text:
User:Egm4313.s12.team4.friedman/R2

=Problem 2.2=

Problem Statement
Find and plot the solution for L2-ODE-CC (2.2.1) with the given initial conditions (2.2.2, 2.2.3) and excitation function (2.2.4).

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$$\displaystyle y'' - 10y' + 25y = r(x)$$
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 * (2.2.1)
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Initial conditions:
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$$\displaystyle y(0) = 1$$
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 * (2.2.2)
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$$\displaystyle y'(0) = 0$$
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 * (2.2.3)
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No excitation:
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$$\displaystyle r(x) = 0$$
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 * (2.2.4)
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Solution
Using the following:

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$$\displaystyle {y}'=\frac{dy}{dx}$$
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 * (2.2.5)
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$$\displaystyle {y}''=\frac{d^{2}y}{dx^{2}}$$
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 * (2.2.6)
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The given equation can be rewritten as follows, using (2.2.5) and (2.2.6):
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$$\displaystyle \frac{d^{2}y}{dx^{2}}-10\frac{dy}{dx}+25y=0$$
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 * (2.2.7)
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Substituting
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$$\displaystyle D= \frac{d}{dx}$$
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 * (2.2.8)
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The ODE can be expressed as follows:
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$$\displaystyle (D^{2}-10 D+25)y=0$$
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 * (2.2.9)
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Solving
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$$\displaystyle D^{2}-10 D+25=0$$
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 * (2.2.10)
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$$\displaystyle (D-5)^{2}=0$$
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 * (2.2.11)
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$$\displaystyle D=5,5$$
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 * (2.2.12)
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The general solution for this problem is :

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$$\displaystyle y=(c_1 + c_2 x) e^{5 x}$$
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 * (2.2.13)
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Use initial condition (2.2.2) to find the particular solution,
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$$\displaystyle y(0) = 1 = (c_1 + c_2 (0)) e^0$$
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 * (2.2.14)
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$$\displaystyle c_1 = 1$$
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 * (2.2.15)
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Find the derivative to apply the second initial condition (2.2.3).
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$$\displaystyle y' = 5 c_1 e^{5 x} + 5 c_2 x e^{5 x} + c_2 e^{5 x}$$
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 * (2.2.16)
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$$\displaystyle y'(0) = 0 = 5 (1) e^0 + 5 c_2 (0) e^0 + c_2 e^0$$
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(2.2.17)
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$$\displaystyle 0 = 5 + c_2$$
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(2.2.18)
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$$\displaystyle c_2 = -5$$
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(2.2.19)
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The solution to the initial value problem:
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(2.2.20)
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Plot of the solution:
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(2.2.21)
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=Problem 2.9=

Problem Statement
Find and plot the solution for the L2-ODE-CC corresponding to (2.9.1).

In another figure, superpose 3 figures: (a) this figure, (b) the solution to Problem 1, (c) the solution to Problem 6.

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$$\displaystyle \lambda^2 + 4 \lambda + 13 = 0$$
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(2.9.1)
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Initial conditions:
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$$\displaystyle y(0) = 1$$
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(2.9.2)
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$$\displaystyle y'(0) = 0$$
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(2.9.3)
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No excitation:
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$$\displaystyle r(x) = 0$$
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(2.9.4)
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Solution to R2.1:
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$$\displaystyle y = \frac{5}{7} e^{-2 x} + \frac{2}{7} e^{5 x}$$
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(2.9.a)
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Solution to R2.6:
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$$\displaystyle y = e^{-3 x} (c_1 + c_2 x)$$
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(2.9.b)
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Solution
Equation (2.9.1) corresponds to the following differential equation:

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$$\displaystyle y'' + 4 y' + 13 = 0$$
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(2.9.5)
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To find the solution, solve for $$\displaystyle \lambda$$ using the quadratic formula:
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$$\displaystyle \lambda = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a}$$
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(2.9.6)
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$$\displaystyle \lambda = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot13}}{2 \cdot 1}$$
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(2.9.7)
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$$\displaystyle \lambda = -2 \pm 3i$$
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(2.9.8)
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This leads to a general solution of

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$$\displaystyle y = e^{-2 x} (c_1 \cos 3 x + c_2 \sin 3 x)$$
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(2.9.9)
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Use initial condition (2.9.2) to find the particular solution,

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$$\displaystyle y(0) = 1 = e^0 (c_1 \cos 0 + c_2 \sin 0)$$
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(2.9.10)
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$$\displaystyle c_1 = 1$$
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(2.9.11)
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Find the derivative to apply the second initial condition (2.9.3).

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$$\displaystyle y' = e^{-2 x} (c_2 \cos 3 x - c_1 \sin 3 x) - 2 e^{-2x} (c_1 \cos 3 x + c_2 \sin 3 x)$$
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(2.9.12)
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$$\displaystyle y'(0) = 0 = e^0 (c_2 \cos 0 - \sin 0) - 2 e^0 (\cos 0 + c_2 \sin 0)$$
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(2.9.13)
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$$\displaystyle 0 = c_2 - 2$$
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(2.9.14)
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$$\displaystyle c_2 = 2$$
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(2.9.15)
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The solution to the initial value problem:

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(2.9.16)
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Plot of the solution (2.9.16):
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(2.9.17)
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Plot of the solution (2.9.16), plot of (2.9.a), and plot of (2.9.b).
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 *
(2.9.18)
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title: User:Egm4313.s12.team4.friedman/R2


text: File:Pea2.s12.sec43.b (1).djvu

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: File:Pea2.s12.sec43.b (1).djvu


text:
User:Egm4313.s12.team11.sheider/R2.4: K 2011 p.59 pb. 6

Problem Statement
Find a general solution to the given ODE. Check your answer by substituting into the original equation.

Given
$$10{y}''-32{y}'+25.6y = 0\!$$

Solution
Initially we modify the original ODE to put it in the form of a second-order homogenous linear ODE with constant coefficients:

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

Dividing both sides by 10:

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

The characteristic equation of this ODE is now therefore:

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

Evaluating the discriminant:

$$ a^2 - 4b = (-3.2)^2 - 4(2.56) = 10.24 - 10.24 = 0\!$$

Therefore the equation has a real double root and a general homogenous solution of the form:

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

And finally we find the general homogenous solution:

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

Checking:

We found that:

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

Differentiating $$ y\!$$ to obtain $${y}'\!$$ and $${y}''\!$$ respectively:

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

And,

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

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

Substituting these equations into the original ODE yields:

$$ 10[(1.6)^2 c_{1}e^{1.6x}+3.2c_{2}e^{1.6x}+(1.6)^2c_{2}xe^{1.6x}] -32[1.6c_{1}e^{1.6x}+c_{2}e^{1.6x}+1.6c_{2}xe^{1.6x}]+25.6[(c_{1}+c_{2}x)e^{1.6 x}] = 0\!$$

$$ 25.6 c_{1}e^{1.6x}+32c_{2}e^{1.6x}+25.6c_{2}xe^{1.6x} -51.2c_{1}e^{1.6x}-32c_{2}e^{1.6x}-51.2c_{2}xe^{1.6x}+25.6c_{1}e^{1.6 x}+25.6c_{2}xe^{1.6 x} = 0\!$$

$$ (25.6-51.2+25.6) c_{1} e^{1.6x} + (32-32)c_{2}e^{1.6x} + (25.6-51.2+25.6)c_{2}xe^{1.6x} = 0\!$$

$$ 0 \equiv 0 \!$$

Therefore this solution is correct.


text:
User:Egm4313.s12.team11.sheider/R2.4: K 2011 p.59 pb. 5

Problem Statement
Find a general solution to the given ODE. Check your answer by substituting into the original equation.

Given:

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

Solution
The characteristic equation of this ODE is therefore:

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

Evaluating the discriminant:

$$ a^2 - 4b = (2\pi)^2 - 4(\pi^2) = 4\pi^2 - 4\pi^2 = 0\!$$

Therefore the equation has a real double root and a general homogenous solution of the form:

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

And finally we find the general homogenous solution:

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

Checking:

We found that:

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

Differentiating $$ y\!$$ to obtain $${y}'\!$$ and $${y}''\!$$ respectively:

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

And,

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

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

Substituting these equations into the original ODE yields:

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

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

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

$$ 0 \equiv 0 \!$$

Therefore this solution is correct.


text:
File:Team14R22diagram1.png

Licensing
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title: File:Team14R22diagram1.png


text:
File:Team14R22diagram2.png

Licensing
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title: File:Team14R22diagram2.png


title: File:R2.1.png


text:
User:Egm4313.s12.team8.kyle/R2.4

Problem Statement
Find a general solution. Check your answer by substitution. (Advanced Engineering Mathematics, Tenth Edition, Kresyzig, Problem set 2.2 pg.59)

5. $$y''+2{\pi}y'+{\pi}^2y=0$$
Find the characteristic equation:

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

A single real root exists at $$r={-\pi}$$ leading to a general solution of:

$$y=C_1e^{rx}+C_2xe^{rx}$$

Substituting $$r={-\pi}$$ gives the general solution:

To check by substitution first express $$y$$, $$y'$$, and $$y''$$:

$$y=C_1e^{{-\pi}x}+C_2xe^{{-\pi}x}$$

$$y'=-C_1{\pi}e^{{-\pi}x}-C_2x{\pi}e^{{-\pi}x}+C_2e^{{-\pi}x}$$

$$y''=C_1{\pi}^2e^{{-\pi}x}+C_2x{\pi}^2e^{{-\pi}x}-2C_2{\pi}e^{{-\pi}x}$$

Substitute into the original equation:

$$C_1{\pi}^2e^{{-\pi}x}+C_2x{\pi}^2e^{{-\pi}x}-2C_2{\pi}e^{{-\pi}x}+2{\pi}(-C_1{\pi}e^{{-\pi}x}-C_2x{\pi}e^{{-\pi}x}+C_2e^{{-\pi}x})+{\pi}^2C_1e^{{-\pi}x}+C_2xe^{{-\pi}x}=0$$

Canceling like terms leaves 0=0, showing the general solution is correct.

6. $$10y''-32y'+25.6y=0$$
First divide each term by the $$y''$$ coefficient of 10 to get the equation in standard form:

$$y''-3.2y'+2.56y=0$$

Find the characteristic equation:

$$r^2-3.2r+2.56=0$$

A single real root exists at $$r=1.6$$ leading to a general solution of:

$$y=C_1e^{rx}+C_2xe^{rx}$$

Substituting $$r=1.6$$ gives the general solution:

To check by substitution first express $$y$$, $$y'$$, and $$y''$$:

$$y=C_1e^{1.6x}+C_2xe^{1.6x}$$

$$y'=C_1{1.6}e^{1.6x}+C_2x{1.6}e^{1.6x}+C_2e^{1.6x}$$

$$y''=C_1{1.6}^2e^{1.6x}+C_2x1.6^2e^{1.6x}+2C_21.6e^{1.6x}$$

Substitute into the original equation:

$$C_1{1.6}^2e^{1.6x}+C_2x1.6^2e^{1.6x}+2C_21.6e^{1.6x}-3.2(C_1{1.6}e^{1.6x}+C_2x{1.6}e^{1.6x}+C_2e^{1.6x})+2.56(C_1e^{1.6x}+C_2xe^{1.6x})=0$$

Canceling like terms leaves 0=0, showing the general solution is correct.

title: User:Egm4313.s12.team8.kyle/R2.4


title: File:R2.2.png


text:
University of Florida/Egm4313/s12.team11.gooding/R1

Problem Statement
Derive the equation of motion of the spring-mass-dashpot, with an applied force $$ r(t)\!$$ on the ball.

Kinematics
$$ y=y_{k}+y_{c}+y_{m}+r(t)\!$$

Kinetics
Spring $$F_{k}=-k*y\!$$

Dashpot $$F_{c}=-c*y'\!$$ Mass $$F_{m}=m*y''\!$$ Forcing Function $$F_{applied}=r(t)\!$$

Sum of Forces $$\sum F_{applied} +F_{m}+ F_{c}+ F_{k}=0\!$$

Substitution $$0=r(t)-m*y''-c*y'-k*y\!$$

Final Answer
$$r(t)=m*y''+c*y'+k*y\!$$

Created By:Kyle Gooding 01:46, 29 January 2012 (UTC)


text:
University of Florida/Egm4313/s12.team11.gooding/R2

R2 2.1 R2 2.8


text:
University of Florida/Egm4313/s12.team11.gooding/R2/2.1

Problem Statement
Given the two roots and the initial conditions: $$\lambda_{1}=-2,\lambda_{2}=5 \!$$ $$y(0)=1,y'(0)=0\!$$ 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: $$ r(x)=0\!$$ Plot the solution

Characteristic Equation:
$$ (\lambda-\lambda_1)(\lambda-\lambda_2)=0 \!$$ $$ (\lambda+2)(\lambda-5)=\lambda^2 +2\lambda-5\lambda-10=0\!$$ $$\lambda^2 -3\lambda-10=0\!$$

Non-Homogeneous L2-ODE-CC
$$y''-3y'-10=r(x)\!$$

Homogeneous Solution:
$$ y_{h}(x)=c_1e^{-2x}+c_2e^{5x}\!$$ $$ y(x)=c_1e^{-2x}+c_2e^{5x}+y_p(x)\!$$ Since there is no excitation, $$y_p(x)=0\!$$ $$ y(x)=c_1e^{-2x}+c_2e^{5x}\!$$

Substituting the given initial conditions:
$$ y(0)=1\!$$ $$1=c_1+c_2\!$$ $$y'(0)=0\!$$ $$0=-2c_1+5c_2\!$$ Solving these two equations for $$ c_1\!$$ and $$ c_2\!$$ yields: $$ c_1=5/4, c_2=-1/4 \!$$

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

Problem Statement
Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the 2 values in (3a) p.3-7 as the 2 roots of the corresponding characteristic equation.

