University of Florida/Egm4313/s12.team11.R1

Kinematics

${\displaystyle y=y_{k}=y_{c}\!}$ (1)

Since the spring and dashpot are placed parallel to each other, they will experience the same deformation when a force ${\displaystyle f(t)}$ is applied on the system

Kinetics

${\displaystyle f(t)-f_{k}-f_{c}=my''\!}$ (2)

Based on Newton's Second Law of Motion ${\displaystyle \sum F=ma}$ where ${\displaystyle a}$ is the second derivative of position with respect to time

Relations

${\displaystyle f_{k}=ky_{k}\!}$ (3)

${\displaystyle f_{c}=cy_{c}'\!}$ (4)

Using (1) in (2)&(3):

${\displaystyle f_{k}=ky\!}$

${\displaystyle f_{c}=cy'\!}$

Then (2) becomes:

                                                     ${\displaystyle f(t)-ky-cy'=my''\!}$

Created by: Francisco Arrieta 22:05, 31 January 2012 (UTC)

Figures

Kinematics

${\displaystyle y=y_{k}+y_{c}+y_{m}+r(t)\!}$

Kinetics

Spring

${\displaystyle F_{k}=-k*y\!}$

Dashpot

${\displaystyle F_{c}=-c*y'\!}$

Mass

${\displaystyle F_{m}=m*y''\!}$

Forcing Function

${\displaystyle F_{applied}=r(t)\!}$

Sum of Forces

${\displaystyle \sum F_{applied}+F_{m}+F_{m}+F_{c}+F_{k}=0\!}$

Substitution

${\displaystyle 0=r(t)-m*y''-c*y'-k*y\!}$

And finally we find:

                                                 ${\displaystyle r(t)=m*y''+c*y'+k*y\!}$

Created By: Kyle Gooding 01:46, 29 January 2012 (UTC)

Figures

Kinetics

${\displaystyle \sum F_{y}=ma=f(t)-f_{I}\!}$

which can be re-written as

                                                        ${\displaystyle my''+f_{I}=f(t)\!}$

Created by: Luca Imponenti 03:01, 30 January 2012 (UTC)

Problem Statement

Find a general solution to the given ODE. Check your answer by substituting into the original equation.

Given:

${\displaystyle {y}''+4{y}'+(\pi ^{2}+4)y=0\!}$

Solution

The characteristic equation of this ODE is therefore:

${\displaystyle \lambda ^{2}+a\lambda +b=\lambda ^{2}+4\lambda +(\pi ^{2}+4)=0\!}$

Evaluating the discriminant:

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

Therefore the equation has two complex conjugate roots and a general homogenous solution of the form:

${\displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))\!}$ ^[1]

Where: ${\displaystyle \omega ={\sqrt {b-{\frac {1}{4}}a^{2}}}={\sqrt {\pi ^{2}+4-{\frac {1}{4}}(4)^{2}}}={\sqrt {\pi ^{2}}}=\pi \!}$

And finally we find the general homogenous solution:

                                                  ${\displaystyle y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$

Checking:

We found that:

${\displaystyle y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$

Differentiating ${\displaystyle y\!}$ to obtain ${\displaystyle {y}'\!}$ and ${\displaystyle {y}''\!}$ respectively:

${\displaystyle {y}'=-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+e^{-2x}(\pi Acos(\pi x)-\pi Bsin(\pi x))\!}$

                                   ${\displaystyle {y}'=-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))\!}$

${\displaystyle {y}''=4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-2\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-2\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$

                 ${\displaystyle {y}''=4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$

Substituting these equations into the original ODE yields:

${\displaystyle 4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$
${\displaystyle +4(-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x)))+(\pi ^{2}+4)(e^{-2x}(Acos(\pi x)+Bsin(\pi x)))=0\!}$

${\displaystyle 4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))-\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!}$
${\displaystyle -8e^{-2x}(Acos(\pi x)+Bsin(\pi x))+4\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))+\pi ^{2}e^{-2x}(Acos(\pi x)+Bsin(\pi x))+4e^{-2x}(Acos(\pi x)+Bsin(\pi x))=0\!}$

${\displaystyle (4-8+4)e^{-2x}(Acos(\pi x)+Bsin(\pi x))+(-4+4)\pi e^{-2x}(Acos(\pi x)-Bsin(\pi x))+(\pi ^{2}-\pi ^{2})e^{-2x}(Acos(\pi x)+Bsin(\pi x))=0\!}$

${\displaystyle 0\equiv 0\!}$

Therefore this solution is correct.

