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User:Egm6321.f10.team2.oztekin/HW5

=Problem 5.1: Finding fifth derivative of y =

From the lecture slide Mtg 26-2

5.1.1 Given
Solving Euler L2_ODE_VC has to methods of solution. For the first method we have 2 stages. In mtheod one the first stage is transformation of variables in which we change dependent variable. The second stage we try to find solution for dependent variable. In second method we directly try to find solution for dependent variable. But know we discuss first stage for the first method.

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$$  \displaystyle \sum_{i=0}^{n}a_{i}x^{i}y^{(i)}=0 $$     (Eq 1.1)
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This equation involves coefficients need to determined.After changing independent variable

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$$  \displaystyle y=y(x(t)) $$     (Eq 1.2)
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Let's take
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$$  \displaystyle x=e^{t} $$     (Eq 1.3)
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$$  \displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} t}\frac{\mathrm{d} t}{\mathrm{d} x} $$ (Eq 1.4)
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$$  \displaystyle y_{x}=y_{t}e^{-t} $$     (Eq 1.5)
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$$  \displaystyle y_{xxxx}=e^{-4t}\left [ y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t} \right ] $$     (Eq 1.6)
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5.1.2 Solution

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$$  \displaystyle y_{xxxxx}=\frac{\mathrm{d} }{\mathrm{d} x}y_{xxxx}=\frac{\mathrm{d} t}{\mathrm{d} x}\frac{\mathrm{d} }{\mathrm{d} t}\left ( e^{-4t}y_{tttt}-6e^{-4t}y_{ttt}+11e^{-4t}y_{tt}-6e^{-4t}y_{t} \right ) $$     (Eq 1.7)
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$$  \displaystyle y_{xxxxx}=e^{-t}\left [ e^{-4t}y_{ttttt}-10e^{-4t}y_{tttt}+35e^{-4t}y_{ttt}-50e^{-4t}y_{tt}+24e^{-4t}y_{t} \right ] $$     (Eq 1.8)
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$$  \displaystyle y_{xxxxx}=e^{-5t}\left [ y_{ttttt}-10y_{tttt}+35y_{ttt}-50y_{tt}+24y_{t} \right ] $$     (Eq 1.9)
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Also note that the coefficients before those derivatives are typical in polynominals like $$\displaystyle a(a-1)(a-2)......$$, then we can have a general form of derivatives of $$\displaystyle y$$ as,

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$$ \displaystyle {{y}^{(n)}}={{e}^{-nt}}D(D-1)(D-2)......(D-n+1)y$$ (Eq 1.10)
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where $$\displaystyle D$$ denotes the operator "$$\displaystyle \frac{d}{dt}$$" and $$\displaystyle n\in \mathbb{N}$$.

We can prove Eq 1.10 by mathematical induction.

When $$\displaystyle n=1$$, Eq 1.5 has prove this case.

Now let's assume Eq 1.10 is valid at some $$\displaystyle n\in \mathbb{N}$$, meaning,

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then,

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$$ \displaystyle

\begin{align} {{y}^{(n+1)}} & =\frac{d}{dx}{{y}^{(n)}} \\ & =\frac{1}\frac{d}{dt}{{y}^{(n)}} \\ & ={{e}^{-t}}\frac{d}{dt}\left[ {{e}^{-nt}}D(D-1)(D-2)......(D-n+1)y \right] \\ & ={{e}^{-t}}\left[ (-n){{e}^{-nt}}D(D-1)(D-2)......(D-n+1)y \right. \\ & \left. {{^ – }^ – }{{^ – }^ – }{{^ – }^ – }{{^ – }^ – }{{^ – }^ – }{{^ – }^ – }{{^ – }^ – }^ – +{{e}^{-nt}}{{D}^{2}}(D-1)(D-2)......(D-n+1)y \right] \\ & ={{e}^{-t}}{{e}^{-nt}}({{D}^{2}}-nD)(D-1)(D-2)......(D-n+1)y \\ & ={{e}^{-(n+1)t}}D(D-1)(D-2)......(D-(n+1)+1)y \\ \end{align}$$

(Eq 1.11)
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So we have proved Eq 1.10 is valid throughout any $$\displaystyle n\in \mathbb{N}$$.

=Problem 5.2:Using Trial Solution to Solve Euler's Equation=

From the lecture slide Mtg 26-3

5.2.1 Given
Given the ODE:
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$$  \displaystyle x^{2}{y}''+2x{y}'+2y=0 $$     (Eq 2.1) and boundary conditions:
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$$  \displaystyle y(1)=-2 $$     (Eq 2.3)
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$$  \displaystyle y(2)=5 $$     (Eq 2.4) solve using
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$$  \displaystyle y=x^{r} $$     (Eq 2.5) and the Trial Solution method.
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5.2.2 Solution
In the trial solution use the given equation and take the first two derivatives:
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$$  \displaystyle y=x^{r} $$     (Eq 2.6)
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$$  \displaystyle y^{'}=rx^{r-1} $$     (Eq 2.7)
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$$  \displaystyle y^{''}=r(r-1)x^{r-2} $$     (Eq 2.8) Substitute these values into Eq. 2.1 above:
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$$  \displaystyle x^{2}r(r-1)x^{r-2}-2xrx^{r-1}+2y=0 $$     (Eq 2.9) From this equation it is easy to eliminate constants and we end up with:
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$$  \displaystyle x^{r}(r^{2}-3r+2)=0 $$     (Eq 2.10) Meaning that our equation of interest is:
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$$  \displaystyle (r^{2}-3r+2)=0 $$     (Eq 2.11) Solving this the answers are:
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$$  \displaystyle r_{1}=1 $$     (Eq 2.12)
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$$  \displaystyle r_{2}=2 $$     (Eq 2.13) Using the equation:
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$$  \displaystyle y(x)=C_{1}x^{r_{1}}+C_{2}x^{r_{2}} $$     (Eq 2.14) we plug in r_{1} and r_{2} to get:
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$$  \displaystyle y(x)=C_{1}x+C_{2}x^{2} $$     (Eq 2.15) Using the above boundary conditions we get two equations:
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$$  \displaystyle -2=C_{1}+C_{2} $$     (Eq 2.16) and
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$$  \displaystyle 5=2C_{1}+4C_{2} $$     (Eq 2.17) These are then solved:
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$$  \displaystyle \frac{9}{2}=C_{2} $$     (Eq 2.18)
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$$  \displaystyle -\frac{13}{2}=C_{1} $$     (Eq 2.19) Plugging back in we get:
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$$  \displaystyle y(x)=-\frac{13}{2}x+\frac{9}{2}x^{2} $$     (Eq 2.20)
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=Problem 5.3: Euler L2_ODE_VC=

From the lecture slide Mtg 27

5.3.1.Given
Characteristic equation is given in meeting 27-1 as:

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$$  \displaystyle \left ( r-\lambda \right )^2=0 $$     (Eq 3.1)
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Where $$\displaystyle \lambda=5$$

The form of equation with unknown coeffficients to be determined

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$$  \displaystyle a_{2}x^{2}{y}''+a_{1}x{y}'+a_{0}y=0 $$     (Eq 3.2)
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And the given first homogeneous solution :

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$$  \displaystyle y_{1}=x^{\lambda} $$     (Eq 3.3)
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5.3.2.1. Homogeneous equation with varying coefficients

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$$  \displaystyle \left ( r-5 \right )^{2}=0 $$     (Eq 3.4)
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$$  \displaystyle r^{2}-10r+25=0 $$     (Eq 3.5)
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$$  \displaystyle y=x^{r} $$     (Eq 3.6)
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$$  \displaystyle {y}'=rx^{(r-1)} $$     (Eq 3.7)
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$$  \displaystyle {y}''=r(r-1)x^{(r-2)} $$     (Eq 3.8) Subtituting the trial solution and its derivatives into Eq 3.2 yields :
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$$  \displaystyle a_{2}x^{2}r(r-1)x^{(r-2)}+a_{1}x^{1}rx^{(r-1)}+a_{0}x^{r}=0 $$     (Eq 3.9)
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$$  \displaystyle x^{r}\left \lfloor a_{2}r(r-1)+a_{1}r+a_{0} \right \rfloor=0 $$     (Eq 3.10)
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$$  \displaystyle a_{2}r^{2}+(a_{1}-a_{2})r+a_{0}=0 $$     (Eq 3.11)
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Comparing with Eq 3.5 we can see that :

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$$  \displaystyle a_{2}=1 ,a_{1}=-9 ,a_{0}=25 $$     (Eq 3.12)
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$$  \displaystyle x^{2}{y}''-9x{y}'+25y=0 $$     (Eq 3.13)
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Let the general solution of homogeneous solution to be,
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$$  \displaystyle y=U(x)x^{5} $$     (Eq 3.14)
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$$  \displaystyle {y}'={U(x)}'x^{5}+5x^{4}U(x) $$     (Eq 3.15)
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$$  \displaystyle {y}={U(x)}x^{5}+10x^{4}{U(x)}'+20x^{3}U(x) $$     (Eq 3.16)
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$$  \displaystyle x^{2}\left [ {U(x)}''x^{5}+10x^{4}{U(x)}'+20x^{3}U(x) \right ]-9x\left [ x^{5}{U(x)}'+5x^{4}U(x) \right ]+25x^{5}U(x)=0 $$     (Eq 3.17)
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$$  \displaystyle {U(x)}''x^{5}+{U(x)}'\left [ 10x^{4}-9x^{4} \right ]+U(x)\left [ 20x^{3}-45x^{3}+25x^{3} \right ]=0 $$     (Eq 3.18)
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$$  \displaystyle {U(x)}''+\frac{1}{x}{U(x)}'=0 $$     (Eq 3.19)
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Now we can use the variable separation to reduce the order.See:Separation Of Variables To decrease our equation order.

