User:Egm6321.f12.team5/Report 2

= Report 2, Team 5 =

Given
(From lecture notes Sec7.)


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$$L_2(y):=(1-x^2)y''-2xy'+n(n+1)y=0$$ $$n=1 \Rightarrow(1-x^2)y''-2xy'+2y=0$$ (2.1.1)
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$$ y^1_H(x) = x =: P_1(x) $$     (2.1.2)
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$$ y^2_H(x) = \frac{x}{2} log \left(\frac{1+x}{1-x}\right) - 1 =: Q_1(x) $$     (2.1.3)
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Find
Verify that


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$$ L_2(y^1_H) = L_2(y^2_H) = 0 $$     (2.1.4)
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Solution
This solution was solved on our own.

To make the problem easier, we can separate the proof into two parts:
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$$L_2(y^1_H)=0$$ (2.1.5)
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$$L_2(y^2_H)=0$$ (2.1.6) Since $$y^1_H(x)=x$$, we can find that:
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$${y^1_H}'=1$$ (2.1.7)
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$${y^1_H}''=0$$ (2.1.8) Thus:
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$$ \begin{align} L_2(y^1_H)|_{n=1}&=(1-x^2){y^1_H}''-2x{y^1_H}'+2y^1_H \\\\ &=(1-x^2)\times 0-2x\times 1+2x \\\\ &=0 \end{align} $$ (2.1.9)
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Since
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$$ \begin{align} y^2_H(x)&=\frac{x}{2} \log \left(\frac{1+x}{1-x} \right)-1 \\\\ &=\frac{x}{2} \log (1+x)-\frac{x}{2} \log (1-x)-1 \\ \end{align} $$ (2.1.10) we can find that:
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$$ \begin{align} {y^2_H}'&=\frac{1}{2}log(1+x)-\frac{1}{2}log(1-x)+\frac{x}{2(x+1)}-\frac{x}{2(x-1)} \\ &=\frac{1}{2}log \left(\frac{1+x}{1-x}\right) + \frac{x}{1-x^2} \\ \end{align} $$ (2.1.11)
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$$ \begin{align} {y^2_H}''&=\frac{1}{x+1}-\frac{1}{x-1}+\frac{x}{2(x-1)^2}-\frac{x}{2(x+1)^2} \\ &=\frac{1}{1-x^2} + \frac{1+x^2}{(1-x^2)^2} \\ \end{align} $$ (2.1.12) Thus:
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$$ \begin{align} L_2(y^2_H)|_{n=1} &= (1-x^2){y^2_H}''-2x{y^2_H}'+2y^2_H \\\\ &= (1-x^2)\left(\frac{1}{1-x^2} + \frac{1+x^2}{(1-x^2)^2}\right) - 2x\left(\frac{1}{2}log \left(\frac{1+x}{1-x}\right) + \frac{x}{1-x^2} \right) + 2\left(\frac{x}{2}log \left(\frac{1+x}{1-x}\right)\right) = 0 \\\\ &= 1 + \frac{1+x^2}{1-x^2} - x\;log\left(\frac{1+x}{1-x}\right) - \frac{2x^2}{1-x^2} + x\;log\frac{1+x}{1-x} - 2 \\ &= 1 + \cancelto{1}{\left(\frac{1+x^2}{1-x^2} - \frac{2x^2}{1-x^2}\right)} + \cancelto{0}{\left(x\;log\left(\frac{1+x}{1-x}\right) - x\;log\left(\frac{1+x}{1-x}\right)\right)} - 2 \\\\ &= 2 - 2 = 0 \\ \end{align} $$ (2.1.13)
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So we get:
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$$L_2(y^1_H)=L_2(y^2_H)=0$$
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Comments
- changed n=2 to n=1 in equations (2.1.9) and (2.1.13) - simplified equations (2.1.11) and (2.1.12) -> shorter equations in (2.1.13) --Egm6321.f12.team5.kim (talk) 07:47, 18 September 2012 (UTC)

Given
(From lecture notes Sec7.)