Solutions
$$ 2(\lambda+2)(\lambda-5)=2\lambda^2 +4\lambda-10\lambda-20=0\!$$ $$2\lambda^2 -6\lambda-20=0\!$$ $$ 3(\lambda+2)(\lambda-5)=3\lambda^2 +6\lambda-15\lambda-30=0\!$$ $$3\lambda^2 -9\lambda-30=0\!$$ $$ 4(\lambda+2)(\lambda-5)=4\lambda^2 +8\lambda-20\lambda-40=0\!$$ $$4\lambda^2 -12\lambda-40=0\!$$ --Egm4313.s12.team11.gooding 02:01, 7 February 2012 (UTC)


text: Category:Survey research and design in psychology/Assessment/Lab reports


title: File:2.1fig1.jpg


title: File:F1.png


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text:
University of Florida/Egm4313/s12.team11.gooding/R2/2.8

Problem Statement
Find a general solution. Check your answer by substitution.

Problem 8
$$y''+ y'+3.25y=0\!$$

Let: $$ \lambda=d/dx\!$$

Characteristic Equation
$$ \lambda^2+\lambda+3.25=0\!$$ Using the quadratic equation to find roots we get: $$\lambda_1=\frac{-1+i\sqrt(12)}{2}\!$$ $$\lambda_2=\frac{-1-i\sqrt(12)}{2}\!$$ Therefore: $$y_h(x)=e^{-\frac{1}{2}x}(c_1\cos(x\sqrt{3})+c_2\sin(x\sqrt{3})\!$$

Check By Substitution
$$y'(x)=-\frac{1}{2}e^{-\frac{1}{2}x}(c_1\cos(x\sqrt{3})+c_2\sin(x\sqrt{3})+e^{-\frac{1}{2}x}(-\sqrt{3}c_1\sin{\sqrt{3}x}+\sqrt{3}c_1\cos{\sqrt{3}x})\!$$ $$y''(x)=\frac{1}{4}e^{-\frac{1}{2}x}(c_1\cos(x\sqrt{3})+c_2\sin(x\sqrt{3})-\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{3}c_1\sin{\sqrt{3}x}+\sqrt{3}c_1\cos{\sqrt{3}x})-\!$$ $$\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{3}c_1\sin{\sqrt{3}x}+\sqrt{3}c_1\cos{\sqrt{3}x})e^{-\frac{1}{2}x}(-3c_1\cos(x\sqrt{3})-3c_2\sin(x\sqrt{3})\!$$ Substituting $$y,y',y\!$$ into the original equation, the result is  $$y+ y'+3.25y=0\!$$

Problem 15
$$y''+0.54y'+(0.0729+\pi)y=0\!$$ Let: $$\frac{d}{dx}=\lambda\!$$

Characteristic Equation
$$ \lambda^2+0.54\lambda+(0.0729+\pi)=0\!$$ Using the quadratic equation to find roots we get: $$\lambda_1=\frac{-0.27+i\sqrt(\pi)}{2}\!$$ $$\lambda_2=\frac{-0.27-i\sqrt(\pi)}{2}\!$$ Therefore: $$y_h(x)=e^{-0.27x}(c_1\cos(x\sqrt{\pi})+c_2\sin(x\sqrt{\pi})\!$$

Check By Substitution
$$y'(x)=-0.27e^{-0.27x}(c_1\cos(x\sqrt{\pi})+c_2\sin(x\sqrt{\pi})+e^{-0.27x}(-\sqrt{\pi}c_1\sin{\sqrt{\pi}x}+\sqrt{\pi}c_1\cos{\sqrt{\pi}x})\!$$

$$y''(x)=0.0729e^{-0.27x}(c_1\cos(x\sqrt{\pi})+c_2\sin(x\sqrt{\pi})-0.27e^{-0.27x}(-\sqrt{\pi}c_1\sin{\sqrt{\pi}x}+\sqrt{\pi}c_1\cos{\sqrt{\pi}x})-\!$$

$$0.27e^{-0.27x}(\sqrt{\pi}c_1\sin{\sqrt{\pi}x}+\sqrt{\pi}c_1\cos{\sqrt{\pi}x})+e^{-0.27x}(-\pi(c_1\cos(x\sqrt{\pi}))-\pi(c_2\sin(x\sqrt{\pi})))\!$$

Substituting $$y,y',y\!$$ into the original equation, the result is  $$y+0.54y'+(0.0729+\pi)y=0\!$$ Egm4313.s12.team11.gooding 03:41, 7 February 2012 (UTC)


text:
User:Egm4313.s12.team6.jagolinzer.r2

My problems for Report 2: 2.7 and 2.8

Given:
$$e^t$$ $$\cos t$$ $$\sin t$$

Find:
Develop the MacLaurin Series which is the Taylor Series at $$t=0$$  for the given equations.

Solution:
The MacLaurin series is of the form: $$f(x)=f(0)+f'(0)+\frac{f''(0)}{2!}+\frac{f^{(3)}(0)}{3!}+...$$

For  $$e^t$$   :
$$f(x)=e^0+xe^0+\frac{x^2e^0}{2!}+\frac{x^3e^0}{3!}+...$$

This simplifies to:

$$f(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$

And a Summation for this series can be derived as:

$$\sum_{i=0}^{n}\frac{x^i}{i!}$$

For  $$\cos t$$   :
$$f(x)=\cos 0-x\sin0+\frac{x^2\cos 0}{2!}+\frac{x^3\sin 0}{3!}+...$$

This simplifies to:

$$f(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$$

And a Summation for this series can be derived as:

$$\sum_{i=0}^{n}\frac{x^{2i}}{(2i)!}$$

For  $$\sin t$$   :
$$f(x)=\sin 0-x\cos0+\frac{x^2\sin 0}{2!}+\frac{x^3\cos 0}{3!}+...$$

This simplifies to:

$$f(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$

And a Summation for this series can be derived as:

$$\sum_{i=0}^{n}\frac{x^{2i+1}}{(2i+1)!}$$

R2.8
This problem consist of problems 8 and 15 from the tenth edition of Erwin Kreyszig's Advanced Engineering Mathematics in section 2.2 on page 59.

Given:
The ODE:    $$y''+y'+3.25y=0$$

Find:
The general solution. Check by substitution.

Solution:
Let:    $$y=e^{\lambda x}$$

Taking the first derivative we get:    $$y'=\lambda e^{\lambda x}$$

Taking the second derivative we get:    $$y''=\lambda^2 e^{\lambda x}$$

Plugging this into   $$y''+y'+3.25y=0$$    we get:     $$e^{\lambda x}(\lambda^2+\lambda+3.25)=0$$

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

Using the Quadratic formula $$\frac{1}{2}(-a\pm \sqrt{a^2-4ab})$$ we get  $$a^2-4b<0$$   and therefore it is complex conjugate roots.

Our roots are therefore  $$y_{1}=e^{\frac{-x}{2}}(\cos \omega x)$$   and   $$y_{1}=e^{\frac{-x}{2}}(\sin \omega x)$$

Using $$\omega^2=b-\frac{a^2}{4}=3.25-0.25=3$$ we get  $$\omega=\sqrt{3}$$

Therefore the general solution is:

 * {| style="width:100%" border="0"

$$y=e^{\frac{-x}{2}}(A\cos \sqrt{3}x+B\sin \sqrt{3}x)$$ Now we must check our solution by taking the derivatives of the general solution and plugging them into the ODE.
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
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 *
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$$y'=e^{\frac{-x}{2}}(-\omega A\sin \omega x+\omega B\cos \omega x-\frac{A}{2}\cos \omega x-\frac{B}{2}\sin \omega x)$$

$$y''=e^{\frac{-x}{2}}(\frac{\omega A}{2}\sin \omega x-\frac{\omega B}{2}\cos \omega x+\frac{A}{4}\cos \omega x+\frac{B}{4}\sin \omega x-\omega^2 A\cos \omega x-\omega^2 B\sin \omega x+\frac{\omega A}{2}\sin \omega x-\frac{\omega B}{2}\cos \omega x)$$

plugging into  $$y''+y'+3.25y=0$$   with the above we get:

$$0=e^{\frac{-x}{2}}(0)$$  which confirms our solution.

Given:
The ODE:    $$y''+0.54y'+(0.0729+\pi)y=0$$

Find:
The general solution. Check by substitution.

Solution:
Let:    $$y=e^{\lambda x}$$

Taking the first derivative we get:    $$y'=\lambda e^{\lambda x}$$

Taking the second derivative we get:    $$y''=\lambda^2 e^{\lambda x}$$

Plugging this into   $$y''+0.54y'+(0.0729+\pi)y=0$$    we get:     $$e^{\lambda x}(\lambda^2+0.54\lambda+(0.0729+\pi))=0$$

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

Using the Quadratic formula $$\frac{1}{2}(-a\pm \sqrt{a^2-4ab})$$ we get  $$a^2-4b<0$$   and therefore it is complex conjugate roots.

Our roots are therefore  $$y_{1}=e^{-0.27x}(\cos \omega x)$$   and   $$y_{1}=e^{-0.27x}(\sin \omega x)$$

Using $$\omega^2=b-\frac{a^2}{4}=(0.0729+\pi)-\frac{0.54^2}{4}=\pi$$ we get  $$\omega=\sqrt{\pi}$$

Therefore the general solution is:

 * {| style="width:100%" border="0"

$$y=e^{-0.27x}(A\cos \sqrt{\pi}x+B\sin \sqrt{\pi}x)$$ Now we must check our solution by taking the derivatives of the general solution and plugging them into the ODE.
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 *
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 * }

$$y'=e^{-0.27x}(-\omega A\sin \omega x+\omega B\cos \omega x-0.27A\cos \omega x-0.27B\sin \omega x)$$

$$y''=e^{-0.27x}(0.27\omega A\sin \omega x-0.27\omega B\cos \omega x+0.0729\omega A\cos \omega x+0.0729\omega A\sin \omega x-\omega^2 A\cos \omega x-\omega^2 B\sin \omega x+0.27\omega A\sin \omega x-0.27\omega B\cos \omega x)$$

plugging into  $$y''+0.54y'+(0.0729+\pi)y=0$$   with the above we get:

$$0=e^{-0.27x}(0)$$  which confirms our solution.

title: User:Egm4313.s12.team6.jagolinzer.r2


text:
University of Florida/Egm4313/s12.team11.R2

Problem Statement
Given the two roots and the initial conditions: $$\lambda_{1}=-2,\lambda_{2}=5 \!$$ $$y(0)=1,y'(0)=0\!$$ 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: $$ r(x)=0\!$$ Plot the solution

Characteristic Equation:
$$ (\lambda-\lambda_1)(\lambda-\lambda_2)=0 \!$$ $$ (\lambda+2)(\lambda-5)=\lambda^2 +2\lambda-5\lambda-10=0\!$$ $$\lambda^2 -3\lambda-10=0\!$$

Non-Homogeneous L2-ODE-CC
$$y''-3y'-10=r(x)\!$$

Homogeneous Solution:
$$ y_{h}(x)=c_1e^{-2x}+c_2e^{5x}\!$$ $$ y(x)=c_1e^{-2x}+c_2e^{5x}+y_p(x)\!$$ Since there is no excitation, $$y_p(x)=0\!$$ $$ y(x)=c_1e^{-2x}+c_2e^{5x}\!$$

Substituting the given initial conditions:
$$ y(0)=1\!$$ $$1=c_1+c_2\!$$ $$y'(0)=0\!$$ $$0=-2c_1+5c_2\!$$ Solving these two equations for $$ c_1\!$$ and $$ c_2\!$$ yields: $$ c_1=5/4, c_2=-1/4 \!$$

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

Problem Statement
Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the 2 values in (3a) p.3-7 as the 2 roots of the corresponding characteristic equation.