Problem Statement

Find a general solution to the given ODE. Check your answer by substituting into the original equation.

Given:

${\displaystyle {y}''+2\pi {y}'+\pi ^{2}y=0\!}$

Solution

The characteristic equation of this ODE is therefore:

${\displaystyle \lambda ^{2}+a\lambda +b=\lambda ^{2}+2\pi \lambda +\pi ^{2}=0\!}$

Evaluating the discriminant:

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

Therefore the equation has a real double root and a general homogenous solution of the form:

${\displaystyle y=(c_{1}+c_{2}x)e^{-ax/2}\!}$ ^[2]

And finally we find the general homogenous solution:

                                                       ${\displaystyle y=(c_{1}+c_{2}x)e^{-\pi x}\!}$

Checking:

We found that:

${\displaystyle y=(c_{1}+c_{2}x)e^{-\pi x}\!}$

Differentiating ${\displaystyle y\!}$ to obtain ${\displaystyle {y}'\!}$ and ${\displaystyle {y}''\!}$ respectively:

                                                  ${\displaystyle {y}'=c_{2}e^{-\pi x}-\pi (c_{1}+c_{2}x)e^{-\pi x}\!}$

And,

${\displaystyle {y}''=-\pi c_{2}e^{-\pi x}-\pi c_{2}e^{-\pi x}+\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}\!}$

                                                 ${\displaystyle {y}''=-2\pi c_{2}e^{-\pi x}+\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}\!}$

Substituting these equations into the original ODE yields:

${\displaystyle -2\pi c_{2}e^{-\pi x}+\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}+2\pi (c_{2}e^{-\pi x}-\pi (c_{1}+c_{2}x)e^{-\pi x})+\pi ^{2}((c_{1}+c_{2}x)e^{-\pi x})=0\!}$

${\displaystyle -2\pi c_{2}e^{-\pi x}+\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}+2\pi c_{2}e^{-\pi x}-2\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}+\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}=0\!}$

${\displaystyle (-2\pi +2\pi )c_{2}e^{-\pi x}+(\pi ^{2}-2\pi ^{2}+\pi ^{2})(c_{1}+c_{2}x)e^{-\pi x}=0\!}$

${\displaystyle 0\equiv 0\!}$

Therefore this solution is correct.
Created by: Jonathan Sheider 16:27, 31 January 2012 (UTC)

Order of ODEs

To find the order of Ordinary Differential Equations, (ODEs) simply check the differential order of the dependent variable. If the dependent variable is differentiated once, then it is a first order ODE. If it is differentiated twice, it is a second order ODE.

Linearity

To find the linearity of the ODE, check the power of the dependent variable. If the dependent variable is raised to the power of anything one than 1, then it is not linear. If the dependent variable is raised to the power of 1, then it is linear.

Superposition Principle

For a homogeneous linear ODE (2), any linear combination of two solutions on an open interval I is again a solution of (2) on I. In particular, for such an equation, sums and constant multiples of solutions are again solutions.^[3]

The function satisfies the superposition principle if the sum of the homogeneous solution and the particular solution is equal to ${\displaystyle {\overline {y}}(x)\!}$, with ${\displaystyle {\overline {y}}=(y_{h}+y_{p})\!}$
${\displaystyle {\overline {y}}(x):=y_{h}(x)+y_{p}(x)\!}$

R1.6a

${\displaystyle y''=g=const\!}$

Check superposition principle:

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

${\displaystyle y_{h}''=0\!}$

${\displaystyle y_{p}''=g\!}$

${\displaystyle (y_{h}''+y_{p}'')\equiv {\overline {y}}''\!}$

Order of ODE: 2nd Order
Linearity: Linear
Superposition Principle: Yes

R1.6b

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

Check superposition principle:

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

${\displaystyle mv_{h}'+bv_{h}^{2}=0\!}$

${\displaystyle mv_{p}'+bv_{p}^{2}=mg\!}$

${\displaystyle m(v_{h}'+v_{p}')+b(v_{h}^{2}+v_{p}^{2})\not \equiv m{\overline {v}}'+b({\overline {v}}^{2})\!}$

Order of ODE: 1st Order
Linearity: Non-linear
Superposition Principle: No

R1.6c

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

Check superposition principle:

