User:Eml5526.s11.team7.jin/HW5

=5.1=

Exact solution
The governing partial differential equation (PDE):
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$$ \displaystyle
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\frac{\partial }\left[ {\left( {2 + 3x} \right)\frac} \right] + 5x = 0{\text{ }}\forall {\text{x}} \in \left] {0,1} \right[

$$     (Eq ) The natural and essential B.C.:
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$$ \displaystyle
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\frac\left( 1 \right) = \frac{5}

$$     (Eq )
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$$ \displaystyle
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u\left( 0 \right) = 4

$$     (Eq ) Integrating
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$$ \displaystyle
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\int {\frac{\partial }\left[ {\left( {2 + 3x} \right)\frac} \right]} dx = \int { - 5xdx}

$$     (Eq )
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$$ \displaystyle
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\left( {2 + 3x} \right)\frac + {C_1} = - \frac{5}{2}{x^2}{\text{ where }}{C_1}{\text{ is a constant}}

$$     (Eq ) By the natural B.C.
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$$ \displaystyle
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{C_1} = - \frac{2}

$$     (Eq ) Then
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$$ \displaystyle
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\frac = \frac

$$     (Eq )
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$$ \displaystyle
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u = - \frac{5}{x^2} + \frac{5}{9}x + \frac\log \left( {3x + 2} \right) + {C_2}{\text{ where }}{C_2}{\text{ is a constant}}

$$     (Eq ) By the essential B.C.
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$$ \displaystyle
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{C_2} = 4 - \frac\log \left( 2 \right)

$$     (Eq ) Then
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$$ \displaystyle
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u = - \frac{5}{x^2} + \frac{5}{9}x + \frac\log \left( {3x + 2} \right) + 4 - \frac\log \left( 2 \right)

$$     (Eq )
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=5.2=

Exact solution
The governing partial differential equation (PDE):
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$$ \displaystyle
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\frac{\partial }\left[ {\left( {2 + 3x} \right)\frac} \right] + 5x = 0{\text{ }}\forall {\text{x}} \in \left] {0,1} \right[

$$     (Eq ) The natural and essential B.C.:
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$$ \displaystyle
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\frac\left( 0 \right) = - 6

$$     (Eq )
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$$ \displaystyle
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u\left( 1 \right) = 4

$$     (Eq ) Then
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$$ \displaystyle
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\int {\frac{\partial }\left[ {\left( {2 + 3x} \right)\frac} \right]} dx = \int { - 5xdx}

$$     (Eq )
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 * }
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$$ \displaystyle
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\left( {2 + 3x} \right)\frac + {C_1} = - \frac{5}{2}{x^2}{\text{ where }}{C_1}{\text{ is a constant}}

$$     (Eq ) By the natural B.C.
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$$ \displaystyle
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{C_1} = 12

$$     (Eq ) Then
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$$ \displaystyle
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\frac = \frac

$$     (Eq )
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$$ \displaystyle
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u = - \frac{5}{x^2} + \frac{5}{9}x - \frac\log \left( {3x + 2} \right) + {C_2}{\text{ where }}{C_2}{\text{ is a constant}}

$$     (Eq ) By the essential B.C.
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$$ \displaystyle
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{C_2} = \frac{5} - \frac{5}{9} + \frac\log \left( 5 \right) + 4

$$     (Eq ) Then
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$$ \displaystyle
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u = - \frac{5}{x^2} + \frac{5}{9}x - \frac\log \left( {3x + 2} \right) + \frac{5} - \frac{5}{9} + \frac\log \left( 5 \right) + 4

$$     (Eq )
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u=-5/12*x.^2+5/9*x+241/54*log(3*x+2)-241/54*log(2)+4;

u=-5/12*x.^2+5/9*x-118/27*log(3*x+2)+5/12-5/9+118/27*log(5)+4;

=5.6=



$$\begin{align} & b1=\frac{\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)}{\left( {{x}_{1}}-{{x}_{2}} \right)\left( {{x}_{1}}-{{x}_{3}} \right)}=\frac{\left( x-2 \right)\left( x-3 \right)}{\left( 1-2 \right)\left( 1-3 \right)} \\ & b2=\frac{\left( x-{{x}_{1}} \right)\left( x-{{x}_{3}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)\left( {{x}_{2}}-{{x}_{3}} \right)}=\frac{\left( x-1 \right)\left( x-3 \right)}{\left( 2-1 \right)\left( 2-3 \right)} \\ & b3=\frac{\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)}{\left( {{x}_{3}}-{{x}_{1}} \right)\left( {{x}_{3}}-{{x}_{2}} \right)}=\frac{\left( x-1 \right)\left( x-2 \right)}{\left( 3-1 \right)\left( 3-2 \right)} \\ & \text{Assume }{{x}_{1}}=1,{{x}_{2}}=2,{{x}_{3}}=3 \\ \end{align}$$

=5.4=