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$$  \displaystyle {Z}'+\frac{1}{x}Z=0 $$     (Eq 3.20)
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Where $$ \displaystyle Z={U}'$$

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$$  \displaystyle \int\frac{dZ}{Z}=-\int \frac{dx}{x} $$     (Eq 3.21)
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$$  \displaystyle logz=logk_{1}-logx $$     (Eq 3.22)
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$$  \displaystyle z=\frac{k_{1}}{x} $$     (Eq 3.23)
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$$  \displaystyle U=\int Z dx+k_{2} $$     (Eq 3.24)
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$$  \displaystyle U=k_{1}logx+k_{2} $$     (Eq 3.25)
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$$  \displaystyle y=U(x)y_{1}(x)=\left ( k_{1}logx+k_{2} \right )x^{5} $$     (Eq 3.26)
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$$  \displaystyle y_{2}(x)=k_{1}x^{5}logx $$     (Eq 3.27)
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5.3.2.2Homogeneous function with constant coefficients
Euler L2_ODE_CC is given as follows ;
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$$  \displaystyle b_{2}{y}''+b_{1}{y}'+b_{0}y=0 $$     (Eq 3.28) Let the trial solution to be
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$$  \displaystyle y=e^{rx} $$     (Eq 3.29)
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$$  \displaystyle {y}'=re^{rx} $$     (Eq 3.30)
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$$  \displaystyle {y}''=r^{2}e^{rx} $$     (Eq 3.31)
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$$  \displaystyle e^{rx}\left [ b_{2}r^{2}+b_{1}r+b_{0} \right ]=0 $$     (Eq 3.32)
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$$  \displaystyle b_{2}r^{2}+b_{1}r+b_{0}=r^{2}-10r+25 $$     (Eq 3.33) It can obciously seen that the unknowns are
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$$  \displaystyle b_{2}=1 ,b_{1}=-10 ,b_{0}=25 $$     (Eq 3.34)
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$$  \displaystyle y=U(x)e^{5x} $$     (Eq 3.35)
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$$  \displaystyle {y}'={U(x)}'e^{5x}+5U(x)e^{5x} $$     (Eq 3.36)
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$$  \displaystyle {y}=e^{5x}{U(x)}+10e^{5x}{U(x)}'+25e^{5x}U(x) $$     (Eq 3.37)
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$$  \displaystyle e^{5x}{U(x)}''+10e^{5x}{U(x)}'+25e^{5x}U(x)-10\left [ e^{5x}{U(x)}'+5e^{5x}U(x) \right ]+25e^{5x}U(x)=0 $$     (Eq 3.38)
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$$  \displaystyle {U(x)}''+{U(x)}'\left [ 10-10 \right ]+U(x)\left [ 25+25-50 \right ]=0 $$     (Eq 3.39)
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$$  \displaystyle {U(x)}''=0 $$     (Eq 3.40)
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$$  \displaystyle U=k_{1}x+k_{2} $$     (Eq 3.41)
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$$  \displaystyle y(x)=U(x)y_{1}(x)=k_{1}xe^{5x}+k_{2}e^{5x} $$     (Eq 3.42)
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$$  \displaystyle y_{2}=k_{1}xe^{5x} $$     (Eq 3.43)
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= Problem 5.4 Particular Solution by using Variation of Parameters Method =

From the lecture slide Mtg 27-2

Given
Linear, nonhomogeneous, first order differential equation

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$$ \displaystyle \frac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$$ (Eq 4.1)
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Find
Particular solution using $$y\displaystyle \left( x \right)=A\left( x \right){{y}_{h}}(x)$$

To find homogeneous solution:
Homogeneous differential equation

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$$ \displaystyle \frac{d{{y}_{h}}}{dx}+P\left( x \right){{y}_{h}}=0$$ (Eq 4.2)
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and the solution of homogeneous differential equation is

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where A is a constant of integration (King, page 512).

To find particular solution:
Assuming

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$$ \displaystyle \left( x \right)=A\left( x \right){{y}_{h}}(x)=A(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}$$ (Eq 4.3)
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is solution of nonhomogeneous ODE (Eq4.1).

We determine A(x) by substituting Eq4.3 and its first derivative into Eq4.1. Differentiating Eq4.3, we obtain,

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$$ \displaystyle \frac{d}{dx}=={A}'(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}-A(x)P(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}$$ (Eq 4.4)
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Now $$\displaystyle $$ must satisfy (Eq4.1). We need to substitute $$\displaystyle $$ and $$\displaystyle $$ according to Eq4.3 and Eq4.4 into Eq4.1.

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$$ \displaystyle \frac{d}{dx}+P\left( x \right){{y}_{p}}=Q\left( x \right)$$ (Eq 4.5)
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$$ \displaystyle \left[ {A}'(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}-A(x)P(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}} \right]+P(x)\left( A(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}} \right)=Q(x)$$ (Eq 4.6)
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Second term and third term in the left side of Eq4.7 eliminate each other and we obtain

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$$ \displaystyle {A}'(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}=Q(x)$$ (Eq 4.7)
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We rearrange Eq4.7

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$$ \displaystyle {A}'(x)=Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}$$ (Eq 4.8)
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If we integrate both sides Eq4.8, A(x) will be

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$$ \displaystyle A(x)=\int{Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}dx}$$ (Eq 4.9)
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If we put Eq4.9 into Eq4.3 in order to get the particular solution, we obtain

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$$ \displaystyle \left( x \right)=A(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}={{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}\left( \int{Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}dx+C} \right)$$ (Eq 4.10)
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Eliminate the constant $$\displaystyle C$$ we can derive particular solution,

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$$ \displaystyle {{y}_{P}}={{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}\int Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}dx$$ (Eq 4.11)
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Now we can solve homework equation in Mtg10-4 as an example,

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$$ \displaystyle \frac{dy}{dx}+xy=2x+3$$ (Eq 4.12)
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Thus we have,

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& P\left( x \right)=x \\ & Q\left( x \right)=2x+3 \\ \end{align} \right.$$

If we solve the homogenous equation for $$\displaystyle P\left( x \right)=x$$,

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$$ \displaystyle {{y}_{h}}=A{{e}^{-\frac{2}}}$$
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then we have(see Kreyszig's book ),

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$$ \displaystyle ={{e}^{-\int\limits_ – ^{x}{tdt}}}\int\limits_ – ^{x}{(2s+3){{e}^{\int\limits_ – ^{s}{tdt}}}ds}={{e}^{-\frac{2}}}\left[ \int\limits_ – ^{x}{2s{{e}^{^{\frac{2}}}}ds}+\int\limits_ – ^{x}{3{{e}^{^{\frac{2}}}}ds} \right]$$ (Eq 4.13)
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Integrating inside of bracket by using Wolfram Alpha, we obtain $$\displaystyle $$ shown below

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$$ \displaystyle =2+3\sqrt{\frac{\pi }{2}}.{{e}^{-\frac{2}}}.erfi\left( \frac{x}{\sqrt{2}} \right)+{{C}_{1}}{{e}^{-\frac{2}}}$$ (Eq 4.14)
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where erfi(x) is the imaginary error function.

Then we extract particular solution from above,

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$$ \displaystyle {{y}_{p}}=2+3\sqrt{\frac{\pi }{2}}.{{e}^{-\frac{2}}}.erfi\left( \frac{x}{\sqrt{2}} \right)$$ (Eq 4.15)
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= Problem 5.5 Solving Non-homogeneous Linear Second Order ODE with Constant Coefficients=

From the lecture slide Mtg 28 and Mtg 29-1

Given
The spring-dashpot system has the Equation of Motion like follows:

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$$ \displaystyle {{a}_{2}}{y}''+{{a}_{1}}{y}'+{{a}_{0}}y=f(t)$$ (Eq 5.1)
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where $$\displaystyle y$$ is a function of $$\displaystyle t$$, and $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{0}}$$ are all constants making Eq 5.1 a linear second-order ODE with constant coeffiients.

Find
(1)Find PDEs for integrating factor $$\displaystyle H(t,y)$$ using condition 2nd of exactness stated in Mtg15

(2)Solve Eq 5.1 using trial solution for integrating factor such that $$\displaystyle h(t)={{e}^{\alpha t}}$$ where $$\displaystyle \alpha $$ is some constant. .

(1) Finding PDEs for integrating factor $$\displaystyle H(t,y)$$
Multiply $$\displaystyle H(t,y)$$ to both sides of the Eq 5.1,

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$$ \displaystyle H(t,y){{a}_{2}}{y}''+H(t,y){{a}_{1}}{y}'+H(t,y){{a}_{0}}y=H(t,y)f(t)$$ (Eq 5.2)
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If we assume firstly $$\displaystyle H(t,y)$$ to be a function of only $$\displaystyle t$$(see Mtg 10), then we can just find some $$\displaystyle H(t)$$ to make the L.H.S of Eq 5.2 exact in order to change its form as,

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$$ \displaystyle \frac{d\phi }{dt}=H(t)f(t)$$ (Eq 5.3)
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then we can integrate both sides to reduce the order of Eq 5.1. However, if the assumption "$$\displaystyle H(t,y)$$ is a function of only $$\displaystyle t$$" is not satisfid, we can't integrate the R.H.S of Eq 5.3. In this case, we should move $$\displaystyle f(t)$$ from R.H.S to L.H.S and try to find any $$\displaystyle H(t,y)$$ to make the whole thing exact.

