for RLC circuit in series=

Problem Statement
Derive equations (4.1) and (4.2) from equation (4.3)


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$$\displaystyle V = LQ'' + RQ' + \frac{1}{C} Q $$ (4.1)
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$$\displaystyle V' = LI'' + RI' + \frac{1}{C} I $$ (4.2)
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$$\displaystyle V = LC \frac{d^2 v_c}{dt^2} + RC \frac{dv_c}{dt} +v_c $$     (4.3) For more information and notes relating to this problem, see [[media:iea.s12.sec2.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 2]]
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Problem Solution
Given:
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$$\displaystyle V = v_R +v_L+v_C $$     (4.4)
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$$\displaystyle v_R = RI $$ (4.5)
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$$\displaystyle v_L = L \frac{dI}{dt} = LI' $$     (4.6)
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$$\displaystyle Q = C v_C \rightarrow v_C = \frac{Q}{C} $$     (4.7)
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$$\displaystyle I = \frac{dQ}{dt} = Q' \therefore Q = \int I dt $$ (4.8)
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Part A
Derivation of Equation (4.1) From Equation (4.7)
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$$\displaystyle v_C = \frac{Q}{C} $$     (4.9) Therefore, taking the first derivative...
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$$\displaystyle \frac{dv_c}{dt} = \frac{d}{dt} (\frac{Q}{C}) = \frac{Q'}{C} $$     (4.10) and taking the second derivative...
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$$\displaystyle \frac{d^2 v_c}{dt^2} = \frac{d}{dt} (\frac{Q'}{C}) = \frac{Q''}{C} $$     (4.11) Substituting Equations (4.7), (4.10), and (4.11) into Equation (4.3)...
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$$\displaystyle V = LC (\frac{Q''}{C}) + RC (\frac{Q'}{C}) + (\frac{Q}{C}) $$     (4.12) which simplifies to...
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$$\displaystyle V = LQ'' + RQ' + \frac{1}{C}Q $$     (4.1)
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Part B
Derivation of Equation (4.2) Taking the derivative from Equation (4.1) (Previously derived from Equation (4.3))
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$$\displaystyle V' = LQ' + RQ + \frac{1}{C}Q' $$     (4.13) Taking the first derivative of Equation (4.8)...
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$$\displaystyle Q' = I $$ (4.14) and taking the second derivative...
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$$\displaystyle Q'' = I' $$ (4.15) and taking the third derivative...
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$$\displaystyle Q' = I $$     (4.16) Then substituting Equations (4.14), (4.15), and (4.16) into Equation (4.1)...
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$$\displaystyle V' = LI'' + RI' + \frac{1}{C}I $$     (4.2)
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=Problem 5: Solving Ordinary Differential Equations by find the general solution=

Problem Statement
Given ODE's of this kind find the general solution and check the results using substitution. Similar ODE's and classification are shown in [[media:Iea.s12.sec2d.djvu|Sec 2 p. 2-5]].
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$$  \displaystyle y''+4y'+(\pi^2 +4)y=0 $$ $$ \displaystyle y''+2\pi y'+\pi^2 y=0 $$
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Part A