Solutions
$$ 2(\lambda+2)(\lambda-5)=2\lambda^2 +4\lambda-10\lambda-20=0\!$$ $$2\lambda^2 -6\lambda-20=0\!$$ $$ 3(\lambda+2)(\lambda-5)=3\lambda^2 +6\lambda-15\lambda-30=0\!$$ $$3\lambda^2 -9\lambda-30=0\!$$ $$ 4(\lambda+2)(\lambda-5)=4\lambda^2 +8\lambda-20\lambda-40=0\!$$ $$4\lambda^2 -12\lambda-40=0\!$$ --Egm4313.s12.team11.gooding 02:01, 7 February 2012 (UTC)

Problem Statement
Find and plot the solution for the homogeneous L2-ODE-CC
 * $$y''(x) - 10y'(x) + 25y(x)=0\!$$

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

Characteristic Equation
$$\lambda^2-10\lambda+25=0\!$$

$$(\lambda-5)(\lambda-5)=0\!$$

$$\lambda=5\!$$

Homogeneous Solution
The solution to a L2-ODE-CC with real double root is given by $$y(x)=c_1e^{\lambda x}+c_2xe^{\lambda x}\!$$

First initial condition

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

$$y(0)=c_1e^{5*0}+c_2*0*e^{5*0}=1\!$$

$$c_1=1\!$$

Second initial condition

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

$$\frac{d}{dx}y(x)=y'(x)=5e^{5x}+c_2e^{5x}(5x+1)\!$$

$$y'(0)=5e^{5*0}+c_2e^{5*0}(5*0+1)=0\!$$

$$5+c_2=0\!$$

$$c_2=-5\!$$

The solution to our L2-ODE-CC is

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

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

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

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

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

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

The characteristic equation of the given DE is

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

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

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

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

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

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

Therefore, we have:

$$ \lambda_1=\frac{-6 + 0.4}{2}=\frac{-5.8}{2}=-2.8 $$

and

$$ \lambda_1=\frac{-6 - 0.4}{2}=\frac{-6.4}{2}=-3.2 $$

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

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

Check:

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

what we found to be the general solution.

$$ \frac{dy}{dx}=y'=-2.8x c_1 e^{-2.8x} + -3.2x c_2 e^{-3.2x} \! $$

$$ \frac{d^2y}{dx^2}=y''=7.84x c_1 e^{-2.8x} + 10.24 c_2 e^{-3.2x} \! $$

Substituting the values of $$ y, y' $$ and $$ y'' $$ in the given equation, we get:

$$ 7.84c_1e^{-2.8x}+10.24c_2e^{-3.2x}+6(-2.8c_1e^{-2.8x}-3.2c_2e^{-3.2x})+ 8.96(c_1e^{-2.8x}+c_2e^{-3.2x}) = 0 \! $$

$$ 7.84c_1e^{-2.8x}+10.24c_2e^{-3.2x}-16.8c_1e^{-2.8x}-19.2c_2e^{-3.2x}+ 8.96c_1e^{-2.8x}+8.96c_2e^{-3.2x} = 0 \! $$

and thus:

$$ 0 \equiv 0. \! $$

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

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

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

Solution
The characteristic equation of this ODE is therefore:

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

Evaluating the discriminant:

$$ a^2 - 4b = 4^2 - 4(\pi^2 + 4) = -4\pi^2 < 0\!$$

Therefore the equation has two complex conjugate roots and a general homogenous solution of the form:

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

Where: $$ \omega = \sqrt{b - \frac{1}{4}a^2} = \sqrt{\pi^2 + 4 - \frac{1}{4}(4)^2} = \sqrt{\pi^2} = \pi\!$$

And finally we find the general homogenous solution:

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

Check:

We found that:

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

Differentiating $$ y\!$$ to obtain $${y}'\!$$ and $${y}''\!$$ respectively:

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

$$ {y}'' = 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -2 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x))- 2 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$$

$$ {y}'' = 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -4 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$$

Substituting these equations into the original ODE yields: $$ 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -4 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$$

$$ + 4(-2e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + \pi e^{-2x}( Acos(\pi x)- Bsin(\pi x))) + (\pi^2 + 4)(e^{-2x}(Acos(\pi x )+ Bsin(\pi x))) = 0\!$$

$$ 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -4 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$$

$$ -8e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + 4\pi e^{-2x}( Acos(\pi x)- Bsin(\pi x)) + \pi^2 e^{-2x}(Acos(\pi x )+ Bsin(\pi x)) + 4e^{-2x}(Acos(\pi x )+ Bsin(\pi x)) = 0\!$$ $$ (4-8+4)e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + (-4+4) \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) + (\pi^2-\pi^2) e^{-2x}(Acos(\pi x)+Bsin(\pi x)) = 0\!$$

$$ 0 \equiv 0 \!$$ Therefore, the solution is correct.

--Gonzalo Perez

Problem Statement
Find a general solution to the given ODE. Check your answer by substituting into the original equation.

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

Solution
The characteristic equation of this ODE is therefore: $$\lambda^2 + a\lambda + b = \lambda^2 + 2\pi\lambda  + \pi^2= 0\!$$ Evaluating the discriminant: $$ a^2 - 4b = (2\pi)^2 - 4(\pi^2) = 4\pi^2 - 4\pi^2 = 0\!$$ Therefore the equation has a real double root and a general homogenous solution of the form: $$ y = (c_{1}+c_{2}x)e^{-ax/2}\!$$ And finally we find the general homogenous solution: $$ y = (c_{1}+c_{2}x)e^{-\pi x}\!$$ Checking: We found that: $$ y = (c_{1}+c_{2}x)e^{-\pi x}\!$$ Differentiating $$ y\!$$ to obtain $${y}'\!$$ and $${y}''\!$$ respectively: $$ {y}' = c_{2}e^{-\pi x} - \pi (c_{1}+c_{2}x)e^{-\pi x}\!$$ And, $$ {y}'' = -\pi c_{2} e^{-\pi x}- \pi c_{2}e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x}\!$$ $$ {y}'' = -2\pi c_{2} e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x}\!$$ Substituting these equations into the original ODE yields: $$ -2\pi c_{2} e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x} + 2\pi(c_{2}e^{-\pi x} - \pi (c_{1}+c_{2}x)e^{-\pi x}) + \pi^2((c_{1}+c_{2}x)e^{-\pi x}) = 0\!$$ $$ -2\pi c_{2} e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x} + 2\pi c_{2}e^{-\pi x} - 2\pi^2 (c_{1}+c_{2}x)e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x} = 0\!$$ $$ (-2\pi + 2\pi) c_{2} e^{-\pi x} + (\pi^2 - 2\pi^2 + \pi^2)(c_{1}+c_{2}x)e^{-\pi x} = 0\!$$

$$ 0 \equiv 0 \!$$ Therefore this solution is correct.

Problem Statement
Find a general solution to the given ODE. Check your answer by substituting into the original equation.

Given
$$10{y}''-32{y}'+25.6y = 0\!$$

Solution
Initially we modify the original ODE to put it in the form of a second-order homogenous linear ODE with constant coefficients: $$10{y}''-32{y}'+25.6y = 0\!$$ Dividing both sides by 10: $${y}''-3.2{y}+2.56y = 0\!$$ The characteristic equation of this ODE is now therefore: $$\lambda^2 + a\lambda + b = \lambda^2 -3.2\lambda  + 2.56= 0\!$$ Evaluating the discriminant: $$ a^2 - 4b = (-3.2)^2 - 4(2.56) = 10.24 - 10.24 = 0\!$$ Therefore the equation has a real double root and a general homogenous solution of the form: $$ y = (c_{1}+c_{2}x)e^{-ax/2}\!$$ And finally we find the general homogenous solution: $$ y = (c_{1}+c_{2}x)e^{1.6 x}\!$$ Checking: We found that: $$ y = (c_{1}+c_{2}x)e^{1.6x}\!$$ Differentiating $$ y\!$$ to obtain $${y}'\!$$ and $${y}''\!$$ respectively: $$ {y}' = 1.6c_{1}e^{1.6x}+c_{2}e^{1.6x}+1.6c_{2}xe^{1.6x}\!$$ And, $$ {y}'' = (1.6)^2 c_{1}e^{1.6x}+1.6c_{2}e^{1.6x}+1.6c_{2}e^{1.6x}+(1.6)^2c_{2}xe^{1.6x}\!$$ $$ {y}'' = (1.6)^2 c_{1}e^{1.6x}+3.2c_{2}e^{1.6x}+(1.6)^2c_{2}xe^{1.6x}\!$$ Substituting these equations into the original ODE yields: $$ 10[(1.6)^2 c_{1}e^{1.6x}+3.2c_{2}e^{1.6x}+(1.6)^2c_{2}xe^{1.6x}] -32[1.6c_{1}e^{1.6x}+c_{2}e^{1.6x}+1.6c_{2}xe^{1.6x}]+25.6[(c_{1}+c_{2}x)e^{1.6 x}] = 0\!$$ $$ 25.6 c_{1}e^{1.6x}+32c_{2}e^{1.6x}+25.6c_{2}xe^{1.6x} -51.2c_{1}e^{1.6x}-32c_{2}e^{1.6x}-51.2c_{2}xe^{1.6x}+25.6c_{1}e^{1.6 x}+25.6c_{2}xe^{1.6 x} = 0\!$$ $$ (25.6-51.2+25.6) c_{1} e^{1.6x} + (32-32)c_{2}e^{1.6x} + (25.6-51.2+25.6)c_{2}xe^{1.6x} = 0\!$$

$$ 0 \equiv 0 \!$$ Therefore this solution is correct.

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

ODE Solutions
Two Real Roots: $$y(x)=c_{1}e^{\lambda_{1}x}+c_{2}e^{\lambda_{2}x}\!$$ Double Real Roots: $$y(x)=c_{1}e^{\lambda_{1}x}+c_{2}xe^{\lambda_{2}x}\!$$

Reverse Engineering
$$e^{\lambda x}(\lambda^2+a\lambda+b)=0\!$$ Case 1: $$a^2-4b>0\!$$ $$\lambda^2-(\lambda_{1}+\lambda_{2})\lambda+\lambda_{1}\lambda_{2}\!$$ $$y''-(\lambda_{1}+\lambda_{2})y'+\lambda_{1}\lambda_{2}y=0\!$$ Case 2: $$a^2-4b=0\!$$ $$\lambda^2-(\lambda_{1}+\lambda_{2})\lambda+\lambda_{1}\lambda_{2}\!$$ $$y''-(\lambda_{1}+\lambda_{2})y'+\lambda_{1}\lambda_{2}y=0\!$$

R2.5 K2011 p.59 pbs.16
$$e^{2.6x}, e^{-4.3x}\!$$ $$(\lambda-2.6)(\lambda+4.3)\!$$ $$\lambda^2+1.7\lambda-11.18\!$$

ODE Form: $$y''+1.7y'-11.18=0\!$$

R2.5 K2011 p.59 pbs.17
$$e^{-\sqrt{5}x}, xe^{-\sqrt{5}x}\!$$ $$(\lambda+\sqrt{5})^2\!$$ $$\lambda^2+2\sqrt{5}\lambda+5\!$$

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

Created by [| Daniel Suh] 20:57, 7 February 2012 (UTC)

Problem 2.6
Solved by: Andrea Vargas

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

Solution
Previously, we have derived the following equation for such a system: (From [[media:iea.s12.sec1.djvu|Sec 1 (d)]], (3) p.1-5) $$ m\left ( y_k''+\frac{k}{c}y_k' \right )+ ky_k=f(t)\!$$ We can write this equation in standard form by diving through by $$ m \!$$: $$ y_k''+\frac{k}{c}y_k'+ \frac{k}{m}y_k=\frac{f(t)}{m} \!$$ Here, we can take the coefficients of $$y_k'\!$$ and $$y_k\!$$ as $$a\!$$ and $$b\!$$:

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

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

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

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

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.

Example of Solution
An example of fixing one of the constants to find the other two is provided here. By solving the simple equations above, we can illustrate how to find $$ k, c ,m \!$$. We had: $$ \frac{k}{c}=6 \; and \; \frac{k}{m}=9\!$$

If we fix the mass to $$ m=10 \; kg \!$$. We find: $$\frac{k}{10}=9\!$$

$$k=90\!$$

Then, $$\frac{90}{c}=6\!$$

$$c=15\!$$

Finally, we obtain:

$$m=10, \; k=90, \; and \; c=15 \!$$ --Andrea Vargas 21:44, 7 February 2012 (UTC)

Problem Statement

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

Solution

 * $$ e^t= \sum_{n=0}^{\infty }\frac{t^n}{n!} \!$$

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

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

 * $$ \cos t= \sum_{k=0}^{\infty }\frac{(-1)^kt^{2k}}{(2k)!} \!$$

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

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

 * $$ \sin t= \sum_{k=0}^{\infty }\frac{(-1)^kt^{2k+1}}{(2k+1)!} \!$$

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

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

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

Problem Statement
Find a general solution. Check your answer by substitution.