Thus let

 * $$\displaystyle F=H(t){{a}_{2}}{y}+H(t){{a}_{1}}{y}'+H(t){{a}_{0}}y=f(t,y,p){y}+g(t,y,p)$$

in which,

 * $$\displaystyle \begin{align}

& f=H(t){{a}_{2}} \\ & g=H(t){{a}_{1}}{y}'+H(t){{a}_{0}}y \\ & p={y}' \\ \end{align}$$

Then we have,

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$$ \displaystyle \begin{align} & {{f}_{tt}}={{a}_{2}}{{H}_{tt}} \\ & {{f}_{ty}}={{a}_{2}}{{H}_{ty}} \\ & {{f}_{tp}}={{f}_{yp}}=0 \\ & {{f}_{y}}={{a}_{2}}{{H}_{y}} \\ & {{f}_{yy}}={{a}_{2}}{{H}_{yy}} \\ & {{g}_{tp}}={{a}_{1}}{{H}_{t}} \\ & {{g}_{yp}}={{a}_{1}}{{H}_{y}} \\ & {{g}_{pp}}=0 \\ & {{g}_{y}}={{a}_{1}}p{{H}_{y}}+{{a}_{0}}H+{{a}_{0}}y{{H}_{y}} \\ \end{align}$$ (Eqs 5.4)
 *

 * }

Substituting Eqs 5.4 into (1)&(2) in Mtg15, we have,

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

$$ \displaystyle {{a}_{2}}{{H}_{tt}}+2p{{a}_{2}}{{H}_{ty}}+{{p}^{2}}{{a}_{2}}{{H}_{yy}}={{a}_{1}}{{H}_{t}}-{{a}_{0}}H-{{a}_{0}}y{{H}_{y}}$$ (Eq 5.5)
 *

 * }

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

$$ \displaystyle 2{{a}_{2}}{{H}_{y}}=0$$ (Eq 5.6)
 *

 * }

From Eq 5.6 we can see that the basic assumption is satisfied and if we substitute Eq 5.6 in to Eq 5.5, we have,

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

$$ \displaystyle {{a}_{2}}{{H}_{tt}}-{{a}_{1}}{{H}_{t}}+{{a}_{0}}H=0$$ (Eq 5.7)
 *

 * }

or

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

$$ \displaystyle {{a}_{2}}{H}''-{{a}_{1}}{H}'+{{a}_{0}}H=0$$ (Eq 5.8)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Since the integrating factor $$\displaystyle H(t,y)$$ turns out to be $$\displaystyle H(t)$$, the PDE governing $$\displaystyle H(t,y)$$ become an ODE given by Eq 5.8.

(2) Solve Eq 5.1 using trial solution for integrating factor $$\displaystyle h(t)={{e}^{\alpha t}}$$
Let's try an integrating factor like $$\displaystyle h(t)={{e}^{\alpha t}}$$ where $$\displaystyle \alpha$$ is a constant.

Multiply Eq 5.1 by $$\displaystyle h(t)={{e}^{\alpha t}}$$, we have,

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

$$ \displaystyle {{e}^{\alpha t}}\left( {{a}_{2}}{y}''+{{a}_{1}}{y}'+{{a}_{0}}y \right)={{e}^{\alpha t}}f(t)$$ (Eq 5.9)
 *

 * }

In order to reduce the order, we integrating Eq 5.9 according to $$\displaystyle t$$ and assume that the result can be written in such form:

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

$$ \displaystyle {{e}^{\alpha t}}\left( {{{\bar{a}}}_{1}}{y}'+{{{\bar{a}}}_{2}}y \right)=\int{{{e}^{\alpha t}}f(t)dt}$$ (Eq 5.10)
 *

 * }

where $$\displaystyle {{{\bar{a}}}_{1}}$$ and $$\displaystyle {{{\bar{a}}}_{2}}$$ are assumed to be constants.

(2.1) Determine $$\displaystyle {{{\bar{a}}}_{1}}$$ and $$\displaystyle {{{\bar{a}}}_{1}}$$ according to $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$ and $$\displaystyle {{a}_{0}}$$
Differentiate Eq 5.10 we have,

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

$$ \displaystyle {{e}^{\alpha t}}\left[ {{{\bar{a}}}_{1}}{y}''+(\alpha {{{\bar{a}}}_{1}}+{{{\bar{a}}}_{0}}){y}'+\alpha {{{\bar{a}}}_{0}}y \right]={{e}^{\alpha t}}f(t)$$ (Eq 5.11)
 *

 * }

Compare 5.11 and 5.9 we have,

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

$$ \displaystyle \begin{align} & {{{\bar{a}}}_{1}}={{a}_{2}} \\ & {{{\bar{a}}}_{0}}={{a}_{1}}-\alpha {{a}_{2}} \\ & {{{\bar{a}}}_{0}}={{a}_{0}}/\alpha \\ \end{align}$$ (Eqs 5.12)
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 *

 * }

from which we can see that the validity of Eq 5.10 relies on the value of $$\displaystyle \alpha $$.

(2.2) Find quadratic equation for $$\displaystyle \alpha $$
From Eqs 5.12 we have,

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

$$ \displaystyle \begin{align} & {{a}_{1}}-\alpha {{a}_{2}}={{a}_{0}}/\alpha \\ \Rightarrow & {{a}_{2}}{{\alpha }^{2}}-{{a}_{1}}\alpha +{{a}_{0}}=0 \\ \end{align}$$ (Eq 5.13)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

(2.3) Derive a reduced - order equation from Eq 5.10
From Eq 5.10 we have,

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

$$ \displaystyle {y}'+\beta y=\frac{1}\int{{{e}^{\alpha t}}f(t)dt}$$ (Eq 5.14)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

where $$\displaystyle \beta ={{{\bar{a}}}_{0}}/{{{\bar{a}}}_{1}}$$

(2.4) Solve Eq 5.14 using Integrating Factor Method
From Mtg10 we know the integrating factor(for Eq 5.14) is,

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

$$ \displaystyle i(t)=\exp \int{\beta dt}={{e}^{\beta t}}$$ (Eq 5.15)
 *

 * }

and the solution is:

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

$$ \displaystyle \begin{align} y & =\frac{1}{i(t)}\int_ – ^{t}{\frac{i(\tau )}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \\ & =\frac\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \end{align}$$ (Eq 5.16)
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

(2.5) Show the relation between $$\displaystyle \alpha $$ and $$\displaystyle \beta $$
Since,

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

$$ \displaystyle \begin{align} & \beta ={{{\bar{a}}}_{0}}/{{{\bar{a}}}_{1}} \\ & {{{\bar{a}}}_{0}}={{a}_{1}}-\alpha {{a}_{2}} \\ & {{{\bar{a}}}_{0}}={{a}_{0}}/\alpha \\ & {{{\bar{a}}}_{1}}={{a}_{2}} \\ \end{align}$$ (Eqs 5.17)
 *

 * }

So we have,

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

$$ \displaystyle \begin{align} & \alpha \beta =\frac \\ & \alpha +\beta =\frac \\ \end{align}$$ (Eq 5.18)
 *

 * }

which means $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ are the two roots of,

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

$$ \displaystyle {{\lambda }^{2}}-(\alpha +\beta )\lambda +\alpha \beta =0$$
 *

 * }

or

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

$$ \displaystyle {{\lambda }^{2}}-\frac\lambda +\frac=0$$ (Eq 5.19)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

which is identical to Eq 5.13. And from Eq 5.17~Eq 5.19 we can conclude that $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ are inseparable pairs so that we can't distiguish them from each other through the their governing equation and the sequense of two integrating factor $$\displaystyle h(t)$$ and $$\displaystyle i(t)$$ is arbitrary. We can also conclude that the final structure of solution must be symetric with respect to $$\displaystyle \alpha $$ and $$\displaystyle \beta $$, meaning exchanging the position of $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ won't change the result.

When $$\displaystyle \alpha \ne \beta $$
Eq 5.16 can be expanded as,

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

$$ \displaystyle \begin{align} y & =\frac\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \\ & =\frac\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau }+{{k}_{1}} \right)d\tau } \\ & =\frac\left[ \int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau } \right)d\tau }+\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}{{k}_{1}}d\tau } \right] \\ & =\frac\left[ \int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}+{{k}_{2}}+\frac{\beta -\alpha }{{e}^{(\beta -\alpha )t}} \right] \\ & =\frac{(\beta -\alpha ){_{1}}}{{e}^{-\alpha t}}+\frac{{e}^{-\beta t}}+\frac{{{{\bar{a}}}_{1}}}\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \end{align}$$ (Eq 5.20)
 *

 * }

Note that Eq 5.20 is under the assumption that $$\alpha \ne \beta $$.

After conbination of constants, we have,

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

$$ \displaystyle y={{C}_{1}}{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\beta t}}+\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}$$ (Eq 5.21)
 *

 * }

In which we can derive,

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

$$ \displaystyle \begin{align} & y_{H}^{1}={{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\beta t}} \\ & {{y}_{P}}=\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \\ \end{align}$$
 *

 * }

However, $$\displaystyle {{y}_{P}}$$ can be furthuer deduced as,

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

$$ \displaystyle \begin{align} {{y}_{P}} & =\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\int{\left( \int{{{e}^{\alpha t}}f(t)dt} \right)d\left( {{e}^{(\beta -\alpha )t}} \right)} \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\left[ {{e}^{(\beta -\alpha )t}}\int{{{e}^{\alpha t}}f(t)dt}-\int{{{e}^{(\beta -\alpha )t}}{{e}^{\alpha t}}f(t)dt} \right] \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\alpha t}}f(t)dt}-\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\beta t}}f(t)dt} \end{align}$$
 *

 * }

This proved my statement "the final structure of solution must be symetric with respect to $$\displaystyle \alpha $$ and $$\displaystyle \beta $$".

