University of Florida/Egm4313/s12.team14.report2

Problem 1

Find

Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $r(x)$

Consider no excitation:

$r(x)=0$

$(Eq.2)$

Plot the solution.

Solution

Characteristic equation:

$(\lambda -{\lambda }_1)(\lambda -{\lambda }_2)=0$

$(Eq.3)$

Substituting $(Eq.1a)$ into $(Eq.3)$:

$(\lambda +2)(\lambda -5)=0$

$(Eq.4)$

${\lambda }^2-3\lambda -10=0$

$(Eq.5)$

Non-homogeneous solution:

$y^{″}-3y^{\prime }-10y=0$

$(Eq.6)$

Homogeneous solution:

$y_h(x)=C_1{e}^{-2x}+C_2{e}^{5x}$

$(Eq.7)$

Overall solution:

$y(x)=C_1{e}^{-2x}+C_2{e}^{5x}+y_p(x)$

$(Eq.8)$

No excitation:

$r(x)=0=>y_p(x)=0$

$(Eq.9)$

From intitial conditions:

$y(0)=1=C_1+C_2$

$(Eq.10)$

$y^{\prime }(0)=0=-2C_1+5C_2$

$(Eq.11)$

Solving for $C_1$ and $C_2$:

$C_1=\frac{5}{7}$

${\displaystyle }$

and

$C_2=\frac{2}{7}$

${\displaystyle }$

So, the final solution is:

$y(x)=\frac{5}{7}{e}^{-2x}+\frac{2}{7}{e}^{5x}$

File:Team14R22diagram1.png

Problem 2

Find

Find and plot the solution for $(Eq.1)$

Solution

Due to no excitation, $(Eq.1)$ becomes:

$y^{″}-10y^{\prime }+25y=0$

$(Eq.4)$

Substituting $\lambda$ into $(Eq.4)$:

${\lambda }^2-10\lambda +25=0$

$(Eq.5)$

Factoring, and solving for $\lambda$:

$(\lambda -5{)}^2=0$

$(Eq.6)$

${\lambda }_2={\lambda }_1=\lambda =5$

${\displaystyle }$

Since $\lambda$ is a double root, the general solution:

$y=C_1{e}^{-5x}+C_{2}x{e}^{-5x}$

$(Eq.7)$

From intitial conditions:

$y(0)=1=C_1$

$(Eq.8)$

$y^{\prime }(0)=0=5C_1+C_2$

$(Eq.9)$

Solving for $C_1$ and $C_2$:

$C_1=1$

${\displaystyle }$

and

$C_2=-5$

${\displaystyle }$

So, the final solution is:

$y(x)=e^{5x}-e^{5x}$

File:Team14R22diagram2.png

Problem 3

Given

(a)

$y^{″}+6y^{\prime }+8.96y=0$

(b)

$y^{″}+4y^{\prime }+({\pi }^2+4)y=0$

Find

General solution to the differential equation

Solution

(a)

Characteristic equation:

${\lambda }^2+6\lambda +8.96=0$

Using the quadratic equation:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where:

$a=1,b=6,c=8.96$

Place values into quadratic equation:

$x=\frac{-6\pm \sqrt{6^2-4*1*8.96}}{2*1}$

$x=\frac{-6\pm \sqrt{36-35.84}}{2}$

$x=\frac{-6\pm 0.4}{2}$

$x=\frac{-6.4}{2},\frac{-5.6}{2}$

$x=-3.2,-2.8$

$y=c_1{e}^{-3.2x}+c_2{e}^{-2.8x}$

(b)

Characteristic equation:

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

Using the quadratic equation:

$\lambda =\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where:

$a=1,b=4,c=({\pi }^2+4)$

Place values into quadratic equation:

$\lambda =\frac{-4\pm \sqrt{4^2-4*1*({\pi }^2+4)}}{2*1}$

$\lambda =\frac{-4\pm \sqrt{16-4{\pi }^2-16)}}{2}$

$\lambda =\frac{-4\pm 2\pi i}{2}$

$\lambda =-2\pm \pi i$

$y=e^{-2x}(C_1\cos \pi x+C_2\sin \pi x)$

Problem 4

Given

(5)

$y{\text{'}}^{\prime }+2\pi y^{\prime }+{\pi }^{2}y=0$

$(Eq.1)$

(6)

$10y^{″}-32y^{\prime }+25.6y=0$

$(Eq.2)$

h

Find

For both (5) and (6), find a general solution. Then, check answers by substitution.