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$$  \displaystyle y''+4y'+(\pi^2 +4)y=0 $$ (5.1) Using the general form
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$$  \displaystyle \lambda^2+a\lambda+b=0 $$
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Equation (5.1) becomes
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$$  \displaystyle \lambda^2+4\lambda+(\pi^2+4)=0 $$ $$   \displaystyle \lambda_1=\frac{-4+(\sqrt{16-4(\pi^2+4)})}2 \; \; \; \; \; \; \; \; \; \; \; \lambda_2=\frac{-4-(\sqrt{16-4(\pi^2+4)})}2 $$ $$   \displaystyle \lambda_1=\frac{-4+(\sqrt{-4\pi^2)}}2 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \lambda_2=\frac{-4-(\sqrt{-4\pi^2)}}2 $$ $$   \displaystyle \lambda_1=\frac{-4+2\pi i}{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \lambda_2=\frac{-4-2\pi i}{2} $$ $$   \displaystyle \lambda_1=-2+\pi i \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \lambda_2=-2-\pi i $$
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$$  \displaystyle y_1=e^{-2x} \cos{\pi x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; y_2=e^{-2x} \sin{\pi x} $$ Calculating the derivatives:
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$$  \displaystyle y=Ae^{-2x} \cos{\pi x}+Be^{-2x} \sin{\pi x} $$ $$ \displaystyle y'=[-2Ae^{-2x} \cos{\pi x}-A\pi e^{-2x} \sin{\pi x}]+[\pi Be^{-2x} \cos{\pi x}-2Be^{-2x} \sin{\pi x}] $$ $$ \displaystyle y''=[4Ae^{-2x} \cos{\pi x}\;+2\;\pi Ae^{-2x} \sin{\pi x}\;+\;2\pi Ae^{-2x} \sin\pi x\;-\;\pi^2 Ae^{-2x} \cos{\pi x}]\;+\;[-2B\pi e^{-2x} \cos{\pi x}\;-\;\pi^2 Be^{-2x} \sin{\pi x}\;+\;4Be^{-2x} \sin{\pi x}\;-\;2\pi Be^{-2x} \cos{\pi x}] $$ Simplifying some of the terms:
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$$  \displaystyle 4y'=[-8Ae^{-2x} \cos{\pi x}-4A\pi e^{-2x} \sin{\pi x}]+[4\pi Be^{-2x} \cos{\pi x}-8Be^{-2x} \sin{\pi x}] $$ $$ \displaystyle (\pi^2 +4)y=(\pi^2 +4)Ae^{-2x} \cos{\pi x}+(\pi^2 +4)Be^{-2x} \sin{\pi x} $$
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Finally subbing the terms into the original equation:
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$$  \displaystyle y''\;+\;4y'\;+\;(\pi^2\;+\;4)y=[e^{-2x} \cos{\pi x}]\cdot[4A\;-\;\pi^2 A\;-\;2\pi B\;-\;2\pi B\;-\;8A\;+\;4\pi B\;+\;\pi^2 A\;+\;4A]\;+\;[e^{-2x} \sin{\pi x}]\cdot[2\pi A\;+\;2\pi A\;-\;\pi^2 B\;+\;4B\;-\;4\pi A\;-\;8B\;+\;\pi^2 B\;+\;4B] $$ $$ \displaystyle [e^{-2x} \cos{\pi x}]\cdot[0]+[e^{-2x} \sin{\pi x}]\cdot[0]=0 $$
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Part B

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$$  \displaystyle y''+2\pi y'+\pi^2 y=0 $$ (5.2) Using the general form
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$$  \displaystyle \lambda^2+a\lambda+b=0 $$ Equation (5.2) becomes
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$$  \displaystyle \lambda^2+2\pi \lambda+\pi^2=0 $$ $$   \displaystyle \lambda_{1,2}=\frac{-2\pi \pm \sqrt{4\pi^2 - 4\pi^2}}{2} $$ $$   \displaystyle \lambda_{1,2}=-\pi $$
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$$  \displaystyle y_{1,2}=(c_1+c_2x)e^{-\pi x} $$ Calculating the derivatives:
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$$  \displaystyle y=(c_1+c_2x)e^{-\pi x} $$ $$ \displaystyle y'=c_2e^{-\pi x}-\pi (c_1+c_2x)e^{-\pi x} $$ $$ \displaystyle y''=-\pi c_2e^{-\pi x}\;+\;[(-\pi (c_1\;+\;c_2x)\;\cdot\; -\pi e^{-\pi x})(-\pi c_2e^{-\pi x})] =-2\pi c_2e^{-\pi x}\;+\;\pi^2 (c_1\;+\;c_2x)e^{-\pi x} $$ Finally subbing the terms into the original equation:
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$$  \displaystyle y''+2\pi y'+\pi^2 y=\pi^2 (c_1+c_2x)e^{-\pi x}-2\pi c_2e^{-\pi x}+2\pi c_2e^{-\pi x}-2\pi^2 (c_1+c_2x)e^{-\pi x}+\pi^2(c_1+c_2x)e^{-\pi x} = 0 $$
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=Problem 6: Classifying ODE's and Applying the Superposition Principle=

Problem Statement
For each of the following ODE's, determine the order and linearity. Then, determine if the superposition principle can be applied.