In sum we have (when $$\displaystyle \alpha \ne \beta $$),

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

$$ \displaystyle \begin{align} & y={{C}_{1}}y_{H}^{1}+{{C}_{2}}y_{H}^{2}+{{y}_{P}} \\ & y_{H}^{1}={{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\beta t}} \\ & {{y}_{P}}=\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\alpha t}}f(t)dt}-\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\beta t}}f(t)dt} \\ \end{align}$$ (Eq 5.22)
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

When $$\displaystyle \alpha =\beta $$
Substitute the relation $$\displaystyle \alpha =\beta $$ into Eq 5.16,

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

$$ \displaystyle \begin{align} y= & \frac\int_ – ^{t}{{{e}^{(\alpha -\alpha )\tau }}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \\ & =\frac\int_ – ^{t}{\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau }+{{k}_{1}} \right)d\tau } \\ & =\frac\left[ \int_ – ^{t}{\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau } \right)d\tau }+\int_ – ^{t}{{{k}_{1}}d\tau } \right] \\ & =\frac\left[ \int{\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}+{{k}_{2}}+{{k}_{1}}t \right] \\ & =\fract{{e}^{-\alpha t}}+\frac{{e}^{-\alpha t}}+\frac\int{\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \\ & =\fract{{e}^{-\alpha t}}+\frac{{e}^{-\alpha t}}+\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right) \end{align}$$ (Eq 5.20a)
 *

 * }

After conbination of constans we have

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

$$ \displaystyle y={{C}_{1}}t{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\alpha t}}+\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right)$$ (Eq 5.21a)
 *

 * }

In which we can derive,

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

$$ \displaystyle \begin{align} & y_{H}^{1}=t{{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\alpha t}} \\ & {{y}_{P}}=\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right) \\ \end{align}$$
 *

 * }

In sum we have (when $$\displaystyle \alpha =\beta $$),

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

 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

$$ \displaystyle \begin{align} & y={{C}_{1}}y_{H}^{1}+{{C}_{2}}y_{H}^{2}+{{y}_{P}} \\ & y_{H}^{1}=t{{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\alpha t}} \\ & {{y}_{P}}=\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right) \\ \end{align}$$ (Eq 5.22a)
 *

 * }

(2.7) Verify particular solution according to table in Mtg20, Fall2009
If $$\displaystyle \alpha \ne \beta$$, then from Eq 5.22 we know,

 * $$\displaystyle {{y}_{P}}=\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\alpha t}}f(t)dt}-\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\beta t}}f(t)dt}$$

If we substitute $$\displaystyle f(t)={{t}^{2}}$$ into the above equation, we have,

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

$$ \displaystyle \begin{align} {{y}_{P}} & =\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\alpha t}}f(t)dt-\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\beta t}}f(t)dt \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\alpha t}}{{t}^{2}}dt-\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\beta t}}{{t}^{2}}dt \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\left( \frac{{{\alpha }^{2}}{{t}^{2}}-2\alpha t+2} \right)-\frac{(\beta -\alpha ){{a}_{2}}}\left( \frac{{{\beta }^{2}}{{t}^{2}}-2\beta t+2} \right) \\ & =\frac{{{\alpha }^{2}}{{\beta }^{2}}{{t}^{2}}-(2{{\alpha }^{2}}\beta +2\alpha {{\beta }^{2}})t+2{{\alpha }^{2}}+2{{\beta }^{2}}+2\alpha \beta } \\ & =\frac{{{(\alpha \beta )}^{2}}{{t}^{2}}-2(\alpha +\beta )t+2{{(\alpha +\beta )}^{2}}-2\alpha \beta } \\ & =\frac{\frac{{{a}_{0}}^{2}}{{{a}_{2}}^{2}}{{t}^{2}}-2\frac{{{a}_{2}}^{2}}t+2\frac{{{a}_{1}}^{2}}{{{a}_{2}}^{2}}-2\frac{{{a}_{2}}^{2}}}{\frac{{{a}_{0}}^{3}}{{{a}_{2}}^{3}}{{a}_{2}}} \\ & =\frac{1}{{t}^{2}}-\frac{2{{a}_{1}}}{{{a}_{0}}^{2}}t+\frac{2{{a}_{1}}^{2}-2{{a}_{0}}{{a}_{2}}}{{{a}_{0}}^{3}} \\ \end{align}$$ (Eq 5.23)
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 *

 * }

in which we utilize,

 * $$\displaystyle \begin{align}

\int{{{e}^{\alpha t}}{{t}^{2}}dt} & =\frac{1}{\alpha }\int{{{t}^{2}}d{{e}^{\alpha t}}} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{1}{\alpha }\int{2t{{e}^{\alpha t}}dt} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{2}\int{td}{{e}^{\alpha t}} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{2}t{{e}^{\alpha t}}+\frac{2}\int{{{e}^{\alpha t}}dt} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{2}t{{e}^{\alpha t}}+\frac{2}{{e}^{\alpha t}} \\ & =\frac{{{\alpha }^{2}}{{t}^{2}}-2\alpha t+2}{{e}^{\alpha t}} \\ \end{align}$$

Compare Eq 5.23 with Mtg20 slide of Fall 2009, we can see that Eq 5.23 has the same form with that in the slide which is a polynominal with at least order 2 according to $$\displaystyle t$$.

If $$\displaystyle \alpha=\beta$$, then from Eq 5.22a we know,

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

$$ \displaystyle \begin{align} & {{y}_{P}}=\frac\left( t\int {{e}^{\alpha t}}{{t}^{2}}dt-\int {{e}^{\alpha t}}{{t}^{3}}dt \right) \\ & =\frac\left( \frac{{{e}^{\alpha t}}({{\alpha }^{2}}{{t}^{3}}-2\alpha {{t}^{2}}+2t)}-\frac{{{e}^{\alpha t}}({{\alpha }^{3}}{{t}^{3}}-3{{\alpha }^{2}}{{t}^{2}}+6\alpha t-6)} \right) \\ & =\frac{{{\alpha }^{2}}{{t}^{2}}-4\alpha t+6} \\ & =\frac{\alpha \beta }{{t}^{2}}-\frac{2(\alpha +\beta )}t+\frac{2{{(\alpha +\beta )}^{2}}-2\alpha \beta } \\ & =\frac{1}{{t}^{2}}-\frac{2{{a}_{1}}}{{{a}_{0}}^{2}}t+\frac{2{{a}_{1}}^{2}-2{{a}_{0}}{{a}_{2}}}{{{a}_{0}}^{3}} \\ \end{align}$$ (Eq 5.23a)
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 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 *

 * }

We can see that Eq 5.23a still has the same form with Eq 5.23 and that in the slide Mtg20. Thus the particular solution is verified.

(2.8) Solve L2-ODE-CC with $$\displaystyle f(t)=\exp (-{{t}^{2}})$$
If $$\displaystyle f(t)=\exp (-{{t}^{2}})$$, then Eqs 5.22 and Eqs 5.22a we derive

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

$$ \displaystyle \begin{align} & y_{H}^{1}=\left\{ \begin{align} & {{e}^{-\alpha t}}....................................................................................(\alpha \ne \beta ) \\ & t{{e}^{-\alpha t}}...................................................................................(\alpha =\beta ) \\ \end{align} \right. \\ & y_{H}^{2}={{e}^{-\beta t}} \\ & {{y}_{P}}=\left\{ \begin{align} & \frac{\sqrt{\pi }\left[ {{e}^{\frac{4}-\alpha t}}erf(\frac{\alpha }{2}-t)-{{e}^{\frac{4}-\beta t}}erf(\frac{\beta }{2}-t) \right]}{2{{a}_{2}}(\alpha -\beta )}........................(\alpha \ne \beta ) \\ & -\frac{1}{2{{a}_{2}}}\sqrt{\pi }{{e}^{\frac{4}-\alpha t}}\left[ (t-\frac{\alpha }{2})erf(\frac{\alpha }{2}-t)-\frac{1}{\sqrt{\pi }}{{e}^{-\frac{4}}} \right].........(\alpha =\beta ) \\ \end{align} \right. \\ \end{align}$$ (Eq 5.24)
 *

 * }

It has been mentioned that governing equation of $$\displaystyle \alpha $$ is indentical to Eq 5.19, and it has also been mentioned that the sequence of $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ is arbitrary.

Case 1: $$\displaystyle (\gamma +1)(\gamma -2)=0$$
From the characteristic equation we can get information of $$\displaystyle \alpha $$ and $$\displaystyle \beta $$, which are the ratio of $$\displaystyle {{a}_{0}}$$ and $$\displaystyle {{a}_{1}}$$ with respect to $$\displaystyle {{a}_{2}}$$. However we can't derive information of $$\displaystyle {{a}_{2}}$$. In fact $$\displaystyle {{a}_{2}}$$ has nothing to do with characteristic equation since in the homogeneous case $$\displaystyle {{a}_{2}}$$ can always be the divisor of both sides(the R.H.S is zero so $$\displaystyle {{a}_{2}}$$ disappeared only left some ratios) and has no effect on the results if $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ are fixed. But in the non-homogeneous case $$\displaystyle {{a}_{2}}$$ can be seen as part of the R.H.S(non-zero) and do have influence on the final result. As we can't get enough information to determine the value of $$\displaystyle {{a}_{2}}$$, we take it as 1:

 * $$\displaystyle {{a}_{2}}=1$$

From the characteristic equation we have the following L2-ODE-CC by refering Eq 5.13 and Eq 5.1:

 * $$\displaystyle {y}''-{y}'-2y={{e}^{-{{t}^{2}}}}$$

then

 * $$\displaystyle \begin{align}

& \left\{ \begin{align} & \alpha +\beta =\frac=-1 \\ & \alpha \beta =\frac=-2 \\ \end{align} \right. \\ & \Rightarrow \left\{ \begin{align} & \alpha =1 \\ & \beta =-2 \\ \end{align} \right. \\ \end{align}$$

Substitute these into Eq 5.24 we have the solution as,

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

$$ \displaystyle y={{C}_{1}}{{e}^{-t}}+{{C}_{2}}{{e}^{2t}}+\frac{\sqrt{\pi }}{6}\left[ {{e}^{0.25-t}}erf(0.5-t)-{{e}^{1+2t}}erf(-1-t) \right]$$ (Eq 5.25)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

where $$\displaystyle {{C}_{1}}$$ and $$\displaystyle {{C}_{2}}$$ are some constants.

