University of Florida/Egm4313/s12.team11.R2

Part 1

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

 ${\displaystyle y''-3y'-10=r(x)\!}$

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

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

 ${\displaystyle 0=-2c_{1}+5c_{2}\!}$

 ${\displaystyle c_{1}=5/4,c_{2}=-1/4\!}$

 ${\displaystyle y(x)=(5/4)e^{-2x}-(1/4)e^{5x}\!}$

Part 2

 ${\displaystyle 2\lambda ^{2}-6\lambda -20=0\!}$

 ${\displaystyle 3\lambda ^{2}-9\lambda -30=0\!}$

 ${\displaystyle 4\lambda ^{2}-12\lambda -40=0\!}$

Problem Statement

Find and plot the solution for the homogeneous L2-ODE-CC

${\displaystyle y''(x)-10y'(x)+25y(x)=0\!}$

with initial conditions ${\displaystyle y(0)=1\!}$ ,and ${\displaystyle y'(0)=0\!}$

Characteristic Equation

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

${\displaystyle (\lambda -5)(\lambda -5)=0\!}$

${\displaystyle \lambda =5\!}$

Homogeneous Solution

The solution to a L2-ODE-CC with real double root is given by

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

First initial condition

${\displaystyle y(0)=1\!}$

${\displaystyle y(0)=c_{1}e^{5*0}+c_{2}*0*e^{5*0}=1\!}$

${\displaystyle c_{1}=1\!}$

Second initial condition

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

${\displaystyle {\frac {d}{dx}}y(x)=y'(x)=5e^{5x}+c_{2}e^{5x}(5x+1)\!}$

${\displaystyle y'(0)=5e^{5*0}+c_{2}e^{5*0}(5*0+1)=0\!}$

${\displaystyle 5+c_{2}=0\!}$

${\displaystyle c_{2}=-5\!}$

The solution to our L2-ODE-CC is

                       ${\displaystyle y(x)=e^{5x}(1-5x)\!}$

Plot

${\displaystyle y(x)=e^{5x}(1-5x)\!}$

Egm4313.s12.team11.imponenti 00:30, 8 February 2012 (UTC)

Problem Statement (K 2011 p.59 pb. 3)

Find a general solution. Check your answer by substitution.

Given

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

Solution

We can write the above differential equation in the following form:

${\displaystyle {\frac {d^{2}y}{dx^{2}}}+6{\frac {dy}{dx}}+8.96y=0}$

Let ${\displaystyle {\frac {d}{dx}}=\lambda .}$

The characteristic equation of the given DE is

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

Now, in order to solve for ${\displaystyle \lambda }$, we can use the quadratic formula:

${\displaystyle \lambda ={\frac {-(6)\pm {\sqrt {(6)^{2}-4(1)(8.96)}}}{2(1)}}}$

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

${\displaystyle \lambda ={\frac {-6\pm {\sqrt {0.16}}}{2}}}$

${\displaystyle \lambda ={\frac {-6\pm 0.4}{2}}}$

Therefore, we have:

${\displaystyle \lambda _{1}={\frac {-6+0.4}{2}}={\frac {-5.8}{2}}=-2.8}$

and

${\displaystyle \lambda _{1}={\frac {-6-0.4}{2}}={\frac {-6.4}{2}}=-3.2}$

Thus, we have found that the general solution of the DE is actually:

${\displaystyle y=c_{1}e^{-2.8x}+c_{2}e^{-3.2x}\!}$

Check:

To check if ${\displaystyle y}$ is indeed the solution of the given DE, we can differentiate the

what we found to be the general solution.