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$$ p(x) = k_1e^{-x} + x - 1 $$     (2.2.1)
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Find
Verify that equation (2.2.1) is indeed the solution for


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$$ p' + p = x $$ (2.2.2)
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Solution
This solution was solved on our own.

Taking the derivative of equation (2.2.1) w.r.t. $$ x $$ gives


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$$ p(x)' = \frac{d (k_1e^{-x})}{dx} + \frac{d (x)}{dx} - \frac{d (-1)}{dx} $$     (2.2.3)
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$$ p(x)' = -k_1e^{-x} + 1 $$     (2.2.4)
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Adding equations (2.2.1) and (2.2.4) gives


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$$ p(x)' + p(x) = \underbrace{(-k_1e^{-x} + 1)}_{\displaystyle p'(x)} + \underbrace{(k_1e^{-x} + x - 1)}_{\displaystyle p(x)} $$     (2.2.5)
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$$ p(x)' + p(x) = \cancelto{0}{(-k_1e^{-x} + k_1e^{-x})} + x + \cancelto{0}{(1 - 1)} $$     (2.2.6)
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$$ p(x)' + p(x) = x $$ (2.2.7)
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Which shows that equation (2.2.1) is indeed the solution for equation (2.2.2).

Comments
initial solution --Egm6321.f12.team5.kim (talk) 16:34, 15 September 2012 (UTC)

Given
(From lecture notes Sec7.)

General N1-ODEs:


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$$ G(y',y,x) = 0 $$     (2.3.1)
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Particular Class of N1-ODEs:


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$$ M(x,y) + N(x,y)\underbrace{y'}_{\displaystyle \frac{dy}{dx}} = 0 $$     (2.3.2)
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Find
Show that equation (2.3.2) is affine in $$y'$$, and that in general an N1-ODE, even if it is not the most general N1-ODE as represented by equation (2.3.1). Give an example of a more general N1-ODE.

Solution
This solution was solved on our own.

Show $$\mathbf{M(x,y) + N(x,y)y' = 0}$$ is affine in $$\mathbf{y'}$$
Affine function is defined as a linear function that has displacement added onto it.

$$N(x,y)y'$$ in equation (2.3.2) is linear since it satisfies the requirement as shown below:


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$$ N(x,y)(\alpha \nu + \beta u) = N(x,y)(\alpha\nu) + N(x,y)(\beta u) $$ (2.3.3)
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Since equation (2.3.2) also has the $$M(x,y)$$ term, which displaces the term $$N(x,y)y'$$ that is linear in $$y'$$, the equation is affine in $$y'.$$

Show $$\mathbf{M(x,y) + N(x,y)y' = 0}$$ is in general an N1-ODE
It is clear from looking at equation (2.3.2) that it is a function of $$x, y,$$ and $$y'$$. Since $$y'$$ is the highest degree derivative of $$y,$$ the equation is a 1-ODE.

To show that equation (2.3.2) is nonlinear, we substitute $$\alpha \nu + \beta u$$ for $$y',$$


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$$ M(x,y) + N(x,y)(\alpha \nu + \beta u) = M(x,y) + N(x,y)(\alpha\nu) + N(x,y)(\beta u) $$ (2.3.4)
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It is clear that the right side of equation (2.3.4) does not equal (2.3.5) below.


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$$ M(x,y) + N(x,y)(\alpha \nu) + M(x,y) + N(x,y)(\beta u) = 2M(x,y) + N(x,y)(\alpha\nu) + N(x,y)(\beta u) $$ (2.3.5)
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Therefore, equation (2.3.2) is in general an N1-ODE. The most general case of the N1-ODE is the equation (2.3.1),


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$$ G(x,y,y') = 0 $$
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which puts no restriction on the form of the ODE other than that it is a 1-ODE of the variables x, y, and y'.