Solution

(5)

To obtain the general solution for (Eq. 1), we let

$y=e^{\lambda x}$

This yields:

$e^{\lambda x}({\lambda }^2+2\pi \lambda +{\pi }^2)=0$

Therefore, we have:

${\lambda }^2+2\pi \lambda +{\pi }^2=0$

Where $\lambda$ is a solution to the characteristic equation

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

We can see that

$a=2\pi$ and

$b={\pi }^2$.

Next, we must examine the quadratic formula to determine whether the system is over-, under-, or critically-damped. The discriminant is:

$\mathrm{△}=a^2-4b$

$\mathrm{△}=(2{\pi }^2{)}^2-4({\pi }^2)$

$\mathrm{△}=0$

Therefore, the equation is critically-damped.

The general solution is therefore represented by:

$y=c_1{e}^{{\lambda }_{1}x}+c_{2}x{e}^{{\lambda }_{2}x}$

Where ${\lambda }_1={\lambda }_2=\lambda$.

We can determine the value of $\lambda$ by again using the quadratic equation, this time in full:

$\lambda =\frac{1}{2}(-a+\sqrt{a^2-4b})$

for which we find that $\lambda =-\pi$.

Therefore, the general solution to the equation is:

$y=c_1{e}^{-\pi x}+c_{2}x{e}^{-\pi x}$

Verification by substitution:

$y\text{'}=-\pi c_1{e}^{-\pi x}+c_2{e}^{-\pi x}-\pi c_{2}x{e}^{-\pi x}$

$y{\text{'}}^{\prime }={\pi }^2{c}_1{e}^{-\pi x}-\pi c_2{e}^{-\pi x}+{\pi }^2{c}_{2}x{e}^{-\pi x}-\pi c_2{e}^{-\pi x}$

Plugging into Eq. 1 and combining terms, it is found:

$c_1{e}^{-\pi x}({\pi }^2-2{\pi }^2+{\pi }^2)+c_2{e}^{-\pi x}(-\pi -\pi +2\pi )+c_{2}x{e}^{-\pi x}({\pi }^2-2{\pi }^2+{\pi }^2)=0$

All the terms cancel, so the statement is true, and the solution is verified.

(6)

To obtain the general solution for (Eq. 2), let

$y=e^{\lambda x}$

So then, we have

$10{\lambda }^2-32\lambda +25.6=0$

as a solution to the characteristic equation. All terms must be divided by 10, to put the equation in standard form:

${\lambda }^2-3.2\lambda +2.56=0$

Examining the discriminant, we find that

$\mathrm{△}=a^2-4b=(-3.2{)}^2-4(2.56)=0$

Therefore, the equation is critically-damped.

The general solution is represented by:

$y=c_1{e}^{{\lambda }_{1}x}+c_{2}x{e}^{{\lambda }_{2}x}$

and ${\lambda }_1={\lambda }_2=\lambda$.

To determine the value of $\lambda$, we set:

$\lambda =\frac{1}{2}(-a+\sqrt{a^2-4b})$

for which we find that $\lambda =1.6$.