Problem 8
$$y''+ y'+3.25y=0\!$$

Let: $$ \lambda=d/dx\!$$

Characteristic Equation
$$ \lambda^2+\lambda+3.25=0\!$$ Using the quadratic equation to find roots we get: $$\lambda_1=\frac{-1+i\sqrt(12)}{2}\!$$ $$\lambda_2=\frac{-1-i\sqrt(12)}{2}\!$$ Therefore: $$y_h(x)=e^{-\frac{1}{2}x}(c_1\cos(x\sqrt{3})+c_2\sin(x\sqrt{3}))\!$$

Check By Substitution
$$y'(x)=-\frac{1}{2}e^{-\frac{1}{2}x}(c_1\cos(x\sqrt{3})+c_2\sin(x\sqrt{3})+e^{-\frac{1}{2}x}(-\sqrt{3}c_1\sin{\sqrt{3}x}+\sqrt{3}c_1\cos{\sqrt{3}x})\!$$ $$y''(x)=\frac{1}{4}e^{-\frac{1}{2}x}(c_1\cos(x\sqrt{3})+c_2\sin(x\sqrt{3})-\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{3}c_1\sin{\sqrt{3}x}+\sqrt{3}c_1\cos{\sqrt{3}x})-\!$$ $$\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{3}c_1\sin{\sqrt{3}x}+\sqrt{3}c_1\cos{\sqrt{3}x})e^{-\frac{1}{2}x}(-3c_1\cos(x\sqrt{3})-3c_2\sin(x\sqrt{3})\!$$ Substituting $$y,y',y\!$$ into the original equation, the result is  $$y+ y'+3.25y=0\!$$

Problem 15
$$y''+0.54y'+(0.0729+\pi)y=0\!$$ Let: $$\frac{d}{dx}=\lambda\!$$

Characteristic Equation
$$ \lambda^2+0.54\lambda+(0.0729+\pi)=0\!$$ Using the quadratic equation to find roots we get: $$\lambda_1=\frac{-0.27+i\sqrt(\pi)}{2}\!$$ $$\lambda_2=\frac{-0.27-i\sqrt(\pi)}{2}\!$$ Therefore: $$y_h(x)=e^{-0.27x}(c_1\cos(x\sqrt{\pi})+c_2\sin(x\sqrt{\pi})\!$$

Check By Substitution
$$y'(x)=-0.27e^{-0.27x}(c_1\cos(x\sqrt{\pi})+c_2\sin(x\sqrt{\pi})+e^{-0.27x}(-\sqrt{\pi}c_1\sin{\sqrt{\pi}x}+\sqrt{\pi}c_1\cos{\sqrt{\pi}x})\!$$

$$y''(x)=0.0729e^{-0.27x}(c_1\cos(x\sqrt{\pi})+c_2\sin(x\sqrt{\pi})-0.27e^{-0.27x}(-\sqrt{\pi}c_1\sin{\sqrt{\pi}x}+\sqrt{\pi}c_1\cos{\sqrt{\pi}x})-\!$$

$$0.27e^{-0.27x}(\sqrt{\pi}c_1\sin{\sqrt{\pi}x}+\sqrt{\pi}c_1\cos{\sqrt{\pi}x})+e^{-0.27x}(-\pi(c_1\cos(x\sqrt{\pi}))-\pi(c_2\sin(x\sqrt{\pi})))\!$$

Substituting $$y,y',y\!$$ into the original equation, the result is  $$y+0.54y'+(0.0729+\pi)y=0\!$$ Egm4313.s12.team11.gooding 03:41, 7 February 2012 (UTC)

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

$$\lambda^2 + 4 \lambda + 13\!$$

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

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

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

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

$$\lambda=-2 \pm 3i\!$$

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

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

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

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

Solving for A and B
first initial condition $$y(0)=1\!$$

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

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

$$A=1\!$$

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

$$y'(x)=\frac{d}{dx}y(x)=\frac{d}{dx}e^{-2x}[cos(3x)+Bsin(3x)]\!$$

$$y'(x)=e^{-2x}[(-2B-3)sin(3x)+(3B-2)cos(3x)]\!$$

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

$$0=3B-2\!$$

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

so the solution to our L2-ODE-CC is                       $$y(x)=e^{-2x}[cos(3x)+\frac{2}{3}sin(3x)]\!$$

Solution to R2.6
After solving for the constants $$\frac{k}{c}$$ and $$\frac{k}{m}$$ we have the following homogeneous equation

$$y''(x)+6y'(x)+9y=0\!$$

Characteristic Equation and Roots
$$\lambda^2 + 6\lambda + 9=0\!$$

$$(\lambda+3)(\lambda+3)=0\!$$

We have a real double root $$\lambda=-3\!$$

Homogeneous Solution
We know the homogeneous solution to a L2-ODE-CC with a double real root to be

$$y(x)=c_1e^{\lambda x}+c_2xe^{\lambda x}\!$$

Assuming object starts from rest

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

Plugging in $$\lambda$$ and applying our first initial condition

$$y(0)=c_1e^{-3*0}+c_2*0*e^{-3*0}=1\!$$

$$c_1=1\!$$

Taking the derivative and applying our second condition

$$y'(x)=\frac{d}{dx}y(x)=\frac{d}{dx}e^{-3x}+c_2xe^{-3x}\!$$

$$y'(x)=-3e^{-3x}+c_2e^{-3x}-3c_2xe^{-3x}\!$$

$$y'(0)=-3e^{-3*0}+c_2e^{-3*0}-3c_2*0*e^{-3*0}=0\!$$

$$c_2=3\!$$

Giving us the final solution $$y(x)=e^{-3x}+3xe^{-3x}\!$$

Solution to this Equation
$$y(x)=e^{-2x}[cos(3x)+\frac{2}{3}sin(3x)]\!$$

Superimposed Graph
Our solution: $$y(x)=e^{-2x}[cos(3x)+\frac{2}{3}sin(3x)]\!$$ shown in blue

Equation for fig. in R2.1 p.3-7: $$y(x)=\frac{5}{4}e^{-2x}-\frac{1}{4}e^{5x}\!$$ shown in red

Equation for fig. in R2.6 p.5-6:$$y(x)=e^{-3x}+3xe^{-3x}\!$$ shown in green

Egm4313.s12.team11.imponenti 03:38, 8 February 2012 (UTC)


text: File:PlotR2 2.jpg

title: File:PlotR2 2.jpg


text:
University of Florida/Egm4313/s12.team8.dupre/R2.3

Problem Statement
Find a general solution. Check your answer by substitution. a) $$\displaystyle y+6y'+8.96y=0$$    (3-1)  b)$$\displaystyle y+4y'+(\pi^2+4)y=0$$    (3-2)

Solution
The quadratic formula is necessary for these solutions: $$\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

Part a
Plugging into the quadratic formula: $$\displaystyle \frac{-6\pm \sqrt{6^2-(4)(1)(8.96)}}{2(1)}=\frac{-6\pm .4}{2}$$ This shows us that the roots of the equation are: $$\displaystyle \lambda_{1} =-2.8, \lambda _{2}=-3.2$$ Therefore, the general equation is: $$\displaystyle y=c_{1}e^{-2.8x}+c_{2}e^{-3.2x}$$    (3-3)

Substitution
We need to first find the first and second derivatives of equation (3-3): $$\displaystyle y'=-3.2c_{1}e^{-3.2x}-2.8c_{2}e^{-2.8x}$$ $$\displaystyle y''=10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x}$$ Plugging into equation (3-1), we find: $$\displaystyle (10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x})+6(-3.2c_{1}e^{-3.2x}-2.8c_{2}e^{-2.8x})+8.96(c_{1}e^{-3.2x}+c_{2}e^{-2.8x})=0$$    (3-4) Continuing to solve: $$\displaystyle 19.2c_{1}e^{-3.2x}+16.8c_{2}e^{-2.8x} -19.2c_{1}e^{-3.2x}-16.8c_{2}e^{-2.8x}=0$$    (3-5) This shows that the general equation is correct, since everything cancels out to 0.

Part b
Plugging into the quadratic formula: $$\displaystyle \frac{4\pm\sqrt{4^{2}-4(1)(\pi^2 +4)}}{2(1)}=\frac{-4\pm (-4)(1))(pi^2+4)}{2(1)} $$ The roots are, therefore:  $$\displaystyle \lambda _{1}=-2-\pi i, \lambda _{2}=-2+\pi i$$  Therefore, the general solution to (3-2) is:  $$\displaystyle y=c_{1}\cos(\pi x)e^{-2x} + c_{2}\sin(\pi x)e^{-2x}$$     (3-6)

Substitution
We must first find the first and second derivatives of equation (3-6): $$\displaystyle y'= -2(c_{1} \cos (\pi x) + c_{2} \sin (\pi x))e^{-2x} + (-c_{1} \pi \sin (\pi x) + c_{2} \pi \cos (\pi x))e^{-2x}$$ $$\displaystyle y''= - 2(-c_{1} \pi \sin (\pi x) + c_{2}\pi \cos (\pi x))e^{-2x} + (-c_{1} \pi ^2 \cos (\pi x) - c_{2} \pi ^2 \sin (\pi x))e^{-2x}$$ Plugging into equation (3-2): $$\displaystyle - 2(-c_{1} \pi \sin (\pi x) + c_{2}\pi \cos (\pi x))e^{-2x} + (-c_{1} \pi ^2 \cos (\pi x) - c_{2} \pi ^2 \sin (\pi x))e^{-2x}...$$ $$\displaystyle+4[-2(c_{1} \cos (\pi x) + c_{2} \sin (\pi x))e^{-2x} + (-c_{1} \pi \sin (\pi x) + c_{2} \pi \cos (\pi x))e^{-2x}]+(\pi^2+4)(c_{1}\cos(\pi x)e^{-2x} + c_{2}\sin(\pi x)e^{-2x})=0$$ Finally, plugging (3-6) and it's first and second derivatives into equation (3-2), we find: $$\displaystyle 4(c_{1} \cos (\pi x) + c_{2} \sin (\pi x))e^{-2x} - 2(-c_{1} \pi \sin (\pi x) + c_{2} \pi \cos (\pi x))e^{-2x} - 2(-c_{1} \pi \sin (\pi x) + c_{2} \pi \cos (\pi x))e^{-2x}...$$ $$\displaystyle+(-c_{1}\pi ^2 \cos (\pi x) - c_{2} \pi ^2 \sin (\pi x))e^{-2x} +4 -2(c_{1} \cos (\pi x) + c_{2} \sin (\pi x))e^{-2x}) +...$$ $$\displaystyle (-c_{1} \pi \sin (\pi x) + c_{2} \pi \cos (\pi x))e^{-2x}+(\pi^{2}+4)((c_{1}\cos(\pi x) + c_{2}\sin(\pi x))e^{-2x})=0$$  Since this equals 0, we know that the general equation (3-6) is correct.


text:
User:ArdentSeeker

Hello World

title: User:ArdentSeeker


text:
User:Brylie/digital photography notes

Aliasing: Pixelation on the edges of lines, shapes, etc.

Moire Pattern:  An interference pattern caused by multiple different patterns combined.

Angle of View: Width of an image. The human eye (one eye open) has a 49mm angle of view.

Pixel: Picture Element. A single photo site on a sensor.

Crop Factor: A comparison of a digital sensor with 35mm film (measurement). The smaller the sensor the higher the multiplication factor. E.g. if the sensor is 17mm X 13mm (4/3) then it is half the size of the 36mm X 24mm film dimensions and has a crop factor, or multiplication factor, of 2.0. A 50mm lense in conjunction with a 2.0 CF sensor would produce the same effect as a 100mm lense with a 35mm sensor.

Depth of Field: A function of magnification. Small sensors produce higher magnification than large sensors. As aperture changes, so does depth of field.

Depth of field should be 1/3 in front of the subject and 2/3 behind the subject.

Parallax: The affect of having different angles of incedent from the lense. E.g. having a flash above and to the side of a lense will reduce red eye through a parallax effect.

Diopter: A viewfinder that can be adjusted to the photographer's eyesight.

Chromatic aberration: inability of a lense to focus red, green, and blue light on a fixed point. This usually occurs around areas of high contrast towards the edges of an image.

Spherical aberration: distortion of focus towards the edges of an image.

Sensor: silicon is sensetive to light when it is charged.

CCD: Charged Coupling Device, 8-bits of information per channel.