Comments
-initial solution --Egm6321.f12.team5.kim (talk) 21:42, 16 September 2012 (UTC)

-added the part about the more general case of N1-ODE --Egm6321.f12.team5.kim (talk) 00:21, 19 September 2012 (UTC)

Given
{From lecture notes Sec9.}


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$$ y^1_H(x) = x =: P_1(x) $$     (2.4.1)
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$$ y^2_H(x) = \frac{x}{2} log \left(\frac{1+x}{1-x}\right) - 1 =: Q_1(x) $$     (2.4.2)
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Find
Show that equations (2.4.1) and (2.4.2) are linearly independent, i.e.,


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$$ \forall \alpha \in \mathbb R, \, y^1_H(\cdot) \ne \alpha \, y^2_H(\cdot) $$     (2.4.3)
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$$\left[ \right.$$i.e., for any given $$\alpha$$, show that $$\exist \hat x \; $$ such that $$ y^1_H(\hat x) \ne \alpha \, y^2_H(\hat x)\left.\right]$$

and plot $$y^1_H(x)$$ and $$y^2_H(x).$$

Solution
This solution was solved on our own. By definition, If $$ \forall \alpha \in \mathbb R, \, y^1_H(\cdot) \ne \alpha \, y^2_H(\cdot) $$ can be proved. Then we can prove that equations (2.4.1) and (2.4.2) are linearly independent.

So if we choose $$ (\cdot)=0.833556 \; \left[\right. $$ (which is a approximate number to the solution of the equation $$ y^2_H(x) = \frac{x}{2} log \left(\frac{1+x}{1-x}\right) - 1=0 \left.\right] $$

Then $$ y^1_H(\cdot)=0.833556 $$ and $$ \alpha \, y^2_H(\cdot)=0. $$ It can be seen simply that no matter what real number is chosen for $$ \alpha, \; y^1_H(\cdot) \ne \alpha \, y^2_H(\cdot). $$ Also, it is obvious from observation that no constant multiple will equate $$y^1_H$$ and $$y^2_H.$$

Since we have proved that $$ \forall \alpha \in \mathbb R, \, y^1_H(\cdot) \ne \alpha \, y^2_H(\cdot), $$ we can assure that the two equations are linearly independent. Following is the plot of the two equations:

Comments
- made some formatting changes - added the comment: 'Also, it is obvious from observation that no constant multiple will equate...' --Egm6321.f12.team5.kim (talk) 00:44, 19 September 2012 (UTC)

Given
(From lecture notes Sec9.)

Consider the following function


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$$ \phi(x,y) = x^2y^\frac{3}{2} + log(x^3y^2) = k $$ (2.5.1)
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Find
Find


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$$ G(y',y,x) = \frac{d}{dx}\phi(x,y) = 0 $$     (2.5.2)
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and show that equation (2.5.2) is an N1-ODE.

Solution
This solution was solved on our own.

Find the Total Derivative
First set the function equal to zero


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$$\phi(x,y)=x^2y^{\frac{3}{2}}+log(x^3y^2)=k$$ (2.5.3)
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$$\phi(x,y)=x^2y^{\frac{3}{2}}+log(x^3y^2)-k=0$$ (2.5.4)
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To make differentiating simpler, use the following log law to separate out the log function into two separate log functions.
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$$ log(ab) = log(a)+log(b)$$ (2.5.5)
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$$\phi(x,y)=(x^2y^{\frac{3}{2}})+log(x^3)+log(y^2)-k$$ (2.5.6)
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The total derivative of a function can be defined as follows