${\displaystyle {\frac {dy}{dx}}=y'=-2.8xc_{1}e^{-2.8x}+-3.2xc_{2}e^{-3.2x}\!}$

${\displaystyle {\frac {d^{2}y}{dx^{2}}}=y''=7.84xc_{1}e^{-2.8x}+10.24c_{2}e^{-3.2x}\!}$

Substituting the values of ${\displaystyle y,y'}$ and ${\displaystyle y''}$ in the given equation, we get:

${\displaystyle 7.84c_{1}e^{-2.8x}+10.24c_{2}e^{-3.2x}+6(-2.8c_{1}e^{-2.8x}-3.2c_{2}e^{-3.2x})+8.96(c_{1}e^{-2.8x}+c_{2}e^{-3.2x})=0\!}$

${\displaystyle 7.84c_{1}e^{-2.8x}+10.24c_{2}e^{-3.2x}-16.8c_{1}e^{-2.8x}-19.2c_{2}e^{-3.2x}+8.96c_{1}e^{-2.8x}+8.96c_{2}e^{-3.2x}=0\!}$

and thus:

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

Therefore, the solution of the given DE is in fact:

${\displaystyle y=c_{1}e^{-2.8x}+c_{2}e^{-3.2x}\!}$

Problem Statement (K 2011 p.59 pb. 4)

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:

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

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:

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

Check:

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, the solution is correct.

--Gonzalo Perez

K 2011 p.59 pb. 5

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

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

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

K 2011 p.59 pb. 6

                                                       ${\displaystyle y=(c_{1}+c_{2}x)e^{1.6x}\!}$

                                                  ${\displaystyle {y}'=1.6c_{1}e^{1.6x}+c_{2}e^{1.6x}+1.6c_{2}xe^{1.6x}\!}$

                                                 ${\displaystyle {y}''=(1.6)^{2}c_{1}e^{1.6x}+3.2c_{2}e^{1.6x}+(1.6)^{2}c_{2}xe^{1.6x}\!}$

Problem Statement

Problem 2.5 Find an Ordinary Differential Equation. Use ${\displaystyle y''+ay'+by=0\!}$ for the given basis.

ODE Solutions

Two Real Roots: ${\displaystyle y(x)=c_{1}e^{\lambda _{1}x}+c_{2}e^{\lambda _{2}x}\!}$
Double Real Roots: ${\displaystyle y(x)=c_{1}e^{\lambda _{1}x}+c_{2}xe^{\lambda _{2}x}\!}$

Reverse Engineering

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

Case 1: ${\displaystyle a^{2}-4b>0\!}$
${\displaystyle \lambda ^{2}-(\lambda _{1}+\lambda _{2})\lambda +\lambda _{1}\lambda _{2}\!}$
${\displaystyle y''-(\lambda _{1}+\lambda _{2})y'+\lambda _{1}\lambda _{2}y=0\!}$

Case 2: ${\displaystyle a^{2}-4b=0\!}$
${\displaystyle \lambda ^{2}-(\lambda _{1}+\lambda _{2})\lambda +\lambda _{1}\lambda _{2}\!}$
${\displaystyle y''-(\lambda _{1}+\lambda _{2})y'+\lambda _{1}\lambda _{2}y=0\!}$

Problem Solutions

                                                            ${\displaystyle y''+1.7y'-11.18=0\!}$

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

Problem Statement

For the following spring-dashpot-mass system (in series) find the values for the parameters ${\displaystyle k,c,m\!}$ knowing that the system has the double real root ${\displaystyle \lambda =-3\!}$

Solution

Previously, we have derived the following equation for such a system:
(From Sec 1 (d), (3) p.1-5)
${\displaystyle m\left(y_{k}''+{\frac {k}{c}}y_{k}'\right)+ky_{k}=f(t)\!}$
We can write this equation in standard form by diving through by ${\displaystyle m\!}$:
${\displaystyle y_{k}''+{\frac {k}{c}}y_{k}'+{\frac {k}{m}}y_{k}={\frac {f(t)}{m}}\!}$
Here, we can take the coefficients of ${\displaystyle y_{k}'\!}$ and ${\displaystyle y_{k}\!}$ as ${\displaystyle a\!}$ and ${\displaystyle b\!}$:

${\displaystyle y_{k}''+\underbrace {\frac {k}{c}} _{a}y_{k}'+\underbrace {\frac {k}{m}} _{b}y_{k}={\frac {f(t)}{m}}\!}$

Next,considering the double real root:
${\displaystyle \lambda =-3\!}$
We can find the characteristic equation to be:
${\displaystyle \left(\lambda +3\right)^{2}=\lambda ^{2}+6\lambda +9=0\!}$
Which is in the form:
${\displaystyle \lambda ^{2}+a\lambda +b=0\!}$

Then, we know that ${\displaystyle a=6\!}$ and ${\displaystyle b=9\!}$:

Setting ${\displaystyle a\!}$ and ${\displaystyle b\!}$ from the first equation equal to these, we obtain:

                                                             ${\displaystyle {\frac {k}{c}}=6\;and\;{\frac {k}{m}}=9\!}$

Clearly, there is an infinite amount of solutions to this problem because we have 2 equations but 3 unknowns. This can be solved by fixing one of the values and finding the other two.