Case 2: $$\displaystyle {{(\gamma -4)}^{2}}=0$$
From the characteristic equation we have the following L2-ODE-CC by refering Eq 5.13 and Eq 5.1:

 * $$\displaystyle {y}''-8{y}'+16y={{e}^{-{{t}^{2}}}}$$

then,

 * $$\displaystyle \begin{align}

& \left\{ \begin{align} & \alpha +\beta =\frac=-8 \\ & \alpha \beta =\frac=16 \\ \end{align} \right. \\ & \Rightarrow \alpha =\beta =-4 \\ \end{align}$$

Substitute this into Eq 5.24 we have,

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

$$ \displaystyle y={{C}_{1}}t{{e}^{4t}}+{{C}_{2}}{{e}^{4t}}-\frac{\sqrt{\pi }}{2}{{e}^{4+4t}}\left[ (t+2)erf(2+t)-\frac{1}{\sqrt{\pi }}{{e}^{-{{(2+t)}^{2}}}} \right]$$ (Eq 5.26)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

where $$\displaystyle {{C}_{1}}$$ and $$\displaystyle {{C}_{2}}$$ are some constants.

= Problem 5.6: Show Agreement of Particular Integrals =

From the lecture slide Mtg 30-1

Given
Show that the following equation:

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

$$ \displaystyle

{{y}_{p}}\left( x \right)={{u}_{1}}\left( x \right)\int{\frac{1}{h\left( x \right)}\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)\,dx}

$$ (Eq 6.1)
 *

 * }

is equivalent to the following particular integral found in the Differential Equations text book by A.C.King:
 * {| style="width:100%" border="0"

$$ \displaystyle

{{y}_{p}}\left( x \right)=\oint^x{f\left( s \right)\left\{ \frac{{{u}_{1}}(s) {{u}_{2}}(x)-{{u}_{1}}(x) {{u}_{2}}(s)}{W(s)} \right\}}\,ds

$$ (Eq 6.2)
 *

 * }

given that the following equations holds true:
 * {| style="width:100%" border="0"

$$ \displaystyle

W={{u}_{1}} {{u}_{2}}'-{{u}_{1}}' {{u}_{2}}=\left| \begin{matrix} {{u}_{1}} & {{u}_{2}} \\ {{u}_{1}}' & {{u}_{2}}' \\ \end{matrix} \right|

$$ (Eq 6.3)
 *

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

$$ \displaystyle

(\frac{u_{2}}{u_{1}})'=\frac{1}{h}

$$ (Eq 6.4)
 *

 * }

And considering the following conversion between the notation used in the lectures and A.C. King's notation:
 * {| style="width:100%" border="0"

$$ \displaystyle

\oint^x{f\left( s \right)}\,ds=\int^x{f\left(s\right)}\,ds =\int{f\left(x\right)}\,dx

$$  w/o constant (Eq 6.5)
 *

 * }

Solution
The first step to tackling this problem requires the application of the Integration by Parts formula to equation 6.1

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

$$\displaystyle \int w\,dv=wv-\int v\,dw $$

(Eq 6.6)
 *

 * }

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

$$\displaystyle

{{y}_{p}}\left( x \right)={{u}_{1}}\left( x \right)\int{\underbrace{\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)}_{w}\underbrace{\frac{1}{h\left( x \right)}\,dx}_{dv}}$$

(Eq 6.7) The remaining unknowns in the right hand side of equation 6.6 are shown below:
 *

 * }

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

$$\displaystyle dw=h\left( x \right) f\left( x \right)dx

$$

(Eq 6.8)
 *

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

$$\displaystyle v= \int {\frac{1}{h\left( x \right)}}\,dx

$$

(Eq 6.9) Equations 6.7, 6.8 and 6.9 can be combined and rewritten into the form shown in equation 6.5 to yield the following:
 *

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

$$\displaystyle \begin{align} {{y}_{p}}\left( x \right)= & {{u}_{1}}\left( x \right)\left[ \left( \mathop{\oint }^{x}h\left( s \right)f\left( s \right)\,ds \right)\int \frac{1}{h\left( x \right)}\,dx \right. \\ & \left. -\mathop{\oint }^{x}\left( \mathop{\oint }^{x}\frac{1}{h\left( s \right)}\,ds \right)h\left( s \right)f\left( s \right)\,ds \right] \\ \end{align}$$

(Eq 6.10)
 *

 * }

The relation shown in equation 6.4 can be used to simplify equation 6.10 as follows:
 * {| style="width:100%" border="0"

$$\displaystyle \begin{align} {{y}_{p}}\left( x \right)= & {{u}_{1}}\left( x \right)\left[ \left( \mathop{\oint }^{x}\frac{u_{1}^{2}\left( s \right)}{{{u}_{{{2}'}}}\left( s \right){{u}_{1}}\left( s \right)-{{u}_{2}}\left( s \right){{u}_}\left( s \right)}f\left( s \right)\,ds \right)\frac{{{u}_{2}}\left( x \right)}{{{u}_{1}}\left( x \right)} \right. \\ & \left. -\left[ \mathop{\oint }^{x}\frac{{{u}_{2}}\left( s \right)}{{{u}_{1}}\left( s \right)}\frac{{{u}_{1}}^{2}\left( s \right)}{{{u}_{{{2}'}}}\left( s \right){{u}_{1}}\left( s \right)-{{u}_{2}}\left( s \right){{u}_}\left( s \right)}f\left( s \right)\,ds \right] \right] \\ \end{align}$$ (Eq 6.11) Now, multiplying through by u1(x) and using the relation shown in equation 6.3 yields the following:
 *

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

$$ \displaystyle \begin{align} {{y}_{p}}\left( x \right) & =\left( \mathop{\oint }^{x}\frac{{{u}_{1}}\left( s \right){{u}_{2}}\left( x \right)}{W\left( s \right)}f\left( s \right)\,ds \right)-\left( \mathop{\oint }^{x}\frac{{{u}_{2}}\left( s \right){{u}_{1}}\left( x \right)}{W\left( s \right)}f\left( s \right)\,ds \right) \\ & =\mathop{\oint }^{x}f\left( s \right)\left\{ \frac{{{u}_{1}}(s){{u}_{2}}(x)-{{u}_{1}}(x){{u}_{2}}(s)}{W(s)} \right\}\,ds \\ \end{align}$$ (Eq 6.12)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

= Problem 5.7: Find Two Homogeneous Solutions using a Trial Solution =

From the lecture slide Mtg 31-1

Given
Find u1 and u2 of Equation 7.1 using the trial solution shown in equation 7.2:

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

$$ \displaystyle

(x +1) y'' - (2x+3)y' + 2y = 0

$$ (Eq 7.1)
 *

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

$$ \displaystyle

y(x)=e^{rx}

$$ (Eq 7.2)
 *

 * }

Solution
The first step is to differentiate the trial solution as follows:

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

$$ \displaystyle y(x) = e^{rx} $$ (Eq 7.3)
 *

 * }

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

$$ \displaystyle y'(x) = r e^{rx} $$ (Eq 7.4)
 *

 * }

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

$$ \displaystyle y''(x) = r^2 e^{rx} $$ (Eq 7.5)
 *

 * }

Next, plug the values into equation 7.1 to yield:

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

$$ \displaystyle e^{rx}\left[(x+1)r^2-(2x+3)r+2\right]=0 $$ (Eq 7.6) Since erx will not equal zero for any value of x, divide equation 7.6 by this value and factor out the x to yield:
 *

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

$$ \displaystyle x(r^2-2r)+(r-2)(r-1)=0 $$ (Eq 7.7) In order to satisfy equation 7.7 for any x, the following equation is solved:
 *

 * }

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

$$ \displaystyle r(r-2)=(r-2)(r-1)=0 $$ (Eq 7.8)
 *

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

$$ \displaystyle r=2 $$ (Eq 7.9) By plugging equation 7.9 into the trial solution in equation 7.3 we have obtained the first homogeneous solution:
 *

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

$$ \displaystyle u_1(x)=e^{2x} $$ (Eq 7.10)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

The second homogeneous solution is obtained by the following:

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

$$ \displaystyle u_2(x)=u_1(x)\int{u_1^{-2}(x)e^{-\int{a_1(x)}\,dx}}\,dx$$
 *

(Eq 7.11)
 * }

Equation 7.1 is rearranged into the following form:

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

$$ \displaystyle y''-\underbrace{\frac{2x+3}{x+1}}_{a_1}y'+\underbrace{\frac{2}{x+1}}_{a_0}y=0 $$ (Eq 7.12)
 *

 * }

The value for a1 shown in equation 7.12 is then plugged into equation 7.11 along with the first homogeneous solution to solve for the second solution as shown:

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

$$ \displaystyle u_2(x)=e^{2x}\int{e^{-4x}e^{\int{\frac{2x+3}{x+1}}\,dx}}\,dx$$ (Eq 7.13)
 *

 * }

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

$$ \displaystyle \Rightarrow u_2(x)=e^{2x}\int{(x+1)e^{-2x}}\,dx$$ (Eq 7.14)
 *

 * }

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

$$ \displaystyle \Rightarrow u_2(x)=\frac{-1}{4}(2x+3)$$ (Eq 7.15)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

= Problem 5.8 L2_ODE_VC (Homogeneous) by using Reverse Engineering =

From the lecture slide Mtg 31-4

Given
Trial Solution: $$\displaystyle y=\frac$$ and Characteristic Equation: $$\displaystyle {{r}^{2}}+3=0$$

Find
Find L2_ODE_VC according to a trial solution and the characteristic equation given above

Solution
Wolfram Alpha used to find the derivatives of the trial solution $$\displaystyle y$$

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

$$ \displaystyle \left[ y=\frac \right].{{a}_{0}}(x)$$ (Eq 8.1)
 *

 * }

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

$$ \displaystyle \left[ \frac{d}{dx}\left( \frac \right)=\left( r-\frac{2}{x} \right)\left( \frac \right) \right].{{a}_{1}}(x)$$ (Eq 8.2)
 *

 * }

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

$$ \displaystyle \left[ \frac{d{{x}^{2}}}\left( \frac \right)=\left( {{r}^{2}}-\frac{4r}{x}+\frac{6} \right)\left( \frac \right) \right].{{a}_{2}}(x)$$ (Eq 8.3)
 *

 * }

Adding Eq8.1, Eq8.2 and Eq8.3 each other with taking common parenthesis of $$\displaystyle y$$ will be equal to $$\displaystyle y$$ multiply by characteristic equation.