CMOS: Complementary metal-oxide-semiconductor. Deeper photo wells, 16-bits of information per channel.

Example photo cell array:

R G R G R G

B G B G B G

The chip has a lower sensitivity to green so more green photo cells are added. Conversely, our eyes have a higher sensitivity to green.

Blooming: Photo sites receive too much light and flare out to neigboring sites.

Noise: When the sensor receives too little light the active charge will create visual artifacts.

Blackbody reference: black cells are scattered throughout a sensor array in order for the software to make adjustments to the image.

title: User:Brylie/digital photography notes


text:
User:Egm4313.s12.team19. Code test page

A Code test page 2
 * Bulleted list item

R2.8
Contributed by Vladimir Horwitz.

Problem Statement:
Answer questions 8 and 15 from Kreyszing.

Solution:
Prob. 15)

 * {| style = "width: 73%"

! $$ \ y'' + .54 y' + (.0729 + \pi) y = 0 $$ !! (Eq. 1)
 * $$ \ \lambda^2 + .54 \lambda + .0729 + \pi = 0 $$
 * (Eq. 2)
 * The characteristic equation has these roots.
 * $$ \lambda = -.27 + \sqrt {\pi} i, \bar \lambda = -.27 - \sqrt {\pi} i $$
 * (Eq. 3, a and b)
 * }
 * $$ \lambda = -.27 + \sqrt {\pi} i, \bar \lambda = -.27 - \sqrt {\pi} i $$
 * (Eq. 3, a and b)
 * }

Due to the fact that this roots are complex, the following equations from Kreyszig 2011, page 57 case 3, can be applied.

 * {| class = wikitable style="width: 73%; text-align:center"

! $$ \ y = e^{-.27} * (A \cos (\sqrt \pi x) + B \sin (\sqrt \pi x)) $$ (Eq. 4)
 * }

 * {| style = "width: 73%"

! And following this, we can say !!
 * $$ y' = -.27 e^{-.27} \ (A \cos (\sqrt \pi x) + B \sin (\sqrt \pi x)) \ + \ e^{-.27} \ (-A \sqrt \pi \sin (\sqrt \pi x) + B \sqrt \pi \cos (\sqrt \pi x)) $$
 *  (Eq. 5)
 * and
 * $$ y'' = .0729 e^{-.27} \ (A \cos \sqrt \pi x \ + B \sin \sqrt \pi x) \ + \ e^{-.27} \ (- \pi A \cos \sqrt \pi x \ - \pi B \sin \sqrt \pi x) + \downarrow $$
 * $$ \uparrow -.27 e^{-.27} \ (-A \sqrt \pi \sin \sqrt \pi x \ + B \sqrt \pi \cos \sqrt \pi x) \ - \ .27 e^{-.27} \ (-A \sqrt \pi \sin \sqrt \pi x \ + B \sqrt \pi \cos \sqrt \pi x) $$
 * (Eq. 6)
 * Then, after major calculator work, we find that
 * $$ 0 = (0) * e^{-.27} * (A \cos \sqrt \pi x \ + B \sin \sqrt \pi x) $$
 * (Eq. 7)
 * }
 * Then, after major calculator work, we find that
 * $$ 0 = (0) * e^{-.27} * (A \cos \sqrt \pi x \ + B \sin \sqrt \pi x) $$
 * (Eq. 7)
 * }
 * (Eq. 7)
 * }

title: User:Egm4313.s12.team19. Code test page


text:
User:Egm4313.s12.team8/R2

=R2.1 - Find and plot the solution to an ODE, and write 3 ODEs with the same roots=

Problem Statement
Part i) Find the non-homogeneous L2-ODE-CC in standard form and the solution for the ODE in terms of the initial conditions ($$\displaystyle y(0)=1$$ $$\displaystyle y'(0)=0$$) and the general excitation where $$\displaystyle r(x)=0$$, and plot the solution.

Part ii) Generate 3 non-standard, non-homoegeneous L2-ODE-CC that yield the same two roots of the characteristic equation as in Part i.

Part i
The standard form of a standard non-homogeneous L2-ODE-CC is: $$\displaystyle y''+ay'+by=r(x)$$

First, we use the given lambas to find the characteristic equation for the homogeneous solution:

Since,

then,

So,

The answer will be in the form of:

Differentiating the homogeneous solution gives:

By substituting Eqn. 1a into Eqn. 1b, we get:

By multiplying and combining like terms we get $$\displaystyle c_1$$:

Then we substitute $$\displaystyle c_1$$ back into Eqn 2a to get $$\displaystyle c_1$$:

So the solution to the ODE is:

This is a graph of the solution to the ODE: $$\displaystyle y(x)=\frac{5}{7}e^{-2x}+\frac{2}{7}e^{5x}$$
 * }

Part ii
=R2.2 - Find and plot solution to homogeneous ODE=

Problem Statement
Find and plot the solution for the L2-ODE-CC Eqn. 4 ($$\displaystyle y''-10y'+25y=r(x)$$) on p. 5-5 in the lecture notes. The initial conditions are $$\displaystyle y(0)=1$$,  $$\displaystyle y'(0)=0$$ with no excitation ($$\displaystyle r(x)=0$$)

Solution
Also because $$\displaystyle r(x)=0$$, the ODE becomes:

Which can be written as the characteristic equation:

Factoring the characteristic equation gives:

Which solves giving a double real root:

The double real root means the solution will have the form:

The first homogeneous solution is:

The second homogeneous solution for a double real root takes the form:

The final homogeneous solution is:

Applying the initial condition $$\displaystyle y(0)=1$$ to the solution gives:

The updated solution is: Differentiting the solution gives: Applying the second initial condition, $$\displaystyle y'(0)=0$$ gives:

The final solution is:

The graph above is a MATLAB plot of the solution to the ODE: $$\displaystyle y(x)=e^{5x}-5xe^{5x}$$

=R2.3 - General solution to 2 ODEs with real roots and complex roots=

Problem Statement
Find a general solution. Check your answer by substitution. a) $$\displaystyle y+6y'+8.96y=0$$    (3-1)  b)$$\displaystyle y+4y'+(\pi^2+4)y=0$$    (3-2)

Solution
The quadratic formula is necessary for these solutions: $$\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

Part a
Plugging into the quadratic formula: $$\displaystyle \frac{-6\pm \sqrt{6^2-(4)(1)(8.96)}}{2(1)}=\frac{-6\pm .4}{2}$$ This shows us that the roots of the equation are: $$\displaystyle \lambda_{1} =-2.8, \lambda _{2}=-3.2$$ Therefore, the general equation is: $$\displaystyle y=c_{1}e^{-2.8x}+c_{2}e^{-3.2x}$$    (3-3)

Substitution
We need to first find the first and second derivatives of equation (3-3): $$\displaystyle y'=-3.2c_{1}e^{-3.2x}-2.8c_{2}e^{-2.8x}$$ $$\displaystyle y''=10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x}$$ Plugging into equation (3-1), we find: $$\displaystyle (10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x})+6(-3.2c_{1}e^{-3.2x}-2.8c_{2}e^{-2.8x})+8.96(c_{1}e^{-3.2x}+c_{2}e^{-2.8x})=0$$    (3-4) Continuing to solve: $$\displaystyle 19.2c_{1}e^{-3.2x}+16.8c_{2}e^{-2.8x} -19.2c_{1}e^{-3.2x}-16.8c_{2}e^{-2.8x}=0$$    (3-5) This shows that the general equation is correct, since everything cancels out to 0.

Part b
Plugging into the quadratic formula: $$\displaystyle \frac{4\pm\sqrt{4^{2}-4(1)(\pi^2 +4)}}{2(1)}=\frac{-4\pm (-4)(1))(pi^2+4)}{2(1)} $$ The roots are, therefore:  $$\displaystyle \lambda _{1}=-2-\pi i, \lambda _{2}=-2+\pi i$$  Therefore, the general solution to (3-2) is:  $$\displaystyle y=c_{1}\cos(\pi x)e^{-2x} + c_{2}\sin(\pi x)e^{-2x}$$     (3-6)

Substitution
We must first find the first and second derivatives of equation (3-6): $$\displaystyle y'= -2(c_{1} \cos (\pi x) + c_{2} \sin (\pi x))e^{-2x} + (-c_{1} \pi \sin (\pi x) + c_{2} \pi \cos (\pi x))e^{-2x}$$ $$\displaystyle y''= - 2(-c_{1} \pi \sin (\pi x) + c_{2}\pi \cos (\pi x))e^{-2x} + (-c_{1} \pi ^2 \cos (\pi x) - c_{2} \pi ^2 \sin (\pi x))e^{-2x}$$ Plugging into equation (3-2): $$\displaystyle - 2(-c_{1} \pi \sin (\pi x) + c_{2}\pi \cos (\pi x))e^{-2x} + (-c_{1} \pi ^2 \cos (\pi x) - c_{2} \pi ^2 \sin (\pi x))e^{-2x}...$$ $$\displaystyle+4[-2(c_{1} \cos (\pi x) + c_{2} \sin (\pi x))e^{-2x} + (-c_{1} \pi \sin (\pi x) + c_{2} \pi \cos (\pi x))e^{-2x}]+(\pi^2+4)(c_{1}\cos(\pi x)e^{-2x} + c_{2}\sin(\pi x)e^{-2x})=0$$ Finally, plugging (3-6) and it's first and second derivatives into equation (3-2), we find: $$\displaystyle 4(c_{1} \cos (\pi x) + c_{2} \sin (\pi x))e^{-2x} - 2(-c_{1} \pi \sin (\pi x) + c_{2} \pi \cos (\pi x))e^{-2x} - 2(-c_{1} \pi \sin (\pi x) + c_{2} \pi \cos (\pi x))e^{-2x}...$$ $$\displaystyle+(-c_{1}\pi ^2 \cos (\pi x) - c_{2} \pi ^2 \sin (\pi x))e^{-2x} +4 -2(c_{1} \cos (\pi x) + c_{2} \sin (\pi x))e^{-2x}) +...$$ $$\displaystyle (-c_{1} \pi \sin (\pi x) + c_{2} \pi \cos (\pi x))e^{-2x}+(\pi^{2}+4)((c_{1}\cos(\pi x) + c_{2}\sin(\pi x))e^{-2x})=0$$  Since this equals 0, we know that the general equation (3-6) is correct.

=R2.4 - General solution to 2 ODEs with repeated roots=

Problem Statement
Find a general solution. Check your answer by substitution. (Advanced Engineering Mathematics, Tenth Edition, Kresyzig, Problem set 2.2 pg.59)

5. $$y''+2{\pi}y'+{\pi}^2y=0$$
Find the characteristic equation:

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

A single real root exists at $$r={-\pi}$$ leading to a general solution of:

$$y=C_1e^{rx}+C_2xe^{rx}$$

Substituting $$r={-\pi}$$ gives the general solution:

To check by substitution first express $$y$$, $$y'$$, and $$y''$$:

$$y=C_1e^{{-\pi}x}+C_2xe^{{-\pi}x}$$

$$y'=-C_1{\pi}e^{{-\pi}x}-C_2x{\pi}e^{{-\pi}x}+C_2e^{{-\pi}x}$$

$$y''=C_1{\pi}^2e^{{-\pi}x}+C_2x{\pi}^2e^{{-\pi}x}-2C_2{\pi}e^{{-\pi}x}$$

Substitute into the original equation:

$$C_1{\pi}^2e^{{-\pi}x}+C_2x{\pi}^2e^{{-\pi}x}-2C_2{\pi}e^{{-\pi}x}+2{\pi}(-C_1{\pi}e^{{-\pi}x}-C_2x{\pi}e^{{-\pi}x}+C_2e^{{-\pi}x})+{\pi}^2C_1e^{{-\pi}x}+C_2xe^{{-\pi}x}=0$$

Canceling like terms leaves 0=0, showing the general solution is correct.

6. $$10y''-32y'+25.6y=0$$
First divide each term by the $$y''$$ coefficient of 10 to get the equation in standard form:

$$y''-3.2y'+2.56y=0$$

Find the characteristic equation:

$$r^2-3.2r+2.56=0$$

A single real root exists at $$r=1.6$$ leading to a general solution of:

$$y=C_1e^{rx}+C_2xe^{rx}$$

Substituting $$r=1.6$$ gives the general solution:

To check by substitution first express $$y$$, $$y'$$, and $$y''$$:

$$y=C_1e^{1.6x}+C_2xe^{1.6x}$$

$$y'=C_1{1.6}e^{1.6x}+C_2x{1.6}e^{1.6x}+C_2e^{1.6x}$$

$$y''=C_1{1.6}^2e^{1.6x}+C_2x1.6^2e^{1.6x}+2C_21.6e^{1.6x}$$

Substitute into the original equation:

$$C_1{1.6}^2e^{1.6x}+C_2x1.6^2e^{1.6x}+2C_21.6e^{1.6x}-3.2(C_1{1.6}e^{1.6x}+C_2x{1.6}e^{1.6x}+C_2e^{1.6x})+2.56(C_1e^{1.6x}+C_2xe^{1.6x})=0$$

Canceling like terms leaves 0=0, showing the general solution is correct.