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$$\frac{d\phi (x,y(x))}{dx}=\frac{\partial\phi(x,y(x)))}{\partial x}+\frac{\partial\phi(x,y(x)))}{\partial y}\frac{dy}{dx}$$ (2.5.7)
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Take the derivative of $$\phi(x,y)$$ with respect to $$x$$.
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$$ \begin{align} \frac{\partial\phi(x,y(x))}{\partial x}&=\frac{d}{dx}(x^2y^{\frac{3}{2}}+log(y^2)-k) \\ &=\frac{d}{dx}x^2y^{\frac{3}{2}}+\frac{d}{dx}log(x^3)+\cancelto{0}{\frac{d}{dx}(y^2)}-\cancelto{0}{\frac{d}{dx}k} \\\\ &=2xy^{\frac{3}{2}}+\frac{3}{x} \\ \end{align} $$ (2.5.8)
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Take the derivative of $$\phi(x,y)$$ with respect to $$y$$.
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$$ \begin{align} \frac{\partial\phi(x,y(x))}{\partial y}&=\frac{d}{dy}(x^2y^{\frac{3}{2}}+log(x^3)+log(y^2)-k) \\ &=\frac{d}{dy}x^2y^{\frac{3}{2}}+\cancelto{0}{\frac{d}{dy}log(x^3)}+\frac{d}{dy}log(y^2)-\cancelto{0}{\frac{d}{dy}k)} \\\\ &=\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{y} \\ \end{align} $$ (2.5.9)
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Therefore
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$$ \begin{align} \frac{d\phi (x,y(x))}{dx}&=\frac{\partial\phi(x,y(x)))}{\partial x}+\frac{\partial\phi(x,y(x)))}{\partial y}\frac{dy}{dx} \\ &=(2xy^{\frac{3}{2}}+\frac{3}{x})+(\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{y})\frac{dy}{dx} \\ \end{align} $$ (2.5.10)
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The solution can be written as follows in the particular form
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$$ \begin{align} G(y',y,x)&=\frac{d}{dx}\phi (x,y)=0 \\ &= \underbrace{2xy^{\frac{3}{2}}+\frac{3}{x}}_{\displaystyle M(x,y)}+\underbrace{\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{y}}_{\displaystyle N(x,y)}\underbrace{\frac{dy}{dx}}_{\displaystyle y'}=0 \end{align}$$ (2.5.11)
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Show Equation (2.5.2) is an N1-ODE
As shown in (2.5.11),  $$G(y',y,x)$$  can be expressed in the particular form
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$$G(y',y,x)= M(x,y)+N(x,y)y'=0$$ (2.5.12)
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From inspection, it can be concluded that the equation is first-order.

To show that $$G(y',y,x)$$ is non-linear in  $$y'$$, it can be shown that


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$$G(\alpha u+\beta v, y, x) \neq \alpha G(u,y,x)+\beta G(v,y,x)$$ (2.5.13)
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Substituting $$y'= \alpha u + \beta v$$ into equation (2.5.11),
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$$ G(\alpha u+\beta v,y,x)= 2xy^{\frac{3}{2}}+\frac{3}{x}+\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{y}(\alpha u +\beta v)=0 $$ (2.5.14)
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Substituting $$y'= \alpha u $$ into equation (2.5.11),
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$$ G(\alpha u,y,x)= 2xy^{\frac{3}{2}}+\frac{3}{x}+\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{y}(\alpha u)=0 $$ (2.5.15)
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Substituting $$y'= \beta v $$ into equation (2.5.11),
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$$ G(\beta v,y,x)= 2xy^{\frac{3}{2}}+\frac{3}{x}+\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{y}(\beta v)=0 $$ (2.5.16)
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Since equation (2.5.14) doesn't equal the sum of equations (2.5.15) and (2.5.16),


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$$ \begin{align} 2xy^{\frac{3}{2}}+\frac{3}{x}+\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{y}(\alpha u +\beta v) &\ne 2xy^{\frac{3}{2}}+\frac{3}{x}+\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{y}(\alpha u) + 2xy^{\frac{3}{2}}+\frac{3}{x}+\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{y}(\beta v) \\ &\ne 2\left[2xy^{\frac{3}{2}}+\frac{3}{x}+\frac{3}{2}x^2y^{\frac{1}{2}}\right]+\frac{2}{y}(\alpha u + \beta v)\\ \end{align} $$ (2.5.17)
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$$G(y',y,x)$$ is non-linear.

Conclusion: Equation (2.5.2) is an N1-ODE

Comments
Got rid of the $$\ln$$ in the equations 2602:306:2468:2819:608B:95AA:CE33:14DC 14:42, 17 September 2012 (UTC)Anh

- cleaned up the equations a little --Egm6321.f12.team5.kim (talk) 03:03, 19 September 2012 (UTC)

Given
(From lecture notes Sec9.)

