# User:Egm4313.s12.team1.wyattling

EGM4313 Team 1 USER: Wyatt Ling

Team Page

## SANDBOX

 ${\displaystyle \displaystyle {2Ce^{5x}+20Cxe^{5x}+25Cx^{2}e^{5x}-2K_{2}-20Cxe^{5x}-50Cx^{2}e^{5x}+20K_{2}x+10K_{1}+25Cx^{2}e^{5x}-25K_{2}x^{2}-25K_{1}x-25K_{0}=7e^{5x}-2x^{2}}}$ (3.0)

Therefore, the general form for the general solution is:

 ${\displaystyle \displaystyle 2Ce^{5x}+20Cxe^{5x}+25Cx^{2}e^{5x}}$

${\displaystyle Q=Cv_{c}\ }$

## Problem R1.6

### 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

 ${\displaystyle \displaystyle y''=g=const.}$ (6.0)
 ${\displaystyle \displaystyle mv'=mg-bv^{2}}$ (6.1)
 ${\displaystyle \displaystyle h'=-k{\sqrt {h}}}$ (6.2)
 ${\displaystyle \displaystyle my''+ky=0}$ (6.3)
 ${\displaystyle \displaystyle y''+w_{0}^{2}y=coswt,w_{0}=w}$ (6.4)
 ${\displaystyle \displaystyle LI''+RI'+{\frac {1}{C}}I=E'}$ (6.5)
 ${\displaystyle \displaystyle EIy''''=f(x)}$ (6.6)
 ${\displaystyle \displaystyle L\Theta ''+gsin\Theta =0}$ (6.7)

### Solution

In order to determine the order of any ODE, one must simply find the highest prime term in each equation.

ROUGH ANSWERS IN CASE ANYONE WANTS TO CHECK: (Okay so I think I got the superposition part, but i really want to run it past you all during our meeting before I latex everything etc.)

(6.0) = 2nd order/linear

(6.1) = 1st order/non-linear

(6.2) = 1st order/non-linear

(6.3) = 2nd order/linear

(6.4) = 2nd order/linear

(6.5) = 2nd order/linear

(6.6) = 4th order/linear

(6.7) = 2nd order/non-linear

### Author

Solved and typed by -

Reviewed By -

## Problem R1.5

### Statement

Find a general solution for the following two problems.

#### Given

 ${\displaystyle \displaystyle y''+4y'+(\pi ^{2}+4)y=0}$ (5.0)

#### Solution

The characteristic equation of this linear ODE with constant coefficients (L2-ODE-CC) is:

 ${\displaystyle \displaystyle \lambda ^{2}+4\lambda +(\pi ^{2}+4)=0}$ (5.1)

Note: The discriminant of the characteristic equation is negative.

 ${\displaystyle \displaystyle \lambda =-{\frac {1}{2}}a\pm bi}$ (5.2)

Which yields a general solution of:

 ${\displaystyle \displaystyle y(x)=c_{1}e^{ax}\cos bx+c_{2}e^{ax}\sin bx}$ (5.3)

By applying the quadratic equation to the characteristic equation, we see that the roots are:

 ${\displaystyle \displaystyle \lambda =-2\pm (4\pi )i}$ (5.4)

This yields a general solution of:

 ${\displaystyle \displaystyle y(x)=c_{1}e^{-2x}\cos(4\pi *x)+c_{2}e^{-2x}\sin(4\pi *x)}$ (5.5)

Solved by: Egm4313.s12.team1.durrance 20:23, 25 January 2012 (UTC)

### Kreyszig 2011 p.59 Problem 5

#### Given

 ${\displaystyle \displaystyle y''=2\pi *y'+\pi ^{2}y}$ (5.6)

#### Solution

This ODE is a linear ODE with constant coefficients (L2-ODE-CC).