=R2.5 - Determine ODE from provided basis=

Problem Statement
Kreyszig 2011 p.59 pbs.16-17

For the given basis, find an ODE of the form:
 * $$\displaystyle y'' + ay' + by = 0 $$    (1)

16. $$ \displaystyle e^{2.6x}, e^{-4.3x} $$

17. $$ \displaystyle e^{- \sqrt{5}x}, xe^{- \sqrt{5}x} $$

Problem 16
Note: Given $$ \displaystyle y = e^{\lambda x} $$ as a solution of the second-order homogeneous linear ODE, the characteristic equation (or auxiliary equation) is defined as

$$ \displaystyle \lambda^2 + a\lambda + b = 0. $$  [K2011 p54]

The problem gives us two distinct real-roots:

$$ \displaystyle \lambda_{1} = 2.6 $$ and $$ \displaystyle \lambda_{2} = -4.3. $$

Solving the quadratic equation above gives us $$ \displaystyle \lambda_{1} = \frac {-a + \sqrt{a^2 - 4b}} 2 = \frac {1}{2} (-a + \sqrt{a^2 - 4b}) = 2.6, $$ and $$ \displaystyle \lambda_{2} = \frac {-a - \sqrt{a^2 - 4b}} 2 = \frac {1}{2} (-a - \sqrt{a^2 - 4b}) = -4.3 $$

or $$ \displaystyle \lambda_{1} = (-a + \sqrt{a^2 - 4b}) = 5.2, $$ and $$ \displaystyle \lambda_{2} = (-a - \sqrt{a^2 - 4b}) = -8.6. $$

Summing $$ \displaystyle \lambda_{1} $$ and $$ \displaystyle \lambda_{2} $$ we can eliminate the discriminant $$ -a + \sqrt{a^2 - 4b} $$ and solve for $$ \displaystyle a $$ $$ \displaystyle -2a = -3.4 $$ $$ \displaystyle a = 1.7 $$

Plugging $$ \displaystyle a $$ back into either $$ \displaystyle \lambda_1 $$ or $$ \displaystyle \lambda_2, $$ we can solve for $$ \displaystyle b $$

$$ \displaystyle -1.7 + \sqrt{1.7^2-4b} = 5.2$$ $$ \displaystyle     \sqrt{2.89-4b} = 6.9$$ $$ \displaystyle 2.89 - 4b = 44.61$$ $$ \displaystyle -4b = 44.72$$ $$ \displaystyle b = -11.18$$

Now plugging $$ \displaystyle a $$ and $$ \displaystyle b $$ back into the general ODE, we have $$ \displaystyle y'' + 1.7y' - 11.18y = 0 $$

Problem 17
Given the basis: $$ \displaystyle e^{- \sqrt{5}x}, xe^{- \sqrt{5}x}, $$ we know we get only one root $$ \displaystyle \lambda = \lambda_1 = \lambda_2 = - \frac {a}{2}. $$ [K2011 p.55]

Thus, the discriminant $$ \displaystyle a^2 - 4b = 0. $$    (2)

Knowing $$ \displaystyle \lambda = - \frac {a}{2} $$ for the given basis, we can say $$ \displaystyle e^{- \frac {a}{2} x} = e^{- \sqrt{5}x} $$

to solve for unknown $$ \displaystyle a. $$

$$ \displaystyle - \frac {a}{2} = - \sqrt {5} $$ $$ \displaystyle a = 2 \sqrt {5} $$

Plugging $$ \displaystyle a $$ into the discriminant gives us $$ \displaystyle ({2 \sqrt {5}})^2 - 4b = 0 $$ $$ \displaystyle b = \frac {20}{4} = 5. $$

Now substituting $$ \displaystyle a $$ and $$ \displaystyle b $$ into the general ODE gives us our solution: $$ \displaystyle y'' + 2 \sqrt{5} y' + 5y = 0 $$

=R2.6 - Solve for parameters of a spring-dashpot system with given characteristic equation=

Problem Statement
Realize spring-dashpot-mass systems in series as shown in Fig. 1 below with the given characteristic equation and double real roots $$ \lambda = -3 $$. Find the values for the parameters k,c,m.

Given
from [[media:iea.s12.sec1.djvu|R1.1 in Sec 1 p. 1-5]]

The equation of motion of the spring-dashpot-mass system is as follows: $$\displaystyle m(y_k''+\frac{k}{c}y_k')+ky_k=f(t) $$

Double real roots: $$\displaystyle \lambda=-3 $$

Solution
Since we know the roots of the system we obtain the following characteristic equation:

$$\displaystyle (\lambda+3)^2=\lambda^2+6\lambda+9$$ (1) The equation of motion is: $$\displaystyle m(y_k''+\frac{k}{c}y_k')+ky_k=f(t) $$ (2)

We can get (1) and (2) to be in the same form by distributing the m in (2). $$\displaystyle my_k''+\frac{mk}{c}y_k'+ky_k=f(t) $$ (2a)

Now we can use coefficient matching to equate the coefficients of (1) and (2):

$$\displaystyle m=1$$ $$\displaystyle \frac{mk}{c}=6$$ $$\displaystyle k=9$$

Therefore: $$\displaystyle m=1,$$ $$\displaystyle c=\frac{3}{2},$$ $$\displaystyle k=9$$

=R2.7 - Maclaurin Series of $$e^t$$, $$\cos{t}$$ and $$\sin{t}$$=

Problem Statement
Develop the Maclaurin series (Taylor Series at t=0) for $$\displaystyle e^{t},cos(t),sin(t)$$ a) $$\displaystyle e^{t}$$    (7-1)  b)$$\displaystyle cos(t)$$    (7-2) c)$$\displaystyle sin(t)$$    (7-3)

Solution
For a Taylor series at any point t, the general form is as follows (to make it more clear, I have changed the usual t in (t-a) to a capital T): $$ \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (T-a)^n=f(a) + \frac{f'(a)}{1!}(T - a) + \frac{f^2(a)}{2!}(T - a)^2 + \frac{f^3(a)}{3!}(T- a)^3 + \frac{f^4(a)}{4!}(T - a)^4 ... (7-4)$$

Part a solution
Using (7-4) as applied to (7-1) at time t=0 gives us the Maclaurin series for (7-1): $$\displaystyle e^t = \sum_{n=0}^{\infty} \frac{T^n}{n!}=e^0 + \frac{e^0}{1!}(T) + \frac{e^0}{2!}(T)^2 + \frac{e^0}{3!}(T)^3 + \frac{e^0}{4!}(T)^4 ... (7-6)$$ Simplifying this result, we obtain our final solution: $$\displaystyle e^t = 1 + T + \frac{T^2}{2} + \frac{T^3}{6} + \frac{T^4}{24} ... (7-7) $$

Part b solution
Using (7-4) as applied to (7-2) at time t=0 gives us the Maclaurin series for (7-2): $$\displaystyle cos(t) = \sum_{n=0}^{\infty} \frac{(-1)^nT^{2n}}{(2n)!} = cos(0) + \frac{-sin(0)}{1!}(T) + \frac{-cos(0)}{2!}(T)^2 + \frac{sin(0)}{3!}(T)^3 + \frac{cos(0)}{4!}(T)^4 + ... (7-8)$$ Simplifying this result, we obtain our final solution: $$\displaystyle cos(t) = \sum_{n=0}^{\infty} \frac{(-1)^nT^{2n}}{(2n)!} = 1-\frac{T^2}{4}+\frac{T^4}{24} ... (7-9)$$

Part c solution
Using (7-5) as applied to (7-3) at time t=0 gives us the Maclaurin series for (7-3): $$\displaystyle sin(t) = \sum_{n=0}^{\infty} \frac{(-1)^nT^{2n + 1}}{(2n + 1)!} = sin(0) + \frac{cos(0)}{1!}(T) + \frac{-sin(0)}{2!}(T)^2 + \frac{-cos(0)}{3!}(T)^3 + \frac{sin(0)}{4!}(T)^4 + ... (7-10)$$ Simplifying this result, we obtain our final solution: $$\displaystyle sin(t) = T - \frac{T^3}{6} + ... (7-11)$$

=R2.8 - General solution to 2 ODEs with complex roots=

Problem Statement
Find a general solution. Check your answer by substitution.

Part A:
 * $$\displaystyle y'' + y' + 3.25y = 0$$

Part B:
 * $$y'' + 0.54 y' + \left(0.0729 + \pi\right) y = 0$$

Finding the General Solution
Assume solutions are proportional to $$e^{rt}$$:
 * $$e^{rt}\left(r^2 + r + 3.25\right) = 0$$

Observe $$e^{rt} \neq 0$$ for any $$t$$:
 * $$\displaystyle r^2 + r + 3.25 = 0$$

Solve for $$r$$:
 * $$r = -\frac{1}{2} \pm \sqrt{3} i$$

Substitute in $$r$$ for initial assumption:
 * $$y(t) = c_1 e^{\left(-\frac{1}{2} + \sqrt{3} i\right) t} + c_2 e^{\left(-\frac{1}{2} - \sqrt{3} i\right) t}$$

Expand the solution using Euler's identity ($$e^{a + bi}=e^{a}\left(\cos{b} + i\sin{b}\right)$$):
 * $$\begin{align}

y(t) &= c_1 e^{-\frac{1}{2} t} \left(\cos{\sqrt{3} t} + i\sin{\sqrt{3} t}\right) \\ &+ c_2 e^{-\frac{1}{2} t} \left(\cos{\sqrt{3} t} - i\sin{\sqrt{3} t}\right) \end{align}$$ Collect like terms:
 * $$\begin{align}

y(t) &= \left(c_1 + c_2\right) e^{-\frac{1}{2} t} \cos{\sqrt{3} t} \\ &+ \left(c_1 - c_2\right) i e^{-\frac{1}{2} t} \sin{\sqrt{3} t} \end{align}$$ Define new constants $$c'_1 = c_1 + c_2$$ and $$c'_2 = \left(c_1 - c_2\right) i$$: $$y(t) = c'_1 e^{-\frac{1}{2} t} \cos{\sqrt{3} t} + c'_2 e^{-\frac{1}{2} t} \sin{\sqrt{3} t}$$

Checking by Substitution
Determine $$3.25 y$$, $$y'$$, and $$y''$$ from $$y(t)$$:
 * $$\begin{array}{rllrlrllcrrlcll}

3.25 & y  & = e^{-\frac{1}{2} t} (   &  \frac{13}{4} & c'_1 &               &        & \cos{\sqrt{3} t} & + &   &           &      &   \frac{13}{4} & c'_2   & \sin{\sqrt{3} t} ) \\ & y' & = e^{-\frac{1}{2} t} ( ( & -\frac{1}{2}  & c'_1 & + \; \sqrt{3} & c'_2 ) & \cos{\sqrt{3} t} & + & ( & -\sqrt{3} & c'_1 & - \frac{1}{2}  & c'_2 ) & \sin{\sqrt{3} t} ) \\ & y'' & = e^{-\frac{1}{2} t} ( ( & -\frac{11}{4} & c'_1 & - \; \sqrt{3} & c'_2 ) & \cos{\sqrt{3} t} & + & ( & \sqrt{3} & c'_1 & - \frac{11}{4} & c'_2 ) & \sin{\sqrt{3} t} ) \end{array}$$ Substitute into original ODE:
 * $$\begin{array}{rrrcrllrlrlll}

( e^{-\frac{1}{2} t} & ( ( & -\frac{11}{4} & c'_1 & - \; \sqrt{3} & c'_2 ) & \cos{\sqrt{3} t} + ( & \sqrt{3} & c'_1 & - \frac{11}{4} & c'_2 ) & \sin{\sqrt{3} t} ) ) & \\ + \; ( e^{-\frac{1}{2} t} & ( ( & -\frac{1}{2} & c'_1 & + \; \sqrt{3} & c'_2 ) & \cos{\sqrt{3} t} + ( & -\sqrt{3} & c'_1 & - \frac{1}{2}  & c'_2 ) & \sin{\sqrt{3} t} ) ) & \\ + \; ( e^{-\frac{1}{2} t} & (  &  \frac{13}{4} & c'_1 &               &        & \cos{\sqrt{3} t} +   &           &      &   \frac{13}{4} & c'_2   & \sin{\sqrt{3} t} ) ) & = 0 \end{array}$$ Reduce: $$\displaystyle 0 = 0$$

Finding the General Solution
Assume solutions are proportional to $$e^{rt}$$:
 * $$e^{rt}\left(r^2 + 0.54 r + \left(0.0729 + \pi\right)\right) = 0$$