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

$$ \displaystyle \frac\left[ {{a}_{2}}\left( {{r}^{2}}-\frac{4r}{x}+\frac{6} \right)+{{a}_{1}}\left( r-\frac{2}{x} \right)+{{a}_{0}} \right]=\frac\left( {{r}^{2}}+3 \right)=0$$ (Eq 8.4)
 *

 * }

First expanding the inside of brackets and then rearranging and taking common parentheses of $$\displaystyle {{r}^{2}}$$, $$\displaystyle r$$ and 1, it will be

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

$$ \displaystyle {{r}^{2}}({{a}_{2}})+r\left( -\frac{4}{x}{{a}_{2}}+{{a}_{1}} \right)+1\left( \frac{6}{{a}_{2}}-\frac{2}{x}{{a}_{1}}+{{a}_{0}} \right)={{r}^{2}}+0.r+3$$ (Eq 8.5)
 *

 * }

From the equality of both sides, we find the values of coefficients $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{0}}$$.

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

$$ \displaystyle \begin{align} & {{a}_{2}}=1 \\ & {{a}_{1}}=\frac{4}{x} \\ & {{a}_{0}}=3+\frac{2} \\ \end{align}$$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

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

$$ \displaystyle {{r}^{2}}+3=0\Rightarrow {{r}_{1,2}}=\mp i\sqrt{3}$$ (Eq 8.6)
 *

 * }

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

$$ \displaystyle {{a}_{2}}(x){y}''+{{a}_{1}}(x){y}'+{{a}_{0}}(x)y=0$$ (Eq 8.7)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

is the general form of a second-order ordinary differential equation with coefficients.

If we substitute $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{0}}$$ into Eq8.7, we obtain

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

$$ \displaystyle {y}''+\frac{4}{x}{y}'+(3+\frac{2})y=0$$ (Eq 8.8)
 *

 * }

Multiply all equation with $$\displaystyle {{x}^{2}}$$ to eliminate denominators

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

$$ \displaystyle {{x}^{2}}{y}''+4x{y}'+\left( 3{{x}^{2}}+2 \right)y=0$$ (Eq 8.9)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

=Contributions Of Team Members=

 * Problem 1:    Solved by: Oztekin,Zou

Author: Oztekin

Proofread by: Oztekin, Dube, Zou

 * Problem 2:    Solved by: Dube

Author: Dube

Proofread by: Oztekin,Ismail

 * Problem 3:    Solved by: Oztekin

Author: Oztekin

Proofread by: Dube

 * Problem 4:    Solved by: Ismail,Yaakov

Author: Zou, Ismail

Proofread by: Zou,Oztekin

 * Problem 5:    Solved by: Zou, Oztekin

Author: Zou

Proofread by: Oztekin,Yaakow

 * Problem 6:    Solved by: Patterson

Author: Patterson

Proofread by: Yaakov, Oztekin

 * Problem 7:    Solved by: Patterson

Author: Patterson

Proofread by: Oztekin,Yaakov,Zou

 * Problem 8:    Solved by: Ismail,Yaakov

Author: Zou, Ismail

Proofread by: Zou,Oztekin,Dube

= References =

title: User:Egm6321.f10.team2.oztekin/HW5


text: File:Mass-spring-damper.JPG

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

title: File:Mass-spring-damper.JPG


text:
User:Davecormier

oh hiz.

title: User:Davecormier


text:
Ancient Greek Chairs

The Ancient Greeks crafted furniture that had a character all its own. Like all aspects of their culture which was infused with drama and dynamism the design of chairs was no different. The chair makers of Ancient Greece were clever and imaginative. It is very fortunate that we have many depictions of their designs to look at. A lot of these can be found on their ceramics. There was some diversity in design with some of them looking pleasing even when seen as a two dimensional picture. As far as having a real example of furniture to look at, we are out of luck. Apparently there are some small fragments left in existence, none of which can be easily accessed. In general Ancient Greek chairs have the characteristic of having outward curved legs. The amount of material in the vicinity of the joints has expanded considerably from that of the lower reaches of the legs where they are much thinner. The major components of Ancient Greek chairs, such as the legs and the backrest posts are mostly quite curved a lot more so than what is seen in modern furniture. Dowels used to tighten the joints, are a visible feature of the design. The front and back rails does not rely on some sort of elaborate joint but is set in place by the forces generated when the seat cord is woven up.

CURVES The sweeping curves are a prominent feature of ancient Greek chairs. The curvatures of some of the components particularly of the legs are considerable. There is no easy way of achieving such curvatures in straight grained wood. It is essential that the grain runs along the length of the component especially the legs otherwise it will look wrong from an engineering point of view as well as being weak structurally. There are several ideas that the Ancient Greeks might have utilized. The most likely is that they used specially selected timber that was the right shape to start with. This could even mean using the trunk and branch of a tree to obtain the shape required. Some of their designs have dramatic curves as seen in a colorful interpretation of an Ancient Greek chair (picture 1). In another colorful interpretation of an Ancient Greek chair (picture 2) the top of the posts have been bent right around which  indicates that steam or heat bending was used to create curves in timber. This would imply that the type of timber selected was amenable to bending. Another way of obtaining the desired timber is to grow a tree and manipulate the shape in which it grows. There is a method of  growing timber at a faster than normal speed called coppicing. Trees are cut down but the sprouts that shoot up from the stump can grow quickly because the mature roots are still in place. Nonetheless it would still take years to grow the desired piece of timber using this method.

JOINTS The joints that holds the legs to the side rail are quite large .The unique style of ancient Greek chairs with it's rapidly expanding amounts of material in the legs culminating where the side rail meet the legs should be thought of being like that for structural reasons as much as aesthetic reasons. Once the general design came into being around 500 B.C. The design lasted for more than 100 years. The substantial sized joints allows ancient Greek chairs to be made without the use of glue. If glue was not used in their designs then that would explain the longevity of that design  because the general design of the chair could not be changed for of structural reasons. The joint that holds the legs to the side rail could be described as a mortise and tenon drawn tight with a dowel or some variation of this joint. On examination of Ancient Greek ceramics with pictures of chairs on them, they usually have a figure sitting on the chair. The position of the figure's legs indicates where the front rail is placed and turns out that it is placed further back than might be expected. It would seem that the dowel is not part of the front rail and each front leg has it's own separate dowel.

STRINGING UP THE SEAT The side rail facing the viewer on ancient Greek ceramics with picture of chairs on them often has lines depicted on it. These lines represents the cord that has be wrapped around the outside of the side rail. To emulate what the Ancient Greeks might of done, an effective way was worked out to string up the seat. The simple criss cross weave with the cord spaced about 10mm apart each way to create an effect similar to that of a tennis racket tuned out well. This type of weave has great utility and is in keeping with what the general design of ancient Greek chairs and stools should look like. Holes are drilled into the rails for the cord to run through. Each rail might have 34 or so holes depending on the design. Each hole is made up from two holes drilled at right angles that meet each other. Long lengths of cord perhaps 10 meters or more can be threaded through these holes and the end result can be seen in picture 3. On completion of the weaving the forces generated on the rails can have them bending inwards. This force is used to keep the rail components of chair or stools together and also is used to keep the seat in square. ARTIST IMPRESSION Because we are only given one view usually the side view it is difficult to come up with a full view where the whole design of an Ancient Greek chair or stool can be seen. Nonetheless pictures 4, 5 and 6 show what can be done by letting what we can see suggest what the parts we can't see should look like. By keeping things unconvoluted but also beautiful does not guarantee success but with a well selected picture and good luck, a good picture may be made. Pictures 4 and 5 represents the chair displayed on Ancient Greek ceramic Harvard 1960.340. For the rest, the name of the picture file gives a clue as to where it was derived from.

THE TOP RAIL The top rail that is depicted in pictures 7 and 8 has the wood grain running up and down. This would mean that the top rail would have to be carved out of a rather large diameter log. This orientation is unorthodox, but that is the only way that short grain and it's weakening effects can be avoided.

title: Ancient Greek Chairs


text:
Paperless office

A Paperless office is a term used to describe an office system where: Documents and information are stored in digital files and databases.

Advantages include:

i. Space saving. (Note a system must be in place so that appropriate persons are aquainted with the digital locations and how to access the information, especially where protected by passwords and encryption)

ii. Mobility of information. (Note a system must be in place to avoid this offering a breach in security)

iii. Security of storage in multiple locations (Note a system must be in place to keep all up to date)

iv. Efficiencies to access, organise, copy, edit cleanly and send by electronic mail. (Note a system must be in place to ensure procedures are maintained, especially where there are multiple users, and more so if there is possibility of a single file which could be altered simultaneously.)

v. Saving of paper and ink/toner.