${\displaystyle h'+kh^{1/2}=0\!}$

${\displaystyle h_{h}'+kh_{h}^{1/2}=0\!}$

${\displaystyle h_{p}'+kh_{p}^{1/2}=0\!}$

${\displaystyle (h_{h}'+h_{p}')+k(h_{h}^{1/2}+h_{p}^{1/2})\not \equiv {\overline {h}}'+k{\overline {h}}^{1/2}\!}$

Order of ODE: 1st Order
Linearity: Non-linear
Superposition Principle: No

R1.6d

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

Check superposition principle:

${\displaystyle y''+{\frac {k}{m}}y=0\!}$

${\displaystyle y_{h}''+{\frac {k}{m}}y_{h}=0\!}$

${\displaystyle y_{p}''+{\frac {k}{m}}y_{p}=0\!}$

${\displaystyle (y_{h}''+y_{p}'')+{\frac {k}{m}}(y_{h}+y_{p})\equiv {\overline {y}}''+{\frac {k}{m}}{\overline {y}}\!}$

Order of ODE: 2nd Order
Linearity: Linear
Superposition Principle: Yes

R1.6e

${\displaystyle y''+\omega _{o}^{2}y=\cos(\omega t)\!}$

Check superposition principle:

${\displaystyle y''+\omega _{o}^{2}y=\cos(\omega t)\!}$

${\displaystyle y_{h}''+\omega _{o}^{2}y_{h}=0\!}$

${\displaystyle y_{p}''+\omega _{o}^{2}y_{p}=\cos(\omega t)\!}$

${\displaystyle (y_{h}''+y_{p}'')+\omega _{o}^{2}(y_{h}+y_{p})\equiv {\overline {y}}''+\omega _{o}^{2}{\overline {y}}\!}$

Order of ODE: 2nd Order
Linearity: Linear
Superposition Principle: Yes

R1.6f

${\displaystyle LI''+RI'+{\frac {1}{C}}I=E'\!}$

Check superposition principle:

${\displaystyle I''+{\frac {R}{L}}I'+{\frac {1}{LC}}I={\frac {E'}{L}}\!}$

${\displaystyle I_{h}''+{\frac {R}{L}}I_{h}'+{\frac {1}{LC}}I_{h}=0\!}$

${\displaystyle I_{p}''+{\frac {R}{L}}I_{p}'+{\frac {1}{LC}}I_{p}={\frac {E'}{L}}\!}$

${\displaystyle (I_{h}''+I_{p}'')+{\frac {R}{L}}(I_{h}'+I_{p}')+{\frac {1}{LC}}(I_{h}+I_{p})\equiv {\overline {I}}''+{\frac {R}{L}}{\overline {I}}'+{\frac {1}{LC}}{\overline {I}}\!}$

Order of ODE: 2nd Order
Linearity: Linear
Superposition Principle: Yes

R1.6g

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

Check superposition principle:

${\displaystyle y''''-{\frac {1}{EI}}y=0\!}$

${\displaystyle y_{h}''''-{\frac {1}{EI}}y_{h}=0\!}$

${\displaystyle y_{p}''''-{\frac {1}{EI}}y_{p}=0\!}$

${\displaystyle (y_{h}''''+y_{p}'''')-{\frac {1}{EI}}(y_{h}+y_{p})\equiv {\overline {y}}''''-{\frac {1}{EI}}{\overline {y}}\!}$

Order of ODE: 4th Order
Linearity: Linear
Superposition Principle: Yes

R1.6h

${\displaystyle L\theta ''+g\sin \theta =0\!}$

Check superposition principle:

${\displaystyle \theta ''+{\frac {g}{L}}\sin \theta =0\!}$

${\displaystyle \theta _{h}''+{\frac {g}{L}}\sin \theta _{h}=0\!}$

${\displaystyle \theta _{p}''+{\frac {g}{L}}\sin \theta _{p}=0\!}$

${\displaystyle (\theta _{h}''+\theta _{p}'')+{\frac {g}{L}}(\sin \theta _{h}+\sin \theta _{p})\not \equiv {\overline {\theta }}''+{\frac {g}{L}}\sin {\overline {\theta }}\!}$

Order of ODE: 2nd Order
Linearity: Non-linear
Superposition Principle: No

R1.6i

${\displaystyle y_{1}^{'}=ay_{1}-by_{1}y_{2}\!}$

${\displaystyle y_{2}^{'}=ky_{1}y_{2}-ly_{1}\!}$

Order of ODE: 1st Order
Linearity: Non-linear
Superposition Principle: N/A

Created by [Daniel Suh] 21:59, 31 January 2012 (UTC)