Assuming that $$ \phi(x,y) $$ is smooth, we have


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$$ \phi_{xy}(x,y) = \phi_{yx}(x,y) \; \; \; \; \; \; \; \left(i.e., \frac{\partial{^2\phi(x,y)}}{\partial{x}\partial{y}} =  \frac{\partial{^2\phi(x,y)}}{\partial{y}\partial{x}} \right) $$     (2.6.1)
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Find
Review calculus, and find the minimum degree of differentiability of the function $$ \phi(x,y) $$ such that equation (2.6.1) is satisfied. State the full theorem and provide a proof.

Solution
This solution was solved on our own.

Theorem:

If a function g(a,b)is twice continuously differentiable, then its mixed partial derivatives are equal; which means


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$$\begin{align} \frac{\partial g^2(a,b)}{\partial a\partial b}=\frac{\partial g^2(a,b)}{\partial b\partial a} \\\end{align}$$
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(2.6.2)
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The mean value theorem:

Imagine an arc in a plane confined to its endpoints, there must be at least one point between arc's endpoints at which the slope of arc equals to the slope of a line passing through the endpoints. For example, if a function f(x) is differentiable on the closed interval [m, n], where m < n, and differentiable on the open interval (m, n), then there exists a point in (m, n) such that


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$$\begin{align} f'(m+k)=\frac{f(n)-f(m)}{n-m} \\\end{align}$$
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(2.6.3)
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where 0 < k < (n-m)

Proving the theorem

Image there are four infinite closed points in the x-y coordinate system. The distance between every 2 points is $$\begin{align}\Delta x \\\end{align}$$ or  $$\begin{align}\Delta y \\\end{align}$$ is infinite small.


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$$\begin{align} F(\Delta x,\Delta y):=\frac{1}{\Delta x\Delta y}[f(x+\Delta x,y+\Delta y)-f(x+\Delta x,y)-f(x,y+\Delta y)+f(x,y)] \\\end{align}$$
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(2.6.4)
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Set $$\begin{align}f ( x,y)\\\end{align}$$ is a smooth function, which means $$\begin{align}f ( x,y)\\\end{align}$$ can be infinitely differentiated.


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$$ \begin{align} F(\Delta x,\Delta y) &=\frac{1}{\Delta x\Delta y}[f(x+\Delta x,y+\Delta y)-f(x+\Delta x,y)-f(x,y+\Delta y)+f(x,y)] \\\\ &=\frac{1}{\Delta x\Delta y}[f(x+\Delta x,y+\Delta y)-f(x+\Delta x,y)]-\frac{1}{\Delta x\Delta y}[f(x,y+\Delta y)-f(x,y)] \\\\ &=\frac{1}{\Delta x}\left[\frac{f(x+\Delta x,y+\Delta y)-f(x+\Delta x,y)}{\Delta y}\right]-\frac{1}{\Delta x}\left[\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y}\right] \\\\ &=\frac{1}{\Delta x}\left[\frac{\partial f(x+\Delta x,y)}{\partial y}\right]-\frac{1}{\Delta x}\left[\frac{\partial f(x,y)}{\partial y}\right] \\\\ &=\frac{\displaystyle \frac{\partial f(x+\Delta x,y)}{\partial y}-\displaystyle \frac{\partial f(x,y)}{\partial y}}{\Delta x} \\\\ &=\frac{\partial f^2(x,y)}{\partial y\partial x} \\\\ \end{align} $$
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(2.6.5)
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On the other hand


