# User:Eml5526.s11.team01.roark/Mtg09

Mtg 9: Fri, 21 Jan 11

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 G1DM1.0: p. 4-4 G = General 1D = 1-dimenional M1 = Model 1 .0 = “simple” boundary conditions PDE = (1) p. 4-4 Boundary conditions: Essential Boundary condition = (3) p. 4-4 Natural Boundary condition = (4) p. 4-4

G1DM1.0 /D1

(D1 – Dataset 1)

 $\displaystyle \text{ }\Omega \text{ }=\left] 0,\underbrace{1}_{L=1} \right[$ (1)
 $\displaystyle {{a}_{2}}\left( x \right)=2+3x$ (2)
 $\displaystyle f\left( x,t \right)=5x$ (3)
 $\displaystyle \frac{{{\partial }^{s}}u(x,t)}{\partial {{t}^{s}}}=0$ (4)

Meaning static or steady state. $\displaystyle {{\text{ }\!\!\Gamma\!\!\text{ }}_{\text{g}}}=\left\{ 1 \right\},\text{g}=4$

 $\displaystyle u\left( x=1 \right)=4$ (5)

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 $\displaystyle {{\text{ }\!\!\Gamma\!\!\text{ }}_{\text{h}}}=\underbrace{\left\{ 0 \right\}}_{x=0},h=12$ (1)
 $\displaystyle \underbrace{n(0)}_{-1}\underbrace{{{a}_{2}}(0)}_{2}\frac{\partial u(x=0)}{\partial x}=12$ (2)

I.e., Solve the following problem:

 $\displaystyle \frac{d}{dx}\left[ \left( 2+3x \right)\frac{du}{dx} \right]+5x=0~\forall x\in \left] 0,1 \right[,~~u\left( 1 \right)=4,~\,-2\frac{du(x=0)}{dx}=12$ (3)

HW 2.4:

Show

 $\displaystyle \mathop{\int }^{}\frac{{{x}^{2}}}{1+x}dx=\frac{{{x}^{2}}}{2}-x+\log \left( 1+x \right)+\underbrace{k}_{Const}$ (4)
1. $\displaystyle \mathop{\int }^{}logxdx=xlogx-x$ (hint: Integration by parts)
2. $\displaystyle \mathop{\int }^{}logxdx=\frac{1}{2}{{x}^{2}}\left[ logx-\frac{1}{2} \right]$
3. Find $\displaystyle \int{\frac{({{x}^{2}}dx)}{(1+cx)}}$ in particular c=1 => (4)
4. Find $\displaystyle \mathop{\int }^{}\frac{{{x}^{2}}dx}{a+bx}\text{ }\!\!~\!\!\text{ }$

End HW 2.4

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Note: Demo Wolfram Alpha (WA) e.g. (debt usa)/(gdp usa) integrate logx, etc. Link WA computational results in HW reports. Avoid plagiarism.