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$$\displaystyle {y}''=g $$     (6.1)
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$$\displaystyle m{v}'=mg-bv^{2} $$     (6.2)
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$$\displaystyle {h}'=-k\sqrt{h} $$     (6.3)
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$$\displaystyle m{y}''+ky=0 $$     (6.4)
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$$\displaystyle {y}''+\omega_0^{2}y=cos(\omega t)\;\;,\;\;\omega_0 \approx \omega $$     (6.5)
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$$\displaystyle L{I}''+R{I}'+\frac{1}{C}I={E}' $$     (6.6)
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$$\displaystyle EI{y}=f(x) $$     (6.7)
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$$\displaystyle L{\theta}''+gsin\theta =0 $$     (6.8)
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Problem Solution
For information regarding the order and linearity of ordinary differential equations, see [[media:iea.s12.sec1.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 1]]. For information regarding the application of the superposition principle, see [[media:iea.s12.sec2.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 2]].

Part A

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$$\displaystyle {y}''=g $$     (6.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

This is a second order, linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

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

$$\displaystyle {y_h}''=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$\displaystyle {y_p}''=g $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We define the sum of the homogeneous solution and particular solution:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{y}=y_h+y_p $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Summing (1) and (2), we obtain the following:


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

$$\displaystyle {y_h}+{y_p}=g $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle {({y_h}'+{y_p}')}'=g $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle {(y_h+y_p)}''=g $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle {\bar{y}}''=g $$     (7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

From equation (7), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Part B

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

$$\displaystyle m{v}'=mg-bv^{2} $$     (6.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

This is a second order, non-linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

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

$$\displaystyle m{v_h}'+bv_h^{2}=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$\displaystyle m{v_p}'+bv_p^{2}=mg $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We define the sum of the homogeneous solution and particular solution:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{v}=v_h+v_p $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Summing (1) and (2), we obtain the following:


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

$$\displaystyle m({v_h}'+{v_p}')+b(v_h^{2}+v_p^{2})=mg $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle m{(v_h+v_p)}'+b(v_h^{2}+v_p^{2})=mg $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle m{\bar{v}}'+b(v_h^{2}+v_p^{2})=mg $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We cannot substitute any further since $$ b(v_h^{2}+v_p^{2}) \neq b(v_h+v_p)^{2} $$.


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

$$ \bar{v} $$ is NOT also a solution to the original ODE, and thus the superposition principle can't be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Part C

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

$$\displaystyle {h}'=-k\sqrt{h} $$     (6.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

This is a first order, non-linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

For a homogeneous equation, the superposition principle applies if a linear combination of any two solutions is also a solution to the original ODE. This is tested below.

Suppose $$h_1$$ is one solution:
 * {| style="width:100%" border="0"

$$\displaystyle {h_1}'+k\sqrt{h_1}=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

And $$h_2$$ is another solution:
 * {| style="width:100%" border="0"

$$\displaystyle {h_2}'+k\sqrt{h_2}=0 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We define any linear combination of these solutions:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{h}=c_1h_1+c_2h_2 $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Summing (1) and (2), we obtain the following:


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

$$\displaystyle {h_1}'+{h_2}'+k\sqrt{h_1}+k\sqrt{h_2}=0 $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle {(h_1+h_2)}'+k(\sqrt{h_1}+\sqrt{h_2})=0 $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle {\bar{h}}'+k(\sqrt{h_1}+\sqrt{h_2})=0 $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We cannot substitute any further since $$ k(\sqrt{h_1}+\sqrt{h_2})) \neq k\sqrt{h_1+h_2} $$.


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

$$ \bar{h} $$ is NOT also a solution to the original ODE, and thus the superposition principle can't be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Part D

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

$$\displaystyle m{y}''+ky=0 $$     (6.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

This is a second order, linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

For a homogeneous equation, the superposition principle applies if a linear combination of any two solutions is also a solution to the original ODE. This is tested below.