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$$ \begin{align} F(\Delta x,\Delta y) &=\frac{1}{\Delta x\Delta y}[f(x+\Delta x,y+\Delta y)-f(x+\Delta x,y)-f(x,y+\Delta y)+f(x,y)] \\\\ &=\frac{1}{\Delta x\Delta y}[f(x+\Delta x,y+\Delta y)-f(x,y+\Delta y)]-\frac{1}{\Delta x\Delta y}[f(x+\Delta x,y)-f(x,y)] \\\\ &=\frac{1}{\Delta y}\left[\frac{f(x+\Delta x,y+\Delta y)-f(x,y+\Delta y)}{\Delta x}\right]-\frac{1}{\Delta y}\left[\frac{f(x+\Delta x,y)-f(x,y)}{\Delta x}\right] \\\\ &=\frac{1}{\Delta y}\left[\frac{\partial f(x,y+\Delta y)}{\partial x}\right]-\frac{1}{\Delta y}\left[\frac{\partial f(x,y)}{\partial x}\right] \\\\ &=\frac{\displaystyle \frac{\partial f(x,y+\Delta y)}{\partial x}-\displaystyle \frac{\partial f(x,y)}{\partial x}}{\Delta y} \\\\ &=\frac{\partial f^2(x,y)}{\partial x\partial y} \\\\ \end{align} $$
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(2.6.6)
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Since $$F(\Delta x,\Delta y)=F(\Delta x,\Delta y),$$
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$$ \frac{\partial f^2(x,y)}{\partial x\partial y}=\frac{\partial f^2(x,y)}{\partial y\partial x} $$ (2.6.7)
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Comments
added some return lines to the format so that it looked less squished on 2602:306:2468:2819:608B:95AA:CE33:14DC 14:45, 17 September 2012 (UTC) Anh

- formatted equations to aid legibility --Egm6321.f12.team5.kim (talk) 03:36, 19 September 2012 (UTC)

Given
(From lecture notes Sec10.)


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$$ \frac{d\phi(x, y(x))}{dx} = 75x^4 + (\cos y)y' = 0 $$     (2.7.1)
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and


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$$ y(x) = \sin ^{-1}(k - 15x^5) $$     (2.7.2)
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Find
Verify that the equation (2.7.2) is indeed a solution of the non-linear 1st order differential equation (N1 - ODE) represented by the equation (2.7.1)

Solution
This solution was solved on our own.

We can find the expression for $$ y' $$ from equation (2.7.2) and substitute it in equation (2.7.1) to verify weather (2.7.2) is indeed the solution for the N1- ODE represented by (2.7.1)


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$$ y(x) = \sin ^{-1}(k - 15x^5) $$     (2.7.3)
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This equation can be written as,


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$$ \sin y = (k - 15x^5) $$     (2.7.4)
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Now differentiating both the sides of the above equation using the chain rule,


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$$ \begin{align} (\cos y)y' &= (0 - 5*15x^4) \\ &= -75x^4 \\ \end{align} $$     (2.7.5)
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Now, substituting the value of $$ (\cos y) y' $$ obtained from (2.7.5) into (2.7.1), we get,


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$$ \begin{align} 75x^4 + (-75x^4) &= 0 \\ 0 &= 0 \\ \end{align} $$     (2.7.6)
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Hence it is proved that $$ y(x) = \sin ^{-1}(k - 15x^5) $$ is indeed a solution of the N1 - ODE represented by the equation (2.7.1)

Comments
// Vishnu 9/14/2012 8:21pm EST

- formatted some of the equations --Egm6321.f12.team5.kim (talk) 03:42, 19 September 2012 (UTC)

Given
(From lecture notes Sec11.)

Consider the following function
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$$\underbrace{(hM)}_{\displaystyle\color{blue}{\bar M}}+\underbrace{(hN)}_{\displaystyle\color{blue}{\bar N}}\,y'=0$$ (2.8.1) Apply 2nd exactness condition to find h:
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$$\bar M_y=\bar N_x$$ (2.8.2)
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$$\begin{cases}&\bar M_y=h_y M+h M_y\\&\bar N_x=h_x N+ h N_x\end{cases}$$ (2.8.3) Thus for (2.8.1) to be exact:
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$$h_x N-h_y M+h(N_x-M_y)=0$$ (2.8.4) Solving (2.8.4) for the integrating factor h(x,y) is usually not easy.
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Find
Explain why solving (2.8.4) for the integrating factor h(x,y) is usually not easy.