The following is an implementation by experience by Alan Cooper Cabinetmaker Ltd in a small business environment:

a. As much mail as possible is received by email or information can be accessed through the internet.

b. Some information still comes through the post. While this is not ideal, some product catalogues are still easier to handle in physical form, and it is not efficient to digitise many pages which may turn out to be superfluous to requirement, especially if new catalogue updates are frequent, so a small area probably needs to be maintained to keep these. In addition to this, some documents are required to be in physical form for legal requirements. For this reason, as yet, a truely paperless office is normally impractical, but we can get close to the ideal. There may also be occasions where fabric or paper thickness real physical samples may be included in a catalogue or physical file.

c. All in-house documents are digitally produced and in most cases can also be sent digitally, not requiring printing onto paper. It is important to have a firewall and updated antivirus protection in place for security, especially when connected to the internet.

d. Where paper documentation from others is received, providing the size will fit, a digital scanner can be used like the Fujitsu Scansnap, which will duplex scan (scan both sides) quickly and produce digital copies, assembling them into PDF files through Adobe Standard for example, (included with some Scansnap models) making them ready to store in a computer storage location of choice or 'cloud' based on an internet based service like Google Apps. Some documents need to be kept secure and an effective way of keeping the information safe from unauthorised viewing is to use Axantum Software's free AxCrypt program which can also be used in conjunction with a USB stick or SD card containing a password in a text file if preferred rather than typing in a password. All files are then accessible just as if they had not been protected, until the end of the session.

e. Careful thought needs to be made when scanning as to the routine for file-naming conventions, and folders to contain various kinds of file. If files are named to begin with YearMonthDayTime this will enable all files to keep in historical order. There are also systems available at cost like Mangofile to make dates of scanning official, recognised by some official organisations. It is possible for filename lists in a folder to be copied and pasted by using, for example, Karens Directory Printer. Double entry can be avoided by naming a file so that it has a hyphen or underscore at any place in the file name where if they are pasted into a spreadsheet, like MS Works, Open Office or MS Excel , they may be used as a command to start a new column in the spreadsheet, having first used notepad [in Windows] to find all hyphens [or underscores] and change all to commas or tabs ready for the spreadsheet. Spreadsheet settings may need to be edited to recognise comma or tab delimiting. This can be very useful for book keeping after scanning receipts and invoices from companies, as the filenames can later be used to generate lists of invoices and receipts, by the method mentioned above, and so save the effort of double-entry. If you change your mind about the naming convention, ReNamer can be a very useful tool for batch renaming of files.

Always remember to:

Backup and keep files in another location, or use an online service. Some Firewall and Antivirus software providers like for example, Bullguard include secure backup services with their package.

Protect files using password protected encryption.

title: Paperless office


text: User:PCock

title: User:PCock


text:
User talk:Alancoopercabinetmaker

 Hello Alancoopercabinetmaker, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 01:03, 6 November 2010 (UTC)

title: User talk:Alancoopercabinetmaker


text:
User talk:PCock

 Hello PCock, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 01:03, 6 November 2010 (UTC)

title: User talk:PCock


text:
User talk:Campbell504

 Hello Campbell504, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 01:03, 6 November 2010 (UTC)

title: User talk:Campbell504


text:
User talk:Arided/Learner Profile

Novel?
What's the plan for your novel? That said, I don't think it applies here. I think writing fiction (as I myself am also doing daily) is separate from a learner profile, though I am intrigued by the ways writing a novel (perhaps the most solo event any person can do) can work with crowdsourcing, though I also feel that might hurt the novel ... --Charles Jeffrey Danoff 03:23, 6 November 2010 (UTC)

I've been writing a combination of stream-of-consciousness-poetry and journal-style notes, experimenting with different media (e.g. typewriter, computer, pen, voice.) and different paces of writing. The long-term attempt is to trace out how I think about things, out of this to develop some realistic material on which to base different "characters", and to stitch the notes and ideas together in a more traditional narrative/storytelling/dialog-based framework later. Both when there's more material, but more importantly mainly when I have written some computer programs to help with the "stitching things together" stuff!

This interjection of computer programming into fiction writing might sound a bit weird... but for me, creative writing is among other things a good "check" on my computer programming work, and vice versa -- bottom up and top down thinking working together, maybe? Anyway, all things considered, I do think the novel as I'm conceiving of it, might fit in with learning, if it was described more as a "learning journal". It just happens that in my learning journal I like to use some techniques of fiction! Via the computer stuff, I definitely plan to look for themes (learning analytics); and it's called "The Penmaster's Apprentice" with reference to a certain kind of learning curve...

The material itself isn't gonna be crowdsourced -- except from the big population of "me at different times"! In a way this sort of simulates "normal" crowdsourcing, however, because the way I'm working on it comes with the challenge of splicing together different snippets into one coherent work. Maybe I'll invite people to submit pictures for an illustrated version of the work... :) And maybe some of the computer stuff I'm working on will help with eventual "real" crowdsourcing for novels.

It did occur to me recently that collaborative (screen-?)playwriting might be easier than collaborative novel-writing, because scripts come broken into very natural and interpersonally obvious chunks of various sizes (speeches and scenes). One of the many programming ideas I want to work on.

What I think is most missing from the Learner Profile as it currently stands is a real sense of my learning goals as opposed to work or productivity goals. The novel is just one example of something I want to do, but the whole profile as it stands isn't doing so well at describing what I need to learn to do the things I want to do. I'll adjust that and hopefully the whole thing will be more coherent.

Thanks kindly for the feedback Arided 12:39, 6 November 2010 (UTC)

 * I like where you're headed with the novel and writing experimentation/intertwining computer programming. Do you have interest in publishing portions before its actually finished, or are you going to keep them all to yourself and drop an atomic bomb on the world of fiction in 2013?
 * I still don't know if the novel belongs, unless, perhaps, are you trying to learn how to write a novel?
 * Regarding your "learning goals" I might have an idea of what you're saying, but I'd like to see you put out what you're imagining so I can know if I'm right, or totally wrong, and so I can follow suit with my profile. --Charles Jeffrey Danoff 07:56, 7 November 2010 (UTC)

title: User talk:Arided/Learner Profile


text:
Motivation and emotion/Textbook/Motivation and emotion/Textbook/Emotion/Antisocial personality disorder

Introduction

Antisocial personality disorder (ASPD) is a relatively rare disorder, which displays some very visible characteristics. It is estimated that as little as 3% of the male population is eligible for a diagnosis of ASPD, and 1% of females, but the effects of the violent and criminal behaviour displayed are felt by the wider community as this is where their crimes are often committed. But what causes people to behave in this way? Do people who behave this way have no feelings, no thought of the consequences of their actions or of the effects on other people? Hopefully those questions will be answered by the end of this chapter. Through the course of this chapter we will discuss the main characteristics of ASPD and also the causes, treatment and prevalence in the population. We will also look more closely at the role emotion, or a lack thereof, plays in ASPD. Additionally, we will look at some specific emotions related to ASPD, such as fear and anxiety.

Focus Questions

 * What are some of the characteristics of ASPD?
 * What role does emotion play in people with ASPD?
 * What are some of the main theories behind ASPD and emotion?

What is Antisocial Personality Disorder?
Antisocial personality disorder, as defined by the Diagnostic and Statistical Manual, Fourth Edition, Text Revision (APA, 2000)(DSM-IV-TR) is an ongoing disregard for the rights and feelings of others. This pattern of behaviour usually begins in childhood or adolescence and continues to develop in adulthood. For a diagnosis of antisocial personality disorder a person must be 18 years old and have displayed symptoms of conduct disorder before the age of 15.

Characteristics
The main characteristic of this disorder is the complete disregard for the wishes, rights or feelings of others. ASPD shares common characteristics with psychopathy, sociopathy and dyssocial personality disorder. Traits typical of ASPD are:

 * Repeated violation of the rights of others
 * Repeatedly performing criminal acts such as destroying property, harassing others, stealing or pursuing illegal occupations.
 * Deceitful or manipulative behaviour used for personal gain or pleasure
 * Lying
 * Impulsivity, and an inability to plan ahead
 * Easily irritated and aggressive, and as a consequence often get into physical fights or physically abuse others (e.g. spouses or children)
 * Recklessness and a disregard for personal safety. This can be displayed in driving behaviours, such as speeding or driving whilst intoxicated. They may also engage in unsafe sexual relationships or use substances at risky levels.
 * Irresponsible- this often leads to an inability to hold a steady job, and difficulty in handling finances.
 * Show no remorse for their behaviour. Often they will rationalise their actions by blaming the victims of their crimes (e.g. "They deserved it", "If I didn't take advantage of them someone else would have")
 * Lack of empathy and a tendency to be callous, cynical and contemptuous of the feelings, rights and suffering of others.
 * Inflated sense of self and arrogant demeanor
 * Superficial charm

For a diagnosis of antisocial personality disorder an individual must display more than three of the above symptoms over a prolonged period of time, and they must significantly impair on an individuals’ daily functioning. If you have a friend who has been in trouble with the law on one or two occasions, or know someone who impulsive, these characteristics alone do not necessarily warrant a diagnosis of ASPD.

Causes
As with many psychological disorders, there are many factors thought to contribute to the development of ASPD. One of the strongest predictors is genetics. ASPD is more common among first degree biological relatives with the disorder than in the general population (APA, 2000). The psychodynamic model proposes that ASPD is caused by a lack of parental love, which leads to a lack of trust in others (Sperry,2003). The behavioural model suggests that antisocial behaviours are learned through imitating the behaviours or those around them (Gaynor & Baird, 2007). These theories will be discussed in more detail later in relation to emotion.

Treatment
Treatment for people with ASPD tends to be very difficult and often ineffective as the patient usually does not have a desire to change, and their lack of empathy leads them to believe that they have done nothing wrong. Patients with ASPD are often difficult to treat, and due to the characteristics of the disorder such a deceit, lying and lack of regard for others they are not desirable patients. Another barrier to treatment is that it is often forced upon them by a judge, so their commitment to the treatment is often half hearted (Duggan, 2009).

Early intervention with children diagnosed with conduct disorder is thought to be one form of effective treatment. Pharmacological treatments (see pharmacology) have been tried with limited effectiveness. Cognitive therapy, which tries to get patients to think about moral issues and attempt to empathise with the needs of others has also shown limited results. Current research suggests that there is no single method of treating ASPD that gives conclusive results (Comer, 2010). The symptoms do appear to become less severe with age, and by middle age many people with ASPD are less likely to be involved in criminal activity, but they continue to have problems with interpersonal relationships, parenting and holding down a job (Paris, 2004).