Therefore, the general solution to the equation is:

$y=c_1{e}^{1.6x}+c_{2}x{e}^{1.6x}$

Verification by substitution:

$y\text{'}=1.6c_1{e}^{1.6x}+c_2{e}^{1.6x}+1.6c_{2}x{e}^{1.6x}$

$y{\text{'}}^{\prime }=(1.6^2)c_1{e}^{1.6x}+3.2c_2{e}^{1.6x}+(1.6^2)c_{2}x{e}^{1.6x}$

Plugging into Eq. 1 and combining terms, it is found:

$c_1{e}^{1.6x}(0)+c_2{e}^{1.6x}(0)+c_{2}x{e}^{1.6x}(0)=0$

All the terms cancel, so the statement is true, and the solution is verified.

Problem 5

Given

$y{\text{'}}^{\prime }+a{y}^{\prime }+by=0$

Given the basis:

(a)

$e^{2.6x},e^{-4.3x}$

(b)

$e^{-\sqrt{5}x},x{e}^{-\sqrt{5}x}$

Find

For the given Information above, Find an ODE for both (a) and (b)

Solution

(a)The general solution can be written as:

$y=C_1{e}^{2.6x}+C_2{e}^{-4.3x}$

The characteristic equation can be written in the following way:

$(\lambda -2.6)(\lambda +4.3)={\lambda }^2+1.7\lambda -11.18=0$

Giving the ODE:

$y{\text{'}}^{\prime }+1.7y^{\prime }-11.18y=0$

(b) The general solution can be written as:

$y=C_1{e}^{-\sqrt{5}x}+C_{2}x{e}^{-\sqrt{5}x}$

The characteristic equation can be written in the following way:

$(\lambda +\sqrt{5})(\lambda +\sqrt{5})={\lambda }^2+2\sqrt{5}\lambda +5=0$

Giving the ODE:

$y{\text{'}}^{\prime }+2\sqrt{5}{y}^{\prime }+5y=0$

Problem 6

Given

alt text

Mass Spring Dashpot system FBD's

Spring-dashpot equation of motion from sec 1-5

$my{}_{k^{″}}+m\frac{k}{c}y{}_{k^{\prime }}+k{y}_k=f(t)$

Find

Spring-dashpot-mass system in series. Find the values for the parameters k,c,m with a double real root of $\lambda =-3$

Solution

Consider the double real root

$\lambda =-3$

Characteristic equation is

$(\lambda +3{)}^2={\lambda }^2+6\lambda +9=0$

$(Eq.1)$

Recall

$my{}_{k^{″}}+m\frac{k}{c}y{}_{k^{\prime }}+k{y}_k=f(t)$

$(Eq.2)$

Thus

$m=1$

$m\frac{k}{c}=6$

$k=9$

Solve for c

$c=\frac{9}{6}=\frac{3}{2}$

Therefore

$c=\frac{3}{2}andk=9andm=1$

Problem 7

Given

Taylor Series at t=0

Find

Develop the MacLaurin Series for $e^t,\cos t,\sin t$

Solution

Taylor series is defined as

$\sum _{n=0}^{\mathrm{\infty }}\frac{f^{(n)}(t)}{n!}(x-t{)}^n$

$(Eq.1)$

The MacLaurin series occurs when t=0

$\sum _{n=0}^{\mathrm{\infty }}\frac{f^{(n)}(0)}{n!}(x{)}^n$

$(Eq.2)$

Development of MacLaurin series for $e^t$

$e^t=\frac{1}{0!}+\frac{1}{1!}t+\frac{1}{2!}{t}^2+\frac{1}{3!}{t}^3...$

$(Eq.3)$

Final Answer

$e^t=1+x+\frac{1}{2}{t}^2+\frac{1}{6}{t}^3...$


Explicit form can be written as


$e^t=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}(t{)}^n$

Development of MacLaurin series for $cos(t)$

$cos(t)=\frac{1}{0!}+\frac{0}{1!}t-\frac{1}{2!}{t}^2+\frac{0}{3!}{t}^3+\frac{1}{4!}{t}^4...$