Solution
This solution was solved on our own.

It is difficult to solve the equation $$h_x N-h_y M+h(N_x-M_y)=0$$, for $$h$$ since it contains the partial differentials in both $$x$$ and $$y$$ of the very function that we are solving for. Even when the simplifying assumption is made that $$h$$ is dependent only in either $$x$$ or $$y$$, the explicit determination of $$h$$ may be difficult if the functions $$M, N, M_y,$$ and $$N_x$$ are complicated.

Comments
- added to the solution --Egm6321.f12.team5.kim (talk) 03:49, 19 September 2012 (UTC)

Given
(From lecture notes Sec11.)

Consider an N1-ODE with the particular form


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$$ M(x,y) + N(x,y)y' = 0 $$     (2.9.1)
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which satisfies the 1st exactness condition, but does not satisfy the 2nd exactness condition,


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$$ M_y(x,y) \ne N_x(x,y) $$     (2.9.2)
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Using the Euler integrating factor method, it is possible to find an integrating factor so that the resulting N1-ODE satisfies both exactness conditions.


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$$ h(x,y)[M(x,y) + N(x,y)y'] = 0 $$     (2.9.3)
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$$ \underbrace{hM}_{\displaystyle \bar M} + \underbrace{hN}_{\displaystyle \bar N}y' = 0 $$     (2.9.4)
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Applying the 2nd exactness condition to equation (2.9.4) gives


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$$ \bar M_y = \bar N_x $$ $$ \begin{cases} & \bar M_y = h_yM + hM_y \\ & \bar N_x = h_xN + hN_x \end{cases} $$     (2.9.5)
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Where rearranging equation (2.9.5) so that all the terms are on the left side of the equation yields,


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$$ h_xN -h_yM + h(N_x - M_y) = 0 $$     (2.9.6)
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If it assumed that $$h_x(x,y) = 0,$$ so that $$h$$ is a function of $$y$$ only, then we can rearrange equation (2.9.6) so that it becomes


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$$ \frac{h_y}{h} = \frac{1}{M}(N_x - M_y) $$     (2.9.7)
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Find
Using equation (2.9.7), solve for $$h.$$

Solution
This solution was solved on our own.

Integrating both sides of equation (2.9.7) w.r.t. $$y$$


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$$ \int \frac{h_y}{h}dy = \int \frac{1}{M}(N_x - M_y)dy $$     (2.9.8)
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$$ log \;h(y) = \int \frac{1}{M}(N_x - M_y)dy $$     (2.9.9)
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Solving for $$h$$ by applying the exponential function on both sides of the equation,


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$$ h(y) = exp\left[\int \frac{1}{M}(N_x - M_y)dy\right] $$     (2.9.10)
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Comments
initial solution --Egm6321.f12.team5.kim (talk) 19:07, 15 September 2012 (UTC)

Given
(From lecture notes Sec11.)

A non - homogenous L1-ODE-VC,
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$$ y' + \frac{1}{x} y = x^2. $$     (2.10.1)
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where $$ a_0(x) = \frac {1}{x} $$ and  $$ b(x) = x^2 $$