Observe $$e^{rt} \neq 0$$ for any $$t$$:
 * $$r^2 + 0.54 r + \left(0.0729 + \pi\right) = 0$$

Rearrange by completing the square:
 * $$\left(r + 0.27\right)^2 + \pi = 0$$

Solve for $$r$$:
 * $$r = -0.27 \pm \sqrt{\pi} i$$

Substitute in $$r$$ for initial assumption:
 * $$y(t) = c_1 e^{\left(-0.27 + \sqrt{\pi} i\right) t} + c_2 e^{\left(-0.27 - \sqrt{\pi} i\right) t}$$

Expand the solution using Euler's identity ($$e^{a + bi}=e^{a}\left(\cos{b} + i\sin{b}\right)$$):
 * $$\begin{align}

y(t) &= c_1 e^{-0.27 t} \left(\cos{\sqrt{\pi} t} + i\sin{\sqrt{\pi} t}\right) \\ &+ c_2 e^{-0.27 t} \left(\cos{\sqrt{\pi} t} - i\sin{\sqrt{\pi} t}\right) \end{align}$$ Collect like terms:
 * $$\begin{align}

y(t) &= \left(c_1 + c_2\right) e^{-0.27 t} \cos{\sqrt{\pi} t} \\ &+ \left(c_1 - c_2\right) i e^{-0.27 t} \sin{\sqrt{\pi} t} \end{align}$$ Define new constants $$c'_1 = c_1 + c_2$$ and $$c'_2 = \left(c_1 - c_2\right) i$$: $$\displaystyle y(t) = c'_1 e^{-0.27 t} \cos{\sqrt{\pi} t} + c'_2 e^{-0.27 t} \sin{\sqrt{\pi} t}$$

Checking by Substitution
Determine $$( 0.0729 + \pi ) y$$, $$0.54 y'$$, and $$y''$$ from $$y(t)$$:
 * $$\begin{array}{rllrlrllcrrlcll}

( 0.0729 + \pi ) & y  & = e^{-0.27 t} (   &  ( 0.0729 + \pi )  & c'_1 &                        &        & \cos{\sqrt{\pi} t} & + &   &                     &      & ( 0.0729 + \pi )  & c'_2   & \sin{\sqrt{\pi} t} ) \\ & y' & = e^{-0.27 t} ( ( &            -0.27   & c'_1 & + \;               \sqrt{\pi} & c'_2 ) & \cos{\sqrt{\pi} t} & + & ( &                -\sqrt{\pi} & c'_1 &       - \; 0.27   & c'_2 ) & \sin{\sqrt{\pi} t} ) \\ 0.54 & y' & = e^{-0.27 t} ( ( & -2 \times  0.0729  & c'_1 & + \; 2 \times 0.27 \sqrt{\pi} & c'_2 ) & \cos{\sqrt{\pi} t} & + & ( & - 2 \times 0.27 \sqrt{\pi} & c'_1 & - \; 2 \times 0.0729 & c'_2 ) & \sin{\sqrt{\pi} t} ) \\ & y'' & = e^{-0.27 t} ( ( & ( 0.0729 - \pi ) & c'_1 & - \; 2 \times 0.27 \sqrt{\pi} & c'_2 ) & \cos{\sqrt{\pi} t} & + & ( &  2 \times 0.27 \sqrt{\pi} & c'_1 & + \; (0.0729 - \pi) & c'_2 ) & \sin{\sqrt{\pi} t} ) \end{array}$$ Substitute into original ODE:
 * $$\begin{array}{rrrlrlllrrllll}

( e^{-0.27 t} & ( ( & ( -\pi + 0.0729 )  & c'_1 & - \; 2 \times 0.27 \sqrt{\pi} & c'_2 ) & \cos{\sqrt{\pi} t} & + \; ( &   2 \times 0.27 \sqrt{\pi} & c'_1 & + \; ( 0.0729 - \pi) & c'_2 ) & \sin{\sqrt{\pi} t} ) ) & \\ + \; ( e^{-0.27 t} & ( ( & -2 \times 0.0729   & c'_1 & + \; 2 \times 0.27 \sqrt{\pi} & c'_2 ) & \cos{\sqrt{\pi} t} & + \; ( & - 2 \times 0.27 \sqrt{\pi} & c'_1 & - \; 2 \times 0.0729 & c'_2 ) & \sin{\sqrt{\pi} t} ) ) & \\ + \; ( e^{-0.27 t} & (  &  ( 0.0729 + \pi  )  & c'_1 &                               &        & \cos{\sqrt{\pi} t} & +   &                            &      & ( 0.0729 + \pi )  & c'_2   & \sin{\sqrt{\pi} t} ) ) & = 0 \end{array}$$ Reduce: $$\displaystyle 0 = 0$$

=R2.9 - Particular solution to a homogeneous ODE=

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

Initial Conditions:

$$ \displaystyle y(0)=1   $$    and    $$ \displaystyle  {y}'(0)=0   $$

No excitation:

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

Solution
Written in standard form:

$$ \displaystyle {y}''+4 {y}'+13 y=0  $$

The equation follows the characteristic equation:

$$ \displaystyle {y}''+a {y}'+b y=0  $$

Therefore:

$$ \displaystyle   a=4   $$  and  $$ \displaystyle    b=13   $$

Calculating the discriminant, we can then decide which case this problem follows.

$$ \displaystyle   a^{2}-4b =  4^{2}-4(13) = 16-52 = -36   $$

Since the discriminant is less than 0, the problem will follow Case III, complex conjugate roots.

The general solution will then take the form of:

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

Where:

$$ \displaystyle   w= \sqrt{b-\frac{1}{4}a^{2}}    $$    and    $$   \displaystyle   a=4   $$ (as previously stated)

Therefore:

$$ \displaystyle   w= \sqrt{13-\frac{1}{4}(4)^{2}} =  \sqrt{9}  = 3 $$

We can now substitute these values back into the general solution, obtaining:

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

We now take the general solution's derivative.

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

$$ \displaystyle  {y}'(x)=e^{-2x}(-3Asin(3x)+3Bcos(3x))-2e^{-2x}(Acos(3x)+Bsin(3x)) $$

We can now use the given initial conditions to solve for the unknown constants A and B.

$$ \displaystyle  y(0)=1,  {y}'(0)=0   $$

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

$$ \displaystyle  {y}'(0)=e^{-2(0)}(-3Asin(3(0))+3Bcos(3(0)))-2e^{-2(0)}(Acos(3(0))+Bsin(3(0)))=0 $$

Reducing to the following system of equations:

$$ \displaystyle  A=1   $$

$$ \displaystyle  3B-2A=0 $$

Solving this system, the constants A and B can be found as:

$$ \displaystyle  A=1, B=\frac{2}{3} $$

Replacing these constants back into the equation, we can obtain the final solution:

Figures
Graph of Solution to 2.9

Matlab code:

>> x=linspace(-3,2);

>> f= exp(-2*x).*(cos(3*x)+.666*sin(3*x))

>> plot(x,f)

>> xlabel('x')

>> ylabel('y')

Graph of 2.9, 2.6, and 2.1

a) from R2.9 (blue) $$ \displaystyle y(x)=e^{-2x}(cos(3x)+\frac{2}{3}sin(3x))   $$

b) from R2.6 (green) $$ \displaystyle y(x)=e^{-3x}+3xe^{-3x}      $$

c) from R2.1 (red) $$ \displaystyle   y(x)=\frac{5}{7}e^{-2x}+\frac{2}{7}e^{5x}     $$

Matlab code:

x = linspace(-3,2);

f = exp(-2*x).*(cos(3*x)+2/3*sin(3*x)); g = exp(-3*x) + 3 * x .* exp(-3*x); h = 5/7 * exp(-2*x) + 2/7*exp(5*x);

fl=sign(f) .* log(abs(f) + 1); gl=sign(g) .* log(abs(g) + 1); hl=sign(h) .* log(abs(h) + 1);

plot(x,f,x,g,x,h) xlabel('x') ylabel('y')

plot(x,fl,x,gl,x,hl) xlabel('x') ylabel('sign(y) * log(abs(y) + 1)')

=R2.10 - Solution to a non-homogeneous ODE=

Problem Statement
Consider the same system as in the Example p.7-3, i.e., with the same L2-ODE-CC (4) p.5-5 and initial conditions (2) p.3-4, but with the following excitation:
 * $$r(x)\;=\;7e^{5x}\,-\,2x^2$$

(4) p.5-5: $$y''\,-\,10y'\,+\,25\;=\;r(x)$$

(2) p.3-4: $$\displaystyle y(0) = 4\text{, }y'(0) = -5$$

Find the homogeneous solution
Solve the characteristic equation:
 * $$\left(\lambda - 5\right)^2 = 0$$

 * $$\displaystyle\lambda = 5$$

The homogeneous solution is then:
 * $$\displaystyle y_h(x) = K_1 e^{5x} + K_2 x e^{5x}$$

Find the particular solution
Following the sum and modification rule from p.7-2:
 * $$\begin{align}

7e^{5x} &\Rightarrow C_1 x^2 e^{5x} \\ -2x^2 &\Rightarrow C_2 x^2 + C_3 x + C_4 \end{align}$$

By the sum rule:
 * $$\displaystyle y_p(x) = C_1 x^2 e^{5x} + C_2 x^2 + C_3 x + C_4$$

Taking derivatives:
 * $$\begin{align}

y'_p(x) &= C_1 e^{5x} \left(5 x^2 + 2 x\right) + 2 C_2 x + C_3 \\ y''_p(x) &= C_1 e^{5x} \left(5\left(5 x^2 + 2 x\right) + 10 x + 2\right) + 2 C_2 \\ &= C_1 e^{5x} \left(25 x^2 + 20 x + 2\right) + 2 C_2 \end{align}$$

Substitute particular solution and its derivatives into the original ODE:
 * $$\begin{align}

&\left(C_1 e^{5x} \left(25 x^2 + 20 x + 2\right) + 2 C_2\right) \\ - 10 &\left(C_1 e^{5x} \left(5 x^2 + 2 x\right) + 2 C_2 x + C_3\right) \\ + 25 &\left(C_1 x^2 e^{5x} + C_2 x^2 + C_3 x + C_4\right) \\ &= 7 e^{5x} - 2 x^2 \end{align}$$

Reduce:
 * $$2 C_1 e^{5x} + 25 C_2 x^2 + \left(-20 C_2 + 25 C_3\right) x + \left(2 C_2 - 20 C_3 + 25 C_4\right) = 7 e^{5x} - 2 x^2$$

Equate coefficients from each side:
 * $$\begin{align}

2 C_1 &= 7 \\ 25 C_2 &= -2 \\ -20 C_2 + 25 C_3 &= 0 \\ 2 C_2 - 20 C_3 + 25 C_4 &= 0 \end{align}$$

Solve for constants:
 * $$\begin{align}

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

Substitute in constants to find particular solution:
 * $$y_p(x) = \frac{7}{2} x^2 e^{5x} - \frac{2}{25} x^2 - \frac{8}{125} x - \frac{12}{625}$$

Finding the solution

 * $$\begin{align}

y(x) &= y_h(x) + y_p(x) \\ &= \left(K_1 e^{5x} + K_2 x e^{5x}\right) + \left(\frac{7}{2} x^2 e^{5x} - \frac{2}{25} x^2 - \frac{8}{125} x - \frac{12}{625}\right) \end{align}$$

Find the first derivative:
 * $$y'(x) = 5 K_1 e^{5x} + K_2 e^{5x} + 5 K_2 x e^{5x} + \frac{7}{2} e^{5x} \left(5 x^2 + 2 x\right) - 2 \frac{2}{25} x - \frac{8}{125}$$

Solve for constants using initial conditions:
 * $$\begin{align}

y(0) = 4 = K_1 - \frac{12}{625} &\Rightarrow K_1 = \frac{2512}{625} \\ y'(0) = -5 = 5 \frac{2512}{625} + K_2 - \frac{8}{125} &\Rightarrow K_2 = -\frac{3129}{125} \end{align}$$

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

=Team Contributions=

title: User:Egm4313.s12.team8/R2


text:
Three-Tier Architecture

Building Mobile Web Applications in a Three-Tier Architecture

Portal:Engineering and Technology —> School:Computer Science —> Topic:Computer Programming —> Topic:Three-Tier Architecture

Audience
This course is written to assist the non-technical person in understanding how the Model-View-Controller pattern is implemented within a Three-Tier Architecture. Or in other words "What is the technology and approach behind building a mobile application for my business?". The course also has a focus on the business value created by well designed and architected technology. Therefore, making the course of interest to middle and senior management. If you are also a beginner to intermediate programmer the information will also be of interest, particularly put into the context of business value.

