University of Florida/Egm4313/s12.team15.r2

<Team 15

REPORT 2

R2.1

Given

${\lambda }_1=-2$

${\lambda }_2=+5$

$y(0)=1$

$y{\prime }^{}(0)=0$

$r(x)=0$

Find

Find the non-homogeneous linear 2nd order ordinary differential equation with constant coefficients in standard form and the solution in terms of the initial conditions and the general excitation $r(x)$.

Then plot the solution.

Solution

Some of the following standard equations were referenced from the textbook Advanced Engineering Mechanics 10th Ed. Kreyszig 2011.

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

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

${\lambda }^2-\lambda {\lambda }_2-\lambda {\lambda }_1+{\lambda }_1{\lambda }_2=0$

${\lambda }^2-5\lambda +2\lambda -10=0$

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

Therfore,
$a=-3$

$b=-10$

Standard Form:
$y{″}^{}+ay{\prime }^{}+by=r(x)=0$

$y{″}^{}-3y{\prime }^{}-10y=0$

Standard Form:
$y(x)=y_h(x)+y_p(x)$

$y_h(x)=c_1{e}^{-2x}+c_2{e}^{5x}$

$y_p(x)=0$

Therefore,
$y_p{\prime }^{}(x)=0$

$y(x)=c_1{e}^{-2x}+c_2{e}^{5x}$

$y{\prime }^{}(x)=-2c_1{e}^{-2x}+5c_2{e}^{5x}$

If
$y(0)=1$

then
$1=c_1{e}^{-2(0)}+c_2{e}^{5(0)}$

$1=c_1+c_2$

If
$y{\prime }^{}(0)=0$

then
$0=c_1{e}^{-2(0)}+c_2{e}^{5(0)}$

$0=c_1+c_2$

$2c_1=5c_2$

$c_1=\frac{5}{2}{c}_2$

$1=\frac{5}{2}{c}_2+c_2$

$1=\frac{7}{2}{c}_2$

$c_2=\frac{2}{7}$

$1=c_1+c_2$

$c_1=1-c_2=1-\frac{2}{7}=\frac{5}{7}$

Final Solution:

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

Plot of Solution:

Author

Solved and typed by Kristin Howe
Reviewed By -

R2.2

Author

Solved and typed by -
Reviewed By -

R2.3

Problems referenced from Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.59 problems.3-4

Given

3a) $y{″}^{}+6y^{\prime }+8.96y=0$

3b) $y^{″}+4y^{\prime }+({\pi }^2+4)y=0$

Find

Find the general solution of the ODEs.

Solution

3a)

$y{″}^{}+6y^{\prime }+8.96y=0$

$a=6$

$b=8.96$

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

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

Using the quadratic formula:

${\lambda }_{1/2}=\frac{-b\pm \sqrt{b^2-4ac}}{2}$ where ${\lambda }_{1/2},$ are the 2 different roots generated using quadratic formula.

Plug-in and solve using known variables (b=6, a=1, c=8.96)

${\lambda }_{1/2}=\frac{-6\pm \sqrt{0.16}}{2}$ $=\frac{-6\pm 0.4}{2}$ $=-3\pm 0.2$

Therefore,

${\lambda }_1=-2.8$

${\lambda }_2=-3.2$

The General Solution formula for Distinct Real roots is
$y(x)=C_1{e}^{{\lambda }_{1}x}+C_2{e}^{{\lambda }_{2}x}$

Now substitute the known roots into the General Solution formula:

  $y(x)=C_1{e}^{-2.8x}+C_2{e}^{-3.2x}$

Check using substitution. First, find the 1st and 2nd derivative of general solution formula:

$y\text{'}=-3.2C_1{e}^{-3.2x}-2.8C_2{e}^{-2.8x}$

$y{\text{'}}^{\prime }=10.24C_1{e}^{-3.2x}+7.84C_2{e}^{-2.8x}$

Then, plug in the known variables into the original equation:

$y{″}^{}+6y^{\prime }+8.96y=0$ (Original Equation)

$10.24c_1{e}^{-3.2x}+7.84c_2{e}^{-2.8x}-19.2c_1{e}^{-3.2x}-16.8c_2{e}^{-2.8x}+8.96c_1{e}^{-3.2x}+8.96c_2{e}^{-2.8x}=0$

By inspection, all of the terms on the left side of the equation cancel out to 0, making the expression correct. Therefore, the general solution is correct.

