University of Florida/Egm4313/s12.team15.r2

<Team 15

REPORT 2

Given
$$\lambda _{1}=-2\!$$

$$\lambda _{2}=+5\!$$

$$y(0)=1\!$$

$$y{}'(0)=0\!$$

$$r(x)=0\!$$

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: $$y{}''+ay{}'+by=r(x)=0\!$$

$$y{}''-3y{}'-10y=0\!$$

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, $$y_{p}{}'(x)=0\!$$

$$y(x)=c_{1}e^{-2x}+c_{2}e^{5x}\!$$

$$y{}'(x)=-2c_{1}e^{-2x}+5c_{2}e^{5x}\!$$

If $$y(0)=1\!$$

then $$1=c_{1}e^{-2(0)}+c_{2}e^{5(0)}\!$$

$$1=c_{1}+c_{2}\!$$

If $$y{}'(0)=0\!$$

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 -

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) $$ y{}''+6{}y'+8.96y=0\!$$

3b) $$ y'' + 4y' + (\pi^2 + 4)y = 0\!$$

Find
Find the general solution of the ODEs.

Solution
3a)

$$ y{}''+6{}y'+8.96y=0\!$$

$$ a=6\!$$

$$ b=8.96\!$$

$$\lambda ^2+a\lambda +b=0 \! $$

$$\lambda ^2+6\lambda +8.96=0 \! $$

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: $$ y(x)= C_1e^{-2.8x}+C_2e^{-3.2x} \! $$

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

$$ {y}' = -3.2C_{1}e^{-3.2x} - 2.8C_{2}e^{-2.8x} \!$$

$$ {y}'' = 10.24C_{1}e^{-3.2x} + 7.84C_{2}e^{-2.8x} \!$$

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

$$ y{}''+6{}y'+8.96y=0\!$$ (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)

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

$$ a=4\!$$

$$ b=(\pi^2 + 4)\!$$

$$\lambda ^2+a\lambda +b=0 \! $$

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

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:

$$ {y}' = e^{-2x}(-C_1\pi sin(\pi x)+C_2\pi cos(\pi x))-2e^{-2x}(C_1cos(\pi x)+C_2sin(\pi x)) \!$$

$$ {y}''=e^{-2x}(-C_1\pi ^{2}cos(\pi x)-C_2\pi ^{2}sin(\pi x)+2C_1\pi sin(\pi x)-2C_2\pi sin(\pi x))-2e^{-2x}(-C_1\pi sin(\pi x)+C_2\pi cos(\pi x)-2C_1cos(\pi x)-2C_2sin(\pi x)) \!$$

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

$$ y'' + 4y' + (\pi^2 + 4)y = 0\!$$ (Original Equation)

$$ e^{-2x}(-C_1\pi ^{2}cos(\pi x)-C_2\pi ^{2}sin(\pi x)+2C_1\pi sin(\pi x)-2C_2\pi sin(\pi x))-2e^{-2x}(-C_1\pi sin(\pi x)+C_2\pi cos(\pi x)-2C_1cos(\pi x)-2C_2sin(\pi x)) \!$$

$$ + 4(e^{-2x}(-C_1\pi sin(\pi x)+C_2\pi cos(\pi x))-2e^{-2x}(C_1cos(\pi x)+C_2sin(\pi x))) \!$$

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

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

4b) $$ 10y{}''-32{}y'+25.6{}y=0\!$$

Find
Find the general solution. Check answer by substitution.

Solution
4a)

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

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:

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

$$y_2=uy_1\!$$

$$ u=c_1x+c_2\!$$         This is due to double integration.

So, $$y_2=(c_1x+c_2)y_1\!$$

Choose $$ c_1= 1 c_2=0\!$$

$$y_2= (x+0)y_1\!$$

$$y_2=xy_1\!$$

$$ y_2=xe^{-\pi{x}}\!$$

$$ y=(c_1+c_2x)e^{-\pi{x}}\!$$ General Solution

Check by substitution

$$ y=(c_1+c_2x)e^{-\pi{x}}\!$$

$$ y'=-\pi{}c_1e^{-\pi{x}}+c_2e^{-\pi{x}}-\pi{}c_2e^{-\pi{x}}\!$$

$$ y''=\pi^{2}c_1e^{-\pi{x}}-\pi{}c_2e^{-\pi{x}}+\pi^{2}c_2xe^{-\pi{x}}-\pi{}c_2e^{-\pi{x}}\!$$

$$\pi^{2}c_1e^{-\pi{x}}-\pi{}c_2e^{-\pi{x}}+\pi^{2}c_2xe^{-\pi{x}}-\pi{}c_2e^{-\pi{x}}+ 2\pi{}(-\pi{}c_1e^{-\pi{x}}+c_2e^{-\pi{x}}-\pi{}c_2e^{-\pi{x}})+ \pi^{2}(e^{-\pi{x}}c_1+e^{-\pi{x}}xc_2)=0 \!$$

Since all terms cancel out then by substitution the general solution $$ y=(c_1+c_2x)e^{-\pi{x}}\!$$ does hold true.