Find
Show that,
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$$ h(x) = x. $$ (2.10.2) and
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$$ y(x) = \frac{x^3}{4} + \frac{k}{x}. $$     (2.10.3)
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Solution
This solution was solved on our own. From equation (3) p.11-4, we get the following expression,
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$$ h(x) = \exp \left[\int^x a_0(s) ds + k_1\right] $$     (2.10.4)
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Now substituting the value of $$ a_0(x) $$ given in the equation (2.10.1) into (2.10.4),
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$$ h(x) = \exp \left[\int^x \frac{1}{s} ds + k_1\right] $$     (2.10.5) Now integrating the equation with respect to s,
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$$ \begin{align} h(x) &= \exp [\ln (x) + k_2 + k_1] \\\\ &= \exp [\ln (x) + C] \\\\ &= \exp [\ln (x)] \exp [C] \\\\ \end{align} $$     (2.10.6) Now considering $$ \exp C $$ as a constant $$ C_1 $$
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$$ h(x) = C_1 x. $$ (2.10.7) Now when $$ C_1 = 1 $$
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$$ h(x) = x. $$ (2.10.8) Now, from equation (1) p.11-5, we get,
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$$ y(x) = \frac{1}{h(x)} \left[\int^x h(s) b(s) ds + k_2\right] $$     (2.10.9) Substituting the value of $$ h(x) $$ and $$ b(x) $$ from equations (2.10.8) and (2.10.1)
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$$ \begin{align} y(x) &= \frac{1}{x} \left[\int^x s * s^2 ds + k_2\right] \\\\ &= \frac{1}{x} \left[\int^x s^3 ds + k_2\right] \\ \end{align} $$     (2.10.10)
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Now integrating the equation with respect to s,
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$$ \begin{align} y(x) &= \frac{1}{x} \left[\frac{x^4}{4} + k_1 + k_2\right] \\\\ &= \frac{1}{x} \left[\frac{x^4}{4} + k\right] \\\\ &= \frac{x^3}{4} + \frac{k}{x} \\ \end{align} $$     (2.10.11) Hence equation (2.10.2) and (2.10.3) are proven.
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Comments
//Vishnu 9/16/2012 9:53am EST

- made formatting changes --Egm6321.f12.team5.kim (talk) 04:16, 19 September 2012 (UTC)

Given
(From lecture notes Sec11.)

The general L1-ODE-VC


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$$ a_1(x)y'+a_0(x)y=b(x). $$     (2.11.1)
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Find
The solution of $$\displaystyle y $$ when:


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$$ 1) a_1(x)=1, \; a_0(x)=x+1, \; b(x)=x^2+4$$.
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$$ 2) $$ In general for $$\displaystyle a_1(x), \; a_0(x), \; b(x)$$.
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$$ 3) a_1(x)=x^2+1, \; a_0(x)=x^3, \; b(x)=x^4$$.
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Solution
This solution was solved on our own. $$ 1)$$ Since $$ a_1(x)=1, a_0(x)=x+1, b(x)=x^2+4 $$

Then we can express the equation $$ a_1(x)y'+a_0(x)y=b(x) $$ as


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$$ \underbrace{(x+1)y}_{\displaystyle M(x,y)}+\underbrace{1}_{\displaystyle N(x,y)}*y'=x^2+4 $$     (2.11.2) Checking with Function $$ \frac{h_x}{h}=-\frac{1}{N(x,y)}(N_x-M_y) $$ We can get
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$$ -\frac{1}{N}(\underbrace{N_x}_{\displaystyle 0}-\underbrace{M_y}_{\displaystyle a_0(x)})=+a_0(x)=x+1 $$     (2.11.3)
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The integrating factor is $$ h(x)=e^{\int^{x}{a_0(s)ds+k_1} } $$

So, the integrating factor is $$ h(x)=e^{\left(\frac{x^2}{2}+x+k_1\right)} $$

Using the integrating factor and eq. $$ y(x)=\frac{1}{h(x)}\left[\int ^{x}{h(s)b(s)ds+k_2}\right] $$ yields


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$$ y(x)= e^{\left(\frac{-x^2}{2}-x-k_1\right)} \left[\int ^{x}{ e^{\left(\frac{s^2}{2}+s+k_1\right)}}*(s^2+4)ds+k_2\right] $$     (2.11.4)
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$$ 2)$$ In general terms of $$\displaystyle a_1(x), \;a_0(x), \; b(x)$$.

The integrating factor is $$ h(x)=e^{\int^{x}{\frac{a_0(x)}{a_1(x)}ds+k_1} } $$.

Using eq. $$ y(x)=\frac{1}{h(x)}\left[\int ^{x}{\frac{h(s)b(s)}{a_1(s)}ds+k_2}\right] $$ and replacing the integrating factor and right side function with the general terms, yields


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$$ y(x)=\frac{1}{e^{\int^