Prerequisites
Before starting the course, we suggest (though it is NOT mandatory) you go over the following: Create a software development environment. This course will not describe the creation of a server and software development environment. The following configuration is recommended to engage in this course;
 * Introduction to Programming
 * Introduction to Computer Science
 * HTML / CSS
 * JavaScript
 * PhP
 * MySQL
 * Firefox and firebug
 * Text editor or programming tool
 * LAMP Server

Course Description
The course will help you learn how to build mobile web applications using a three-tier architecture approach. The learning will be facilitated by having each learner build their own mobile application and going through code and materials step by step and explaining relevant concepts. Reference materials will be complemented by learning guides, which will expand on the reference by elaborating the how and why of each topic.

Lessons

 * 1) Building a Mobile HTML5 page – Start by building a working mobile view
 * 2) MVC Three-Tier Architecture – The Architecture Described
 * 3) Building the Model and Controller - Bringing context to the web page
 * 4) Information Architecture – Designing the user experience
 * 5) Touching the business tier – Thinking about services

Mentoring
I am available to assist in your learning. If you are interested in learning MVC and three-tier architecture with a mentor, list yourself below by adding * ~ (an asterisk, a space, and four tildes) to the bottom of the list. I will follow-up.

title: Three-Tier Architecture


text:
University of Florida/Egm4313/s12.team8.dupre/R2.7

Problem Statement
Develop the Maclaurin series (Taylor Series at t=0) for $$\displaystyle e^{t},cos(t),sin(t)$$ a) $$\displaystyle e^{t}$$    (7-1)  b)$$\displaystyle cos(t)$$    (7-2) c)$$\displaystyle sin(t)$$    (7-3)

Solution
For a Taylor series at any point t, the general form is as follows (to make it more clear, I have changed the usual t in (t-a) to a capital T): $$ \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (T-a)^n=f(a) + \frac{f'(a)}{1!}(T - a) + \frac{f^2(a)}{2!}(T - a)^2 + \frac{f^3(a)}{3!}(T- a)^3 + \frac{f^4(a)}{4!}(T - a)^4 ... (7-4)$$

Part a solution
Using (7-4) as applied to (7-1) at time t=0 gives us the Maclaurin series for (7-1): $$\displaystyle e^t = \sum_{n=0}^{\infty} \frac{T^n}{n!}=e^0 + \frac{e^0}{1!}(T) + \frac{e^0}{2!}(T)^2 + \frac{e^0}{3!}(T)^3 + \frac{e^0}{4!}(T)^4 ... (7-6)$$ Simplifying this result, we obtain our final solution: $$\displaystyle e^t = 1 + T + \frac{T^2}{2} + \frac{T^3}{6} + \frac{T^4}{24} ... (7-7) $$

Part b solution
Using (7-4) as applied to (7-2) at time t=0 gives us the Maclaurin series for (7-2): $$\displaystyle cos(t) = \sum_{n=0}^{\infty} \frac{(-1)^nT^{2n}}{(2n)!} = cos(0) + \frac{-sin(0)}{1!}(T) + \frac{-cos(0)}{2!}(T)^2 + \frac{sin(0)}{3!}(T)^3 + \frac{cos(0)}{4!}(T)^4 + ... (7-8)$$ Simplifying this result, we obtain our final solution: $$\displaystyle cos(t) = \sum_{n=0}^{\infty} \frac{(-1)^nT^{2n}}{(2n)!} = 1-\frac{T^2}{4}+\frac{T^4}{24} ... (7-9)$$

Part c solution
Using (7-5) as applied to (7-3) at time t=0 gives us the Maclaurin series for (7-3): $$\displaystyle sin(t) = \sum_{n=0}^{\infty} \frac{(-1)^nT^{2n + 1}}{(2n + 1)!} = sin(0) + \frac{cos(0)}{1!}(T) + \frac{-sin(0)}{2!}(T)^2 + \frac{-cos(0)}{3!}(T)^3 + \frac{sin(0)}{4!}(T)^4 + ... (7-10)$$ Simplifying this result, we obtain our final solution: $$\displaystyle sin(t) = T - \frac{T^3}{6} + ... (7-11)$$


title: File:MVC3tierArchitecture.jpg


text:
TAO/Creating Personas and Working with Them

Goal of the Learning Module
Explain the concept of personas, give instructions how to create them, give examples for their use in the context of TAO.

Target group(s)
People engaging in product or service design. Users intending to create their own learning module.

Introduction
In marketing and user-centered design, personas are fictional characters created to represent different user types within a targeted demographic, attitude and/or behavior set that might use a service or product in a similar way. Personas are used as a tool or method for market segmentation and user-centered design.

Personas are useful in considering the goals, desires, and limitations of costumers and users in order to help to guide decisions about a service, product or interaction space. By fostering empathy among the design team, they help the team to avoid designing the product or service according to their own needs and expectations. Furthermore, by making assumptions about the users transparent, they allow for an open discussion and questioning of stereotypes, based on empirical research.

Learning Units

 * Unit 1: Create your own personas

Results

 * Personas


text:
Mobile Web Applications

Mobile Web Applications
The best way to learn to develop web pages is to develop web pages. This philosophy is how this course has been designed and built. It treats learning web development more like learning an art than a science. And when attending an art class the opportunity to jump right in starts usually on the first day, during the first hour of class.

Considerations

 * 1) When on the Internet users prefer to use web apps rather than mobile apps on their mobile devices.
 * 2) Learners using this learning resource should find or create their own development environment capable of hosting the technologies used within this course. A review of my series of blog posts describing the setup of a rackspace cloud server can help here; http://criticaltechnology.blogspot.com/search/label/hosting
 * 3) This course will be taught more as an ART course than a SCIENCE course. So be prepared to get dirty and learn with reckless abandon.
 * 4) The creation of this course / book is as an example toward Agile Learner Design

Conventions

 * 1) Each chapter will focus only on what it needs to meet its objective. Within any feature of technology there is often many more attributes than need to be discussed to meet a chapters objectives. It is preferred to only focus on the immediate need, rather than all that is possible with any given feature.
 * 2) Pseudocode will be used to describe all algorithms / programming logic
 * 3) Each chapter will include a business value section to describe the value of the technology will have to the business. This convention is used for two reasons; first, to get away from technology for technologies sake. And second, to provide a laypersons view into why the technology is important. The business value section answers the c-level question, "What's the value of doing this?"

Course concept map
This concept map shows the content covered by this resource. The concept map is a work in progress as it will be added to as the course continues as suggested by the Agile Instructional Design methodology.

Table of Contents

 * ch01. Building a Mobile HTML5 page – Start by building a working mobile view
 * ch02. MVC Three-Tier Architecture – The Architecture Described
 * ch03. Building the Model and Controller - Bringing context to the web page
 * ch04. Information Architecture – Designing the user experience
 * ch05. Touching the business tier – Thinking about services
 * apxA. Debugging with firebug - fixing and improving your code

Related Badges
Mobile Web Application Badges - a set of badges awarded for demonstrating mastery of the different sections within this book.

title: Mobile Web Applications


text:
User:Peterrawsthorne/ALD

Table of Contents

 * ch01. Introduction to Agile Learner Design
 * ch02. Examples of ALD implementation
 * ch03. History of Agile and Lean
 * ch04. History of relevant pedagogies and ID
 * ch05. Know Thyself (exploring your metacognitive abilities)
 * ch06. Envisioning and Personal Curriculum Mapping
 * ch07. Planning and finding community
 * ch08. Building content and content licensing
 * ch09. Stabilization and Assessment
 * ch10. Deployment and reputation management
 * ch11. Tools and technologies
 * ch12. Begin it now
 * ch13. Case studies

Background / Suggestions

 * 1) DIYU
 * 2) Buccaneer Scholar...
 * 3) Book needs to be written to the correct literacy level (Grade 8 - 10?)
 * 4) Book needs images, pictures and online references
 * 5) find online publisher (lulu, amazon...)
 * 6) further investigate mobile / reader formats

Research Tasks

 * 1) Get industry examples of Agile in ID
 * 2) Start building collaborative assessment system
 * 3) Complete PHP MySQL online course
 * 4) Begin data modeling (write blog post on the importance of data modeling even within agile)
 * 5) Study Lean approaches

title: User:Peterrawsthorne/ALD


text:
TAO/Creating Personas and Working with Them/unit 1

Personas are hypothetical archetypes, or “stand-ins” for actual users or customers that drive the decision making for product or service design projects:

 * Personas are not real people, but they represent real people throughout the design process

 * Personas are not "made up"; they are discovered as a by-product of the investigative process into your users or costumers

 * Although personas are imaginary, they are defined with significant rigor and precision

 * Names and personal details for personas are made up to make them more realistic

 * Personas are defined by their needs and goals

 * For each product or service, more than one persona is usually created, but one or two personas should always be the primary focus for the design

You may draw your knowledge of the target group from various sources, including statistics. However, in order to be effective, you will usually need to gather a team of people who know these personas or to organize focus groups with members of the target group.

There are lots of ways to sketch out personas. You may want to adapt one of the example procedures to your purpose:

 * Example Procedure 1 (e.g. for designing e-inclusion measures for a particular fringe of the population):
 * Step 1: Who are they? - Describe their personal characteristics to give them a human touch, turn them into personalities (provide a photo)

 * Step 2: What are their behaviour and goals?

 * Step 3: What are barriers and obstacles regarding these goals?

 * Step 4: What are possible solutions to overcome these barriers and obstacles?

 * Step 5: What difficulties would you expect? Any further comments?

 * Example Procedure 2 (e.g. for designing support offers for professionals or volunteers in a particular function):

 * Step 1: What are their personal characteristics? - name, age, profession, further details from their life which could play a role regarding the situation under analysis

 * Step 2: What are their motivations and personal goals?

 * Step 3: What are their stakeholders? What are their stakeholders' expectations? (identify the expectations they are faced with in their function)

 * Step 4: What are critical events they could face? What challenges do they pose? (identify the challenges they would typically face in their function)

 * Step 5: What are possible support offers to help them living up to the expectations and mastering the challenges they face? (in terms of peer support, coaching, inter-vision, external support, etc.)

 * Example Procedure 3 (e.g. for internet marketing):

 * Step 1: Who are they, every day?

 * Step 2: What do they want, all the time?

 * Step 3: How do they interact with you and your competitors?

 * Step 4: What might make this interaction happen/not happen?

 * Step 5: How may they affect your organization?

 * Source: "Get In Your Customers' Heads: Creating Great Personas", Conversation Marketing, 2007. See also for more detailed instructions and an example.

Do you want to give your persona a face?
You may want to try http://www.pimptheface.com/ to create a phantom picture and upload it to Wikimedia Commons to integrate it in Wikiversity (Example).


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TAO/Personas

This page contains personas representing the different target groups of the TAO Handbook and Community of Practice. You may want to browse already existing persona descriptions in order to discover the various life situations and perspectives of our target public or to add a persona description of your own if you feel that an important group, life situation or perspective is still missing (instructions for persona creation).

Staff members in a managing or co-ordinating function (managers)
This section contains personas of staff members in a managing or co-ordinating function (volunteers or paid staff) at organizations providing or intending to provide:
 * e-learning / blended learning for seniors
 * any other type of online activity for seniors related to online communities or social networks
 * web2.0 introductory courses
 * any type of service involving online communities or virtual teams which heavily rely on the participation of seniors

Personas

 * Heiko (56), program manager at a community centre
 * Guillaume (30), project manager at Wikimedia France

Instructors, moderators, facilitators (instructors)
This section contains personas of instructors, moderators, facilitators (volunteers or paid staff) of:
 * e-learning / blended learning courses for seniors
 * web2.0 introductory courses for seniors
 * online activities for seniors in social networks or online communities
 * activities aiming at developing online services together with seniors in a co-creative approach

Personas

 * Marianne (62), volunteer instructor for new media courses for older people

Consultants
This section contains personas of consultants working with organizations providing or intending to provide:
 * e-learning / blended learning for seniors
 * any other type of online activity for seniors related to online communities or social networks
 * web2.0 introductory courses
 * any type of service involving online communities or virtual teams which heavily rely on the participation of seniors

Personas

 * Simona (39), consultant for adult education association of NRW / vhs-Landesverband Nordrhein Westfalen, Germany
 * Bernhard Hofer (33), researcher in Zurich, interested in consulting online communities for seniors

Senior end users
This section contains personas of seniors affected by the digital divide in various degrees who participate in:
 * e-learning / blended learning activities
 * any other type of online activity related to online communities, social networks, or virtual teams
 * web2.0 introductory courses

Personas
See personas of older end users on the German version of this page (in German).

Other groups
This section contains personas describing stakeholders of the TAO Handbook and Community of Practice that don't fall in one of the categories above.

title: TAO/Personas