Prevalence
The prevalence of ASPD in the general population is about 3% in males and 1% in females. These rates increase drastically in prison settings or substance abuse treatment settings (APA, 2000). Although results vary due to different methods of measurement, the prevalence of ASPD amongst prison inmates has been reported to be as little as 11% and as high as 78% (Rotter et al., 2002). However, Ogloff (2006) believes that the prevalence of ASPD, particularly in the prison population is greatly over estimated, and over diagnosed. Ogloff states that many adults in prison meet the criteria for ASPD based on the diagnostic focus on criminality, but the diagnosis may not be warranted in many cases and has led to overinflated figures from prisons.

Emotion
In the introduction, emotion is defined as the complex psychophysiological experience of an individual’s state of mind as interacting with biochemical and environmental influences (Myers, 2004). Emotion involves mood, temperament, personality and motivation. Emotion plays a large part in our everyday lives, and contributes to our development. There are three main components of emotion:
 * 1) Behaviour- The result of an emotion, such as crying or a facial expression of fear.
 * 2) Physiology- The parts of the brain that are involved in producing emotion.
 * 3) Cognition- Processing external stimuli in relation to emotion. (Durand & Barlow, 2006).

For more information on emotion, see emotion. We will now look at the role emotion plays in people with antisocial personality disorder.

Psychodynamic Theory
The psychodynamic theory of ASPD proposes that the disorder develops due to a lack of parental love and closeness during infancy, which leads to an individual becoming emotionally distant and mistrusting others later in life. Research supporting this theory has found that people with this disorder are more likely to have had significant childhood stress, in the form of family poverty, violence and parental conflict or divorce, indicating that there is an etiological connection between trauma and the development of ASPD (Martens, 2005). Whilst these childhood traumas do create vulnerability for the child to develop later psychological problems, the traumas alone do not cause disorders. Psychodynamic theory is not a leading theory in regard to ASPD, and as such there is little research surrounding it. However, it has been included in this section on ASPD and emotion as the theory relates specifically to emotions, or a lack thereof, experienced in childhood.

Behavioural Theory
Behavioural theorists suggest that antisocial behaviours can be learned through modelling, or imitation. Modelling is when a person imitates the actions or behaviours of those around them. As you have read previously, there is a high genetic relationship with the development of ASPD. Thus it is reasonable to conclude that as well as the genetic predisposition for the disorder, there are also environmental influences. A child who grows up with a parent with ASPD could grow up imitating the actions of that parent, including the antisocial aspects such as aggressiveness, deceitful and criminal behaviours (Paris, 2004).

Other behavioural theorists have suggested that antisocial behaviours displayed in childhood are rewarded and reinforced, often unintentionally by parents. When a child misbehaves or shows aggressive behaviours as a reaction to parents’ requests, the parents may simply give in to restore the peace. Over time the child will learn that by displaying this behaviour they get what they want and will repeatedly react in the same manner. Parent Management Training has been identified as a preventative strategy for children displaying antisocial behaviours, and improved parenting practices following the training have been linked with a reduction in undesirable behaviours in children (Kazdan & Whitley, 2006).

Genetic Influences
Family and adoption studies suggest that genetics may be one of the main predictors in the development of ASPD. One of the first adoption studies conducted by Crowe (1974) involving ASPD investigated the rates of ASPD prevalence in adopted children who had biological mothers with a criminal history. The rate of adoptees displaying signs of ASPD was 13%, compared with just 2% for the control group of matched adoptees. The children of the criminal mothers were removed as newborn, reducing the risk of environmental influences and indicating a high genetic correlation. Although the results were significant, the sample group was quite small (N = 46). Another interesting finding from Crowe’s research was that the adoptees with criminal mothers who went on to commit acts of crime themselves spent a longer period of time in an orphanage than the control group, or the group who did not become criminals later in life, indicating that as well as a genetic predisposition, environmental influences also play a large role (Roth & Finley, 1998).

Subsequent research has supported the gene-environment interaction. An adoption study compared the likelihood of children with a parent with ASPD and the child developing conduct disorder, which is a precursor for ASPD in adulthood. The results found that if the children with the biological predisposition were exposed to chronic stress by their adopted family, in the form of marital, legal or psychiatric problems, the children were at a greater risk of developing conduct disorder. A genetic influence does not necessarily mean that a person will develop ASPD, but evidence shows that the genetic influence matched with environmental influences increases the risk (Cadoret, Yates, Troughton, Woodworth & Stewart, 1995).

Underarousal Hypothesis
The underarousal hypothesis states that people with low levels of cortical arousal often display many of the symptoms of ASPD, such as antisocial and risk taking behaviours. According to the Yerkes-Dodson curve, people at either end of the curve, with very high or very low levels of arousal experience negative affect and have difficulty performing, whereas people within the normal range of the curve experience no negative affect and have no trouble performing tasks. Due to low levels of arousal, people with ASPD appear to become bored very easily engage in risk-seeking behaviour such as taking illicit drugs or getting into fights in an effort to increase their level of stimulation and reduce the feelings of boredom (Durand & Barlow, 2006). This hypothesis could also explain why people with ASPD have trouble with holding down a job or long term planning. Research into the effects of low cortical arousal has shown that people operate best with medium levels of cortical arousal, yet people showing low levels of arousal do not perform as well and will actively try and produce a situation where they are more stimulated. For example, if a person with low arousal levels becomes bored whilst driving they will engage in unsafe behaviours such as speeding and reckless driving to provide stimulation (Fischer, Langner, Birbaumer & Brocke, 2008). The underarousal hypothesis was developed with psychopaths and has been inferred to work with people with ASPD. As many of the characteristics displayed are the same, it can be assumed that those characteristics have the same cause.

Fearlessness Hypothesis
The fearlessness hypothesis proposes that people with ASPD have a higher threshold for fear than most individuals. This lack of fear can be seen in some of the characteristics of ASPD, such as criminality, involvement in violence and other risk taking behaviours. The fearlessness also results in an inability to respond to and learn from the negative consequences of their behaviours. As discussed earlier in regard to treatment of ASPD, many patients who present for treatment do not recognise that they have done anything wrong and show no remorse for their actions. Societal punishments, such as imprisonment or community service, for crimes committed by people with ASPD often have no effect (van Honk & Schutter, 2006). In relation to the fearlessness hypothesis, pioneering research by Lykken (1957) showed that people with ASPD traits displayed significantly lower levels of anxiety than the control group.

The Response Modulation (RM) hypothesis was derived from the fearlessness hypothesis, but investigated other aspects, as well as fear. Newman, Schmitt and Voss (1997) conducted research which supported their claim that people displaying antisocial characteristics would have difficulty processing motivationally neutral stimuli, as well as the fear related stimuli. The research findings supported the RM hypothesis, indicating that people with ASPD not only do not recognise the emotion of fear, but any peripheral stimuli which they do not feel is relevant to them.

These two hypotheses validate the claim that people with ASPD not only are unable to experience fear and display a lack of empathy, but they may in fact be unable to experience a wider range of emotions, including anxiety.

Behavioural Inhibition System
According to Gray’s model of brain functioning, the behavioural inhibition system (BIS) is responsible for an individuals’ ability to stop and recognise when they are faced with possible punishment, non-reward or situations which may lead to frustration and anxiety. The behavioural activation system (BAS) is responsible for a person’s sensitivity to reward and regulates approach behaviour. It is generally believed that people with ASPD have an imbalance between the BIS and the BAS, which leads to the fear and anxiety generally felt by most people in dangerous situations being less prominent. People with weaker BIS not only show a lack of anxiety, but also an inability to inhibit behaviours in the face of punishment. This will also result in strong reward seeking behaviour, also known as impulsivity. An additional result of a weak BIS is the inability to learn from past experiences, or past punishments (Fowles, 1980).

Antisocial Personality Disorder and Stress
Being in involved in combat is a stressful experience, and it is believed that it may lead to the development of antisocial behaviours in some people. Research from Vietnam War veterans found that there was a significant relationship between combat exposure and adult antisocial behaviours. Participants (N = 118) showed signs of antisocial behaviour indicating that there may be a relationship between combat exposure level and adult antisocial behaviours and the development or maintenance of antisocial behaviours (Resnick, Foy, Donahoe & Miller, 1989). However, the relationship between witnessing violence and the development of antisocial behaviours is still unclear.

The relationship between stress and ASPD was also discussed briefly in relation to causes, where it is thought that significant environmental/social stressors experienced during childhood could play a part in the development of the disorder, if a genetic predisposition exists.

Deficits in Emotional Processing
Research has shown that participants with ASPD demonstrated poorer performance on emotion-discrimination tasks, compared to a control group not displaying any signs of ASPD. People with ASPD have less intense emotional reactions to everyday events, and also have trouble distinguishing between positive and negative emotions. Higher levels of emotional detachment were also found, providing an insight as to why people with ASPD can so easily lie and manipulate others (Habel, Kuhn, Salloum, Devos & Schneider 2002).

Conclusion
ASPD is a complex disorder, and it can be difficult for people to understand how those with ASPD can act in such manipulative and deceitful ways and not feel any remorse for their actions. It is one of the few disorders that does not appear to respond to any type of treatment, either in the form of therapy or drug treatment. There are a number of theories regarding the cause of the disorder, but the most generally accepted and tested theory is that of genetic inheritance. Throughout this chapter we looked briefly at various models explaining the lack of emotion in people with ASPD. The underarousal hypothesis and the fearlessness hypothesis are two of the best know models and explain the reasons behind the thrill seeking, impulsive and aggressive behaviours displayed. Finally, the chapter looked at the role of stress in ASPD, and how the characteristic deficits in emotion may be explained. Although there has been a lot of research into ASPD, there are still a lot of gaps particularly in the areas of prevention and treatment. This chapter closes by leaving you with a few questions to consider:

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 * 1) If a person is diagnosed with ASPD, is it possible for them to lead a "normal" life?
 * 2) Do you think ASPD has varying degrees of severity?
 * 3) What do you think is the main area that needs more research in regard to ASPD?


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User:PaulJRichardson

Oh joy. Yet ANOTHER page to maintain. Jeeeze. I guess I've got about a 100 now?

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