4b) $$ 10y{}''-32{}y'+25.6{}y=0\!$$

Divide by 10 to simplify equation :

$$ y{}''-3.2{}y'+2.56{}y=0\!$$

$$ a=-3.2\!$$

$$ b=2.56\!$$

$$ \lambda ^2+a\lambda +b=0 \! $$

$$ \lambda ^2-3.2\lambda +2.56=0 \! $$

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

$$ y_1=e^{1.6x}\!$$

$$y_2=uy_1\!$$

$$ u=c_1x+c_2\!$$  This is due to double integration

So

$$y_2=(c_1x+c_2)y_1\!$$

Choose $$ c_1= 1 c_2=0\!$$

$$y_2= (x+0)y_1\!$$

$$y_2=xy_1\!$$

$$ y_2=xe^{1.6x}\!$$

$$ y=(c_1+c_2x)e^{1.6x}\!$$  General Solution

 Check by Substitution

$$ y''=2.56c_1e^{1.6x}+1.6c_2e^{1.6x}+1.6c_2xe^{1.6x}+2.56c_2xe^{1.6x}\!$$

$$y'=1.6c_1e^{1.6x}+c_2e^{1.6x}+ 1.6c_2xe^{1.6x} \!$$

$$y=c_1e^{1.6x}+c_2xe^{1.6x} \!$$

$$10(2.56c_1e^{1.6x}+1.6c_2e^{1.6x}+1.6c_2xe^{1.6x}+2.56c_2xe^{1.6x})-32(1.6c_1e^{1.6x}+c_2e^{1.6x}+ 1.6c_2xe^{1.6x})+25.6(c_1e^{1.6x}+c_2xe^{1.6x})=0\!$$

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

$$ y=(c_1+c_2x)e^{1.6x}\!$$ does hold true.

Author
Solved and typed by - Cynthia Hernandez Reviewed By -

Given
Problem 16, P 59 Kreyszig:

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

Problem 17, P 59 Kreyszig:

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

Find
Problem 16, P 59 Kreyszig:

ODE in the form:

$${y}''+a{y}'+by=0\!$$

Problem 17, P 59 Kreyszig:

ODE in the form:

$${y}''+a{y}'+by=0\!$$

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:

$$y''+1.7y'-11.18y=0\!$$

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:

$$y''+2\sqrt{5}y'+5y=0\!$$

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

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



Has double real root $$\lambda=-3\!$$

Find
$$k,c,m\!$$

Solution
For spring-mass-dashpot system:

$$m(y_k''+\frac{k}{c} y_k')+ky_k=f(t)\!$$

This simplifies to:

$$y_k''+\frac{k}{mc} y_k'+\frac{k}{m}y_k=f(t)\!$$

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 -

Given
Develop the MacLaurin Series (Taylor Series at $$ t = 0\! $$) for $$ e^t,\cos t,\sin t\! $$ $$ L[y] and  L[w]  then L[y+w] \! $$ $$ L[y+w] = L[y]+ L[w] \! $$ $$ L[cy] = cL[y] $$ $$ L[kw] = kL[w] \! $$

Author
Solved and typed by - Jenny Schulze Reviewed By -

Given
Find a general solution to the equations, and check answers by substitution. Problem 8

$$ y''+y'+3.25y=0 \! $$  (8.0)

Problem 15

$$ y''+0.54y'+(0.0729+\pi )y=0 \!$$  (8.1)

Find
Factor as in the text and solve.

Author
Solved and typed by - Jenny Schulze Reviewed By -

Given
Initial Conditions:

$$ y(0)=1, y'(0)=0\ $$

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:

$$ y''+4y'+13y=r(x) \!$$

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:

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

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

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

$$ A=1 \!$$

$$ y'(t)=e^{-2t}((-2A+3B) cos(3t)-(3A+2B)sin(3t)) \!$$

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

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

thus:

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

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



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



Equation 1 (R2.9):

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

Equation 2 (R2.6):

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

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 -