3b)

$y^{″}+4y^{\prime }+({\pi }^2+4)y=0$

$a=4$

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

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

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

Using the quadratic formula:

${\lambda }_{1/2}=\frac{-b\pm \sqrt{b^2-4ac}}{2}$ where ${\lambda }_{1/2},$ are the 2 different roots generated using quadratic formula.

Plug-in and solve using known variables (b=4, a=1, c=[${\pi }^2$ + 4])

${\lambda }_{1/2}=\frac{-4\pm \sqrt{-4{\pi }^2}}{2}$ $=\frac{-4\pm 2\pi i}{2}$ $=-2\pm \pi i$

Therefore,

${\lambda }_1=-2+\pi i$

${\lambda }_2=-2-\pi i$

The General Solution formula for Complex Conjugate roots is
$y(x)=e^{-\frac{\alpha x}{2}}(C_1\cos \beta x+C_2\sin \beta x)$

Now substitute the known roots into the General Solution formula:

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

Check using substitution. First, find the 1st and 2nd derivative of general solution formula:

$y\text{'}=e^{-2x}(-C_1\pi sin(\pi x)+C_2\pi cos(\pi x))-2e^{-2x}(C_{1}cos(\pi x)+C_{2}sin(\pi x))$

$y{\text{'}}^{\prime }=e^{-2x}(-C_1{\pi }^{2}cos(\pi x)-C_2{\pi }^{2}sin(\pi x)+2C_1\pi sin(\pi x)-2C_2\pi sin(\pi x))-2e^{-2x}(-C_1\pi sin(\pi x)+C_2\pi cos(\pi x)-2C_{1}cos(\pi x)-2C_{2}sin(\pi x))$

Then, plug in the known variables into the original equation:

$y^{″}+4y^{\prime }+({\pi }^2+4)y=0$ (Original Equation)

$e^{-2x}(-C_1{\pi }^{2}cos(\pi x)-C_2{\pi }^{2}sin(\pi x)+2C_1\pi sin(\pi x)-2C_2\pi sin(\pi x))-2e^{-2x}(-C_1\pi sin(\pi x)+C_2\pi cos(\pi x)-2C_{1}cos(\pi x)-2C_{2}sin(\pi x))$

$+4(e^{-2x}(-C_1\pi sin(\pi x)+C_2\pi cos(\pi x))-2e^{-2x}(C_{1}cos(\pi x)+C_{2}sin(\pi x)))$

$+({\pi }^2+4)(e^{-2x}(C_1\cos \pi x+C_2\sin \pi x))=0$

By inspection, all of the terms on the left side of the equation cancel out to 0, making the expression correct. Therefore, the general solution is correct.

Author

Solved and typed by - Tim Pham 19:36, 07 February 2012 (UTC)
Reviewed By -

R2.4

The following problems are referenced from Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.59.

For additional information and/or practice reference Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.53-59 and the lecture notes Lecture 5.

• Note the lecture notes and Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 textbook were used to solve the following.

Given

4a) $y{″}^{}+2\pi y^{\prime }+{\pi }^{2}y=0$

4b) $10y{″}^{}-32y^{\prime }+25.6y=0$

Find

Find the general solution. Check answer by substitution.

Solution

4a)

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

$a=2\pi$

$b={\pi }^2$

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

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

Finding the roots:

${\lambda }_1=\frac{1}{2}(-2\pi +\sqrt{(-2\pi {)}^2-4({\pi }^2)})=-\pi$

${\lambda }_2=\frac{1}{2}(-2\pi -\sqrt{(-2\pi {)}^2-4({\pi }^2)})=-\pi$

Both solutions above show the double root characteristic so the following is applied:

$y_1=e^{-\pi x}$

$y_2=u{y}_1$

$u=c_{1}x+c_2$ This is due to double integration.