First equation (5.6) is rearranged into general form:

 ${\displaystyle \displaystyle y''-(2\pi )y'-(\pi ^{2})y=0}$ (5.7)

The characteristic equation is:

 ${\displaystyle \displaystyle \lambda ^{2}-2\pi \lambda -\pi ^{2}=0}$ (5.8)

Substituting the characteristic equation coefficients into the quadratic formula, we find:

 ${\displaystyle \displaystyle \lambda _{1}=\lambda _{2}=\lambda =\pi }$ (5.9)

Therefore, the general form for the general solution is:

 ${\displaystyle \displaystyle y(x)=c_{1}e^{\pi *x}+c_{2}xe^{\pi *x}}$ (5.10)

Solved by Egm4313.s12.team1.durrance 20:30, 25 January 2012 (UTC)

## Problem R1.4

### Statement

Using the Circuit equation (4.1) (see Sec2 Notes), derive equations (4.2) and (4.3) from it.

### Solution

Given:

Capacitance Equation:

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

Circuit Equation:

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

First Alternate Circuit Equation:

 ${\displaystyle \displaystyle LI''+RI'+{\frac {1}{C}}I=V'}$ (4.2)

Second Alternate Circuit Equation:

 ${\displaystyle \displaystyle LQ''+RQ'+{\frac {1}{C}}Q=V}$ (4.3)

First we are asked to derive the alternate circuit equation (4.2) from the original circuit equation (4.1).

 ${\displaystyle \displaystyle V=LC{\frac {\mathrm {d^{2}} v_{c}}{\mathrm {d} t^{2}}}+RC{\frac {\mathrm {d} v_{c}}{\mathrm {d} t}}+v_{c}\rightarrow LI''+RI'+{\frac {1}{C}}I=V'}$ (4.4)

The first step is to take the derivative of the original equation so there is the V' term on one side:

 ${\displaystyle \displaystyle V'=LC{\frac {\mathrm {d^{3}} v_{c}}{\mathrm {d} t^{3}}}+RC{\frac {\mathrm {d^{2}} v_{c}}{\mathrm {d} t^{2}}}+{\frac {\mathrm {d} v_{c}}{\mathrm {d} t}}}$ (4.5)

L and C are constants and therefore remain unchanged by the derivative.

Next use the capacitance equation which states:

 ${\displaystyle \displaystyle I=C{\frac {\mathrm {d} v_{c}}{\mathrm {d} t}}}$ (4.6)

Take both the first and second derivative of this equation.

First Derivative:

 ${\displaystyle \displaystyle I'=C{\frac {\mathrm {d^{2}} v_{c}}{\mathrm {d} t^{2}}}}$ (4.7)

Second Derivative:

 ${\displaystyle \displaystyle I''=C{\frac {\mathrm {d^{3}} v_{c}}{\mathrm {d} t^{3}}}}$ (4.8)

Now substitute in the the I terms into the appropriate places. The equation below groups terms that need to be substituted.

 ${\displaystyle \displaystyle V'=L(C{\frac {\mathrm {d^{3}} v_{c}}{\mathrm {d} t^{3}}})+R(C{\frac {\mathrm {d^{2}} v_{c}}{\mathrm {d} t^{2}}})+{\frac {({C}{\frac {\mathrm {d} v_{c}}{\mathrm {d} t}})}{C}}}$ (4.9)

The final product after substitution is the alternate circuit equation:

 ${\displaystyle \displaystyle LI''+RI'+{\frac {1}{C}}I=V'}$ (4.2)

We are also asked to derive the second alternate circuit equation

 ${\displaystyle \displaystyle LQ''+RQ'+{\frac {1}{C}}Q=V}$ (4.3)

from the original circuit equation.