                                                             ${\displaystyle m=10,\;k=90,\;and\;c=15\!}$

Problem Statement

Develop the McLaurin Series (Taylor Series at t=0) for ${\displaystyle e^{t},\cos t,\sin t\!}$

Solution

${\displaystyle e^{t}=\sum _{n=0}^{\infty }{\frac {t^{n}}{n!}}\!}$

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

                                      ${\displaystyle e^{t}=1+t+{\frac {t^{2}}{4}}+{\frac {t^{3}}{6}}+...{\frac {t^{n}}{n!}}\!}$

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

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

                                      ${\displaystyle \cos t=1-{\frac {t^{2}}{2}}+{\frac {t^{4}}{24}}+{\frac {t^{6}}{720}}...{\frac {(-1)t^{2k}}{(2k)!}}\!}$

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

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

                                      ${\displaystyle \sin t=t-{\frac {t^{3}}{6}}+{\frac {t^{5}}{120}}-{\frac {t^{7}}{5040}}+...{\frac {(-1)^{k}t^{2k+1}}{(2k+1)!}}\!}$

--Egm4313.s12.team11.arrieta 17:07, 6 February 2012 (UTC)

Problem Statement

Find a general solution. Check your answer by substitution.

Problem 8

${\displaystyle y''+y'+3.25y=0\!}$

Let:
${\displaystyle \lambda =d/dx\!}$

${\displaystyle y_{h}(x)=e^{-{\frac {1}{2}}x}(c_{1}\cos(x{\sqrt {3}})+c_{2}\sin(x{\sqrt {3}}))\!}$

 ${\displaystyle y''+y'+3.25y=0\!}$

Problem 15

${\displaystyle y''+0.54y'+(0.0729+\pi )y=0\!}$
Let: ${\displaystyle {\frac {d}{dx}}=\lambda \!}$

${\displaystyle y_{h}(x)=e^{-0.27x}(c_{1}\cos(x{\sqrt {\pi }})+c_{2}\sin(x{\sqrt {\pi }})\!}$

 ${\displaystyle y''+0.54y'+(0.0729+\pi )y=0\!}$

Problem Statement

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

${\displaystyle \lambda ^{2}+4\lambda +13\!}$

with ${\displaystyle r(x)=0\!}$

and initial conditions ${\displaystyle y(0)=1\!}$, ${\displaystyle y'(0)=0\!}$

In another figure, superimpose 3 figs.:(a)this fig. (b) the fig. in R2.6 p.5-6, and (c) the fig. in R2.1 p.3-7

Quadratic Equation

${\displaystyle \lambda ={\frac {-b\pm {\sqrt {(}}b^{2}-4ac)}{2a}}\!}$ with ${\displaystyle a=1,b=4,c=13\!}$

${\displaystyle \lambda ={\frac {-4\pm {\sqrt {(}}4^{2}-4*1*13)}{2*1}}=-2\pm 3i\!}$

${\displaystyle \lambda =-2\pm 3i\!}$

Homogeneous Solution

The solution to a L2-ODE-CC with two complex roots is given by

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

where ${\displaystyle \lambda =-{\frac {a}{2}}\pm \omega i=-2\pm 3i\!}$

${\displaystyle y(x)=e^{-2x}[Acos(3x)+Bsin(3x)]\!}$

Solving for A and B

first initial condition ${\displaystyle y(0)=1\!}$

${\displaystyle y(x)=e^{-2x}[Acos(3x)+Bsin(3x)]\!}$

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

${\displaystyle A=1\!}$

second initial condition ${\displaystyle y'(0)=0\!}$