$(Eq.4)$

Final Answer

$cos(t)=1-\frac{1}{2}{t}^2+\frac{1}{24}{t}^4...$


Explicit form can be written as


$cos(t)=\sum _{n=0}^{\mathrm{\infty }}\frac{-1}{2n!}(t{)}^{2n}$

Development of MacLaurin series for $sin(t)$

$sin(t)=\frac{0}{0!}+\frac{1}{1!}t-\frac{0}{2!}{t}^2-\frac{1}{3!}{t}^3+\frac{0}{4!}{t}^4+\frac{1}{5!}{t}^5...$

$(Eq.5)$

Final Answer

$sin(t)=1t-\frac{1}{6}{t}^3+\frac{1}{120}{t}^5...$


Explicit form can be written as


$sin(t)=\sum _{n=0}^{\mathrm{\infty }}\frac{-1}{(2n+1)!}(t{)}^{(2n+1)}$

Problem 8

Given

(8)

$y{\text{'}}^{\prime }+y^{\prime }+3.25y=0$

$(Eq.1a)$

(15)

$y{\text{'}}^{\prime }+(0.54)y^{\prime }+(0.0729+\pi )y=0$

$(Eq.1b)$

Solution

(8)

Assume that the solution is of the form

$y_h=e^{\lambda x}$

$(Eq.2)$

$\therefore y{\text{'}}_h=\lambda e^{\lambda x}$

$(Eq.3)$

And

$y^{\prime }{\text{'}}_h={\lambda }^2{e}^{\lambda x}$

$(Eq.4)$

Substituting equations 2, 3, and 4 into equation 1a yields

${\lambda }^2{e}^{\lambda x}+\lambda e^{\lambda x}+3.25e^{\lambda x}=0$
$e^{\lambda x}({\lambda }^2+\lambda +3.25)=0$
Since
$e^{\lambda x}\ne 0$
$\therefore ({\lambda }^2+4\lambda +3.25)=0$

We must use the quadratic formula to obtain values for lambda
$\lambda =\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$\lambda =\frac{-1\pm \sqrt{(1{)}^2-4(1)(3.25)}}{2(1)}$
$\lambda =\frac{-1\pm \sqrt{1-13)}}{2}$
$\lambda =\frac{-1\pm \sqrt{-12}}{2}$
$\lambda =-0.5\pm j\sqrt{3}$
Where $j=\sqrt{-1}$

$y_h=C_1{e}^{(-0.5+j\sqrt{3})x}+C_2{e}^{(-0.5-j\sqrt{3})x}$

In order to verify solution, we must evaluate it's first and second derivative, then plug them into the equation
Let $a=-0.5,b=\sqrt{3}$
$y_h=C_1{e}^{(a+jb)x}+C_2{e}^{(a-jb)x}$
$y_h=C_1{e}^{ax}{e}^{jbx}+C_2{e}^{ax}{e}^{-jbx}$
$y_h\text{'}=C_1[e^{ax}(jb{e}^{jbx})+e^{jbx}(a{e}^{ax})]+C_2[e^{ax}(-jb{e}^{jbx})+e^{jbx}(a{e}^{ax})]$
$y_h{\text{'}}^{\prime }=C_1[e^{ax}((jb{)}^2{e}^{jbx})+(jb{e}^{jbx})(a{e}^{ax})+e^{jbx}((a{)}^2{e}^{ax})+(a{e}^{ax})(jb{e}^{jbx})]+C_2[e^{ax}((jb{)}^2{e}^{-jbx})+(-jb{e}^{-jbx})(a{e}^{ax})+e^{-jbx}((a{)}^2{e}^{ax})+(a{e}^{ax})(-jb{e}^{-jbx})]$