Once again use the capacitance equation which also states:

 ${\displaystyle \displaystyle Q=Cv_{c}}$ (4.12)

Take the first and second derivative:

First Derivative:

 ${\displaystyle \displaystyle Q'=C{\frac {\mathrm {d} v_{c}}{\mathrm {d} t}}}$ (4.13)

Second Derivative:

 ${\displaystyle \displaystyle Q''=C{\frac {\mathrm {d^{2}} v_{c}}{\mathrm {d} t^{2}}}}$ (4.14)

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:

 ${\displaystyle \displaystyle V=L(C{\frac {\mathrm {d^{2}} v_{c}}{\mathrm {d} t^{2}}})+R(C{\frac {\mathrm {d} v_{c}}{\mathrm {d} t}})+{\frac {(Cv_{c})}{C}}}$ (4.15)

After substituting, the final product is the second alternate circuit equation:

 ${\displaystyle \displaystyle LQ''+RQ'+{\frac {1}{C}}Q=V}$ (4.3)

### Author

Solved and typed by - Chris Stewart

Reviewed By -

## Problem R1.2

### Statement

Derive the equation of motion of the spring-mass-dashpot in Fig. 53, in Kreyszig 2011 p.85, with an applied force ${\displaystyle r(t)}$ on the ball.

### Solution

Kinematics:

 ${\displaystyle \displaystyle y=y_{k}=y_{c}}$ (2.0)

Kinetics:

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

Rearranging equations:

 ${\displaystyle \displaystyle r(t)=my''+k{{y}_{k}}+c{{y}_{c}}'}$ (2.4)
 ${\displaystyle \displaystyle y=y_{k}=y_{c}}$ (2.5)

Equation of motion:

 ${\displaystyle \displaystyle r(t)=my''+ky+cy'}$ (2.6)

### Author

Solved and typed by - --Egm4313.s12.team1.essenwein 23:06, 25 January 2012 (UTC)

Reviewed By -

## Problem R1.3

### Statement

For the spring-dashpot system on p.1-4 (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:

 ${\displaystyle \displaystyle a=y''}$ (3.0)

Because only two forces are acting on the dashpot and they are in opposite directions, they are equal in magnitude.

 ${\displaystyle \displaystyle ky=cv=f_{I}}$ (3.1)

The resulting acceleration of the mass, m, must be the difference of the applied force ${\displaystyle f(t)\!}$ and the internal force ${\displaystyle f_{I}\!}$.

 ${\displaystyle \displaystyle y''\propto f(t)-f_{I}}$ (3.2)

Using Newton's second law:

 ${\displaystyle \displaystyle my''=f(t)-f_{I}}$ (3.3)

Rearranging, this yields:

 ${\displaystyle \displaystyle my''+f_{I}=f(t)}$ (3.4)

### Author

Solved and Typed By: Egm4313.s12.team1.armanious 20:09, 25 January 2012 (UTC)

Reviewed By:

## Problem R1.1

### Statement

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

Figure 1

### Solution

Kinematics, Kinetics, and Relations equations as interpreted from picture

Kinematics:

 ${\displaystyle \displaystyle y=y_{k}=y_{c}}$ (1.0)

Kinetics:

 ${\displaystyle \displaystyle f(t)=my''+f_{k}+f_{c}}$ (1.1)

Relations:
-Force of spring relation:

 ${\displaystyle \displaystyle f_{k}=ky_{k}}$ (1.2)

-Force of dash-pot relation:

 ${\displaystyle \displaystyle f_{c}=Cy'_{c}}$ (1.3)

Putting kinematics, kinetics, and relations together:

 ${\displaystyle \displaystyle y''=y''_{k}=y''_{c}}$ (1.4)

Using kinetic relationship

 ${\displaystyle \displaystyle f(t)=my''_{c}+f_{k}+f_{c}}$ (1.5)

<br|>
Using the kinematics and relations equations:

 ${\displaystyle \displaystyle f_{k}=ky_{k}=ky_{c}}$ (1.6)

 ${\displaystyle \displaystyle f(t)=my''_{c}+ky_{c}+f_{c}}$ (1.7)

Using dash-pot relation:

 ${\displaystyle \displaystyle f_{c}=Cy'_{c}}$ (1.8)

 ${\displaystyle \displaystyle f(t)=my''_{c}+ky_{c}+Cy'_{c}}$ (1.9)

Final Equation

 ${\displaystyle \displaystyle f(t)=my''_{c}+Cy'_{c}+ky_{c}}$ (1.10)

### Author

Solved and typed by Egm4313.s12.team1.silvestri 20:44, 25 January 2012 (UTC)

Reviewed By -