So,

$y_2=(c_{1}x+c_2)y_1$

Choose $c_1=1c_2=0$

$y_2=(x+0)y_1$

$y_2=x{y}_1$

$y_2=x{e}^{-\pi x}$

$y=(c_1+c_{2}x)e^{-\pi x}$ General Solution

Check by substitution

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

$y^{\prime }=-\pi c_1{e}^{-\pi x}+c_2{e}^{-\pi x}-\pi c_2{e}^{-\pi x}$

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

${\pi }^2{c}_1{e}^{-\pi x}-\pi c_2{e}^{-\pi x}+{\pi }^2{c}_{2}x{e}^{-\pi x}-\pi c_2{e}^{-\pi x}+2\pi (-\pi c_1{e}^{-\pi x}+c_2{e}^{-\pi x}-\pi c_2{e}^{-\pi x})+{\pi }^2(e^{-\pi x}{c}_1+e^{-\pi x}x{c}_2)=0$

Since all terms cancel out then by substitution the general solution $y=(c_1+c_{2}x)e^{-\pi x}$ does hold true.

4b) $10y{″}^{}-32y^{\prime }+25.6y=0$

Divide by 10 to simplify equation :

$y{″}^{}-3.2y^{\prime }+2.56y=0$

$a=-3.2$

$b=2.56$

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

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

Solving for the roots:

${\lambda }_1=\frac{1}{2}(-(-3.2)+\sqrt{(-3.2{)}^2-4(2.56)})=1.6$

${\lambda }_2=\frac{1}{2}(-(-3.2)-\sqrt{(-3.2{)}^2-4(2.56)})=1.6$

${\lambda }_1={\lambda }_2$ This shows the double root characteristic.

So, the following is applied

$y_1=e^{1.6x}$

$y_2=u{y}_1$

$u=c_{1}x+c_2$ This is due to double integration

So

$y_2=(c_{1}x+c_2)y_1$

Choose $c_1=1c_2=0$

$y_2=(x+0)y_1$

$y_2=x{y}_1$

$y_2=x{e}^{1.6x}$

$y=(c_1+c_{2}x)e^{1.6x}$ General Solution

Check by Substitution

$y^{″}=2.56c_1{e}^{1.6x}+1.6c_2{e}^{1.6x}+1.6c_{2}x{e}^{1.6x}+2.56c_{2}x{e}^{1.6x}$

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

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

$10(2.56c_1{e}^{1.6x}+1.6c_2{e}^{1.6x}+1.6c_{2}x{e}^{1.6x}+2.56c_{2}x{e}^{1.6x})-32(1.6c_1{e}^{1.6x}+c_2{e}^{1.6x}+1.6c_{2}x{e}^{1.6x})+25.6(c_1{e}^{1.6x}+c_{2}x{e}^{1.6x})=0$

Since all terms cancel out then through substitution it can be confirmed that the general solution

$y=(c_1+c_{2}x)e^{1.6x}$ does hold true.

Author

Solved and typed by - Cynthia Hernandez
Reviewed By -

R2.5

Given

Problem 16, P 59 Kreyszig:

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

Problem 17, P 59 Kreyszig:

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

Find

Problem 16, P 59 Kreyszig:

ODE in the form:

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

Problem 17, P 59 Kreyszig:

ODE in the form:

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

Solution

Problem 16, P 59 Kreyszig:

For two distinct real-roots:

$y=c_1{e}^{{\lambda }_{1}x}+c_2{e}^{{\lambda }_{2}x}$

Where:

${\lambda }_1=2.6$ and ${\lambda }_2=-4.3$

The characteristic equation is:

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

Multiplying the roots of the equation together:

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

Therefore the ODE is:

$y^{″}+1.7y^{\prime }-11.18y=0$

Problem 17, P 59 Kreyszig:

For real double root:

$y=(c_1+c_{2}x)e^{\lambda x}$

Where:

$\lambda =-\sqrt{5}$

The characteristic equation is:

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

Multiplying the roots of the equation together:

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

Therefore the ODE is:

$y^{″}+2\sqrt{5}{y}^{\prime }+5y=0$

Author

Solved and typed by - Neil Tidwell 1:55, 06 February 2012 (UTC)
Reviewed By -

R2.6

Given

Spring-Dashpot-Mass System from Sec 1 lecture notes.