Suppose $$y_1$$ is one solution:
 * {| style="width:100%" border="0"

$$\displaystyle m{y_1}''+ky_1=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

And $$y_2$$ is another solution:
 * {| style="width:100%" border="0"

$$\displaystyle m{y_2}''+ky_2=0 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We define any linear combination of these solutions:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{y}=c_1y_1+c_2y_2 $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Summing (1) and (2), we obtain the following:


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

$$\displaystyle m({y_1}+{y_2})+k(y_1+y_2)=0 $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle m{(y_1+y_2)}''+k(y_1+y_2)=0 $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle m{\bar{y}}''+k\bar{y}=0 $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

From equation (6), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Part E

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

$$\displaystyle {y}''+\omega_0^{2}y=cos(\omega t)\;\;,\;\;\omega_0 \approx \omega $$     (6.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

This is a second order, linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

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

$$\displaystyle {y_h}''+\omega_0^{2}y_h=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$\displaystyle {y_p}''+\omega_0^{2}y_p=cos(\omega t) $$ (2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We define the sum of the homogeneous solution and particular solution:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{y}=y_h+y_p $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Summing (1) and (2), we obtain the following:


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

$$\displaystyle {y_h}+{y_p}+\omega_0^{2}y_h+\omega_0^{2}y_p=cos(\omega t) $$ (4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle {(y_h+y_p)}''+\omega_0^{2}(y_h+y_p)=cos(\omega t) $$ (5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle {\bar{y}}''+\omega_0^{2}(\bar{y})=cos(\omega t) $$ (6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

From equation (6), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Part F

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

$$\displaystyle L{I}''+R{I}'+\frac{1}{C}I={E}' $$     (6.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

This is a second order, linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

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

$$\displaystyle L{I_h}''+R{I_h}'+\frac{1}{C}I_h=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$\displaystyle L{I_p}''+R{I_p}'+\frac{1}{C}I_p={E}' $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We define the sum of the homogeneous solution and particular solution:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{I}=I_h+I_p $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Summing (1) and (2), we obtain the following:


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

$$\displaystyle L({I_h}+{I_p})+R({I_h}'+{I_p}')+\frac{1}{C}(I_h+I_p)={E}' $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle L{(I_h+I_p)}''+R{(I_h+I_p)}'+\frac{1}{C}(I_h+I_p)={E}' $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle L{\bar{I}}''+R{\bar{I}}'+\frac{1}{C}\bar{I}={E}' $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

From equation (6), we can see that $$ \bar{I} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Part G

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

$$\displaystyle EI{y}=f(x) $$     (6.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

This is a fourth order, linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

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

$$\displaystyle EI{y_h}=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$\displaystyle EI{y_p}=f(x) $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We define the sum of the homogeneous solution and particular solution:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{y}=y_h+y_p $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Summing (1) and (2), we obtain the following:


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

$$\displaystyle EI({y_h}'+{y_p}')=f(x) $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle EI{(y_h+y_p)}=f(x) $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle EI{\bar{y}}=f(x) $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

From equation (6), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Part H

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

$$\displaystyle L{\theta}''+gsin\theta =0 $$     (6.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

This is a second order, non-linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

For a homogeneous equation, the superposition principle applies if a linear combination of any two solutions is also a solution to the original ODE. This is tested below.

Suppose $$\theta_1$$ is one solution:
 * {| style="width:100%" border="0"

$$\displaystyle L{\theta_1}''+gsin\theta_1 =0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

And $$\theta_2$$ is another solution:
 * {| style="width:100%" border="0"

$$\displaystyle L{\theta_2}''+gsin\theta_2 =0 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We define any linear combination of these solutions:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{\theta}=c_1\theta_1+c_2\theta_2 $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Summing (1) and (2), we obtain the following:


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

$$\displaystyle L({\theta_1}+{\theta_2})+g(sin\theta_1+sin\theta_2)=0 $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle L{(\theta_1+\theta_2)}''+g(sin\theta_1+sin\theta_2)=0 $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$\displaystyle L{\bar{\theta}}''+g(sin\theta_1+sin\theta_2)=0 $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We cannot substitute any further since $$ sin\theta_1+sin\theta_2 \neq sin(\theta_1+\theta_2) $$.


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

$$ \bar{\theta} $$ is NOT also a solution to the original ODE, and thus the superposition principle can't be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

=Contributing Members for This Project=