Plugging $y_h,{y_h}^{\prime },{y_h}^{″}$ into Eq. 1 and collecting terms gives
$0=C_1{e}^{ax}{e}^{jbx}[-b^2+jab+a^2+jab+jb+a+3.25]+C_2{e}^{ax}{e}^{-jbx}[-b^2-jab+a^2-jab-jb+a+3.25]$
$0=C_1{e}^{-0.5x}{e}^{j\sqrt{3}x}[-3-j\sqrt{3}/2+1/4-j\sqrt{3}/2+j\sqrt{3}-1/2+3.25]+C_2{e}^{-0.5x}{e}^{-j\sqrt{3}x}[-3+j\sqrt{3}/2+1/4+j\sqrt{3}/2-j\sqrt{3}-1/2+3.25]$
$0=C_1{e}^{-0.5x}{e}^{j\sqrt{3}x}(0)+C_2{e}^{-0.5x}{e}^{-j\sqrt{3}x}(0)$

Therefore the solution holds for any values of $C_1,C_2,x$

(15)

Assume that the solution is of the form

$y_h=e^{\lambda x}$

$(Eq.2)$

$\therefore y{\text{'}}_h=\lambda e^{\lambda x}$

$(Eq.3)$

And

$y^{\prime }{\text{'}}_h={\lambda }^2{e}^{\lambda x}$

$(Eq.4)$

Substituting equations 2, 3, and 4 into equation 1b yields

${\lambda }^2{e}^{\lambda x}+(0.54)\lambda e^{\lambda x}+(0.0729+\pi )e^{\lambda x}=0$
$e^{\lambda x}({\lambda }^2+(0.54)\lambda +(0.0729+\pi ))=0$
Since
$e^{\lambda x}\ne 0$
$\therefore ({\lambda }^2+(0.54)\lambda +(0.0729+\pi ))=0$

We must use the quadratic formula to obtain values for lambda
$\lambda =\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$\lambda =\frac{-(0.54)\pm \sqrt{(0.54{)}^2-4(1)(0.0729+\pi )}}{2(1)}$
$\lambda =\frac{-0.54\pm j3.545}{2}$
$\lambda =-0.27\pm j(1.772)$
Where $j=\sqrt{-1}$

$y_h=C_1{e}^{(-0.27+j(1.772))x}+C_2{e}^{(-0.27-j(1.772))x}$

In order to verify solution, we must evaluate it's first and second derivative, then plug them into the equation
Let $a=-0.27,b=1.772$
$y_h=C_1{e}^{(a+jb)x}+C_2{e}^{(a-jb)x}$
$y_h=C_1{e}^{ax}{e}^{jbx}+C_2{e}^{ax}{e}^{-jbx}$
$y_h\text{'}=C_1[e^{ax}(jb{e}^{jbx})+e^{jbx}(a{e}^{ax})]+C_2[e^{ax}(-jb{e}^{jbx})+e^{jbx}(a{e}^{ax})]$
$y_h{\text{'}}^{\prime }=C_1[e^{ax}((jb{)}^2{e}^{jbx})+(jb{e}^{jbx})(a{e}^{ax})+e^{jbx}((a{)}^2{e}^{ax})+(a{e}^{ax})(jb{e}^{jbx})]+C_2[e^{ax}((jb{)}^2{e}^{-jbx})+(-jb{e}^{-jbx})(a{e}^{ax})+e^{-jbx}((a{)}^2{e}^{ax})+(a{e}^{ax})(-jb{e}^{-jbx})]$

Plugging $y_h,{y_h}^{\prime },{y_h}^{″}$ into Eq. 1 and collecting terms gives
$0=C_1{e}^{ax}{e}^{jbx}[-b^2+jab+a^2+jab+(0.54)(jb+a)+0.0729+\pi ]+C_2{e}^{ax}{e}^{-jbx}[-b^2-jab+a^2-jab+(0.54)(-jb+a)+0.0729+\pi ]$
$0=C_1{e}^{-0.27x}{e}^{j1.772x}[-(1.772{)}^2+j(-0.27)(1.772)+(0.27{)}^2+j(-0.27)(1.772)+(0.54)(j(1.772)-0.27)+0.0729+\pi ]+C_2{e}^{-0.27x}{e}^{-j1.772x}[-(1.772{)}^2-j(-0.27)(1.772)+(0.27{)}^2-j(-0.27)(1.772)+(0.54)(-j(1.772)-0.27)+0.0729+\pi ]$
$0=C_1{e}^{-0.27x}{e}^{j1.772x}(0)+C_2{e}^{-0.27x}{e}^{-j1.772x}(0)$