File:R1a3a1.jpg

Has double real root $\lambda =-3$

Find

$k,c,m$

Solution

For spring-mass-dashpot system:

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

This simplifies to:

${y_k}^{″}+\frac{k}{mc}{y_k}^{\prime }+\frac{k}{m}{y}_k=f(t)$

Multiply roots together to get characteristic equation:

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

Set coefficients of simplified spring-mass-dashpot equation equal to those of characteristic equation:

$\frac{k}{mc}=6$

$\frac{k}{m}=9$

Have one degree of freedom, so arbitrarily choose $m=1$

Solve system:

$m=1$
$k=9$
$c=1.5$

Author

Solved and typed by - Neil Tidwell 2:14, 06 February 2012 (UTC)
Reviewed By -

R2.7

Given

Develop the MacLaurin Series (Taylor Series at $t=0$) for $e^t,\cos t,\sin t$

$L[y]andL[w]thenL[y+w]$

$L[y+w]=L[y]+L[w]$

$L[cy]=cL[y]$

$L[kw]=kL[w]$

Author

Solved and typed by - Jenny Schulze
Reviewed By -

R2.8

Given

Find a general solution to the equations, and check answers by substitution.

Problem 8

$y^{″}+y^{\prime }+3.25y=0$

(8.0)

Problem 15

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

(8.1)

Find

Factor as in the text and solve.

Author

Solved and typed by - Jenny Schulze
Reviewed By -

R2.9

Given

Initial Conditions:

$y(0)=1,y^{\prime }(0)=0$

No excitation:

$r(x)=0$

Find

Find and plot the solution for the L2-ODE-CC corresponding to:

${\lambda }^2+4\lambda +13=0$

In another Fig., superimpose 3 Figs.: (a) this Fig., (b) the Fig. in R2.6 p.5-6, (c) the Fig. in R2.1 p.3-7.

Solution

The corresponding ODE in standard form:

$y^{″}+4y^{\prime }+13y=r(x)$

To find the roots of the ODE, convert ODE to this form:

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

which in this case is:

${\lambda }^2+4\lambda +13=0$

use the formula to find the roots:

${\lambda }_{1,2}=\frac{-b\pm i\sqrt{b^2-4ac}}{2a}=-\alpha \pm i\omega$

${\lambda }_{1,2}=\frac{-4\pm i\sqrt{4^2-4*1*13}}{2*1}=-2\pm 3i$

because the roots are complex conjugates, the general solution can be given as:

$y(t)=e^{-\alpha t}(Acos\omega t+bsin\omega t)=e^{-2t}(Acos3t+Bsin3t)$

to find A and B, we use the initial conditions and take the first derivative of y(t):

$y(0)=1=e^{-2*0}(Acos(3*0)+Bsin(3*0)=1*A=A$

$A=1$

$y^{\prime }(t)=e^{-2t}((-2A+3B)cos(3t)-(3A+2B)sin(3t))$

$y^{\prime }(0)=0=e^{-2*0}((-2*1+3B)cos(3*0)-(3*1+2B)sin(3*0))=1*(-2+3B)=-2+3B$

$B=\frac{2}{3}$

thus:

$y(t)=e^{-2t}(cos3t+\frac{2}{3}sin3t)$

The graph of the solved equation is shown below in Fig. 1.

Fig. 1

The three graphed equations are shown together below in Fig. 2

Equation 1 (R2.9):

$y_1(t)=e^{-2t}(cos3t+\frac{2}{3}sin3t)$

Equation 2 (R2.6):

$y_2(t)=e^{-3t}+3t{e}^{-3t}$

Equation 3 (R2.1):

$y_3(t)=\frac{5}{7}{e}^{-2t}+\frac{2}{7}{e}^{5t}$

Author

Solved and typed by - --James Moncrief 01:42, 8 February 2012 (UTC)
Reviewed By -

Contributing Members

Template:Center topTeam Contribution TableTemplate:Center bottom
Problem Number	Lecture	Assigned To	Solved By	Typed By	Proofread By
2.1	Lecture 3	Kristin Howe	Kristin Howe	Kristin Howe	Name
2.2	Lecture 5	Jerry Shugart	Jerry Shugart	Jerry Shugart	Name
2.3	Lecture 5	Tim Pham	Tim Pham	Tim Pham	Name
2.4	Lecture 5	Cynthia Hernandez	Cynthia Hernandez	Cynthia Hernandez	Name
2.5	Lecture 5	Neil Tidwell	Neil Tidwell	Neil Tidwell	Name
2.6	Lecture 5	Neil Tidwell	Neil Tidwell	Neil Tidwell	Name
2.7	[Lecture link]	Jenny Schulze	Jenny Schulze	Jenny Schulze	Name
2.8	[Lecture link]	Jenny Schulze	Jenny Schulze	Jenny Schulze	Name
2.9	Lecture 6	James Moncrief	James Moncrief	James Moncrief	Name