Therefore the solution holds for any values of $C_1,C_2,x$

Problem 10

Given

Differential equation, initial conditions, and forcing function as shown:

$y^{″}-10y^{\prime }+25y=r(x),$

$(Eq.1)$

$y(0)=4,y^{\prime }(0)=-5,r(x)=7e^{(}5x)-2x^2.$

$(Eq.1)$

Find

The solution to Eq. 1

Solution

$y_p=C_1{x}^2{e}^{5x}+C_2{x}^2+C_{3}x+C_4$

$(Eq.2)$

${y_p}^{\prime }=5C_1{x}^2{e}^{5x}+2C_{1}x{e}^{5x}+2C_{2}x+C_3$

$(Eq.3)$

${y_p}^{″}=25C_1{x}^2{e}^{5x}+20C_{1}x{e}^{5x}+2C_1{e}^{5x}+2C_2$

$(Eq.4)$

Substituting Equations 2, 3, and 4 into Equation 1 yields

$7e^{5x}-2x^2=2C_1{e}^{5x}+25C_2{x}^2+25C_{3}x-20C_{2}x+2C_2-10C_3+25C_4$

$(Eq.5)$

To solve for the constants:

$7=2C_1$

$(Eq.6)$

$-2=25C_2$

$(Eq.7)$

$0=25C_3-20C_2$

$(Eq.8)$

$0=2C_2-10C_3+25C_4$

$(Eq.9)$

Solving Equations 6-9 yields

$C_1=\frac{7}{2},C_2=-\frac{2}{25},C_3=-\frac{8}{125},C_4=-\frac{12}{625}$

${\displaystyle }$

The homogeneous solution:

$y_h=C_5{e}^{5x}+C_{6}x{e}^{5x}$

$(Eq.10)$

The total solution:

$y=C_5{e}^{5x}+C_{6}x{e}^{5x}+2\frac{7}{2}{x}^2{e}^{5x}-\frac{2}{25}{x}^2-\frac{8}{125}x-\frac{12}{625}$

$(Eq.11)$

Setting $x=0$ yields

$y(0)=C_5-\frac{12}{625}=4$

$(Eq.12)$

Solving for $C_5$ yields

$C_5=\frac{2512}{625}$

$(Eq.13)$

Finding $y^{\prime }$ yields

$y^{\prime }=\frac{35}{2}{x}^2{e}^{5x}+5C_{6}x{e}^{5x}+7x{e}^{5x}+\frac{2512}{125}{e}^{5x}+C_6{e}^{5x}-\frac{4}{25}{x}^2-\frac{8}{125}$

$(Eq.14)$

Setting $x=0$ yields

$y^{\prime }(0)=\frac{2512}{125}+C_6-\frac{8}{125}=-5$

$(Eq.15)$

Solving for $C_6$ yields

$C_6=-\frac{3129}{125}$

$(Eq.16)$

Therefore, the total solution is

$y=\frac{2512}{625}{e}^{5x}-\frac{3129}{125}x{e}^{5x}+2\frac{7}{2}{x}^2{e}^{5x}-\frac{2}{25}{x}^2-\frac{8}{125}x-\frac{12}{625}$

$(ANSEq.17)$

Team Member Tasks

Name	Responsibilities	Checked by
Bo Turano	Problem R2.	--
David Parsons	Problem R2.4	--
Dean Pickett	Problem R2.8	--
Giacomo Savardi	Problem R2.	--
Isaac Kimiagarov	Problem R2.3	--
Kyle Steiner	Problem R2.	--
Tony Han	Problem R2.6, R2.7	--

All team members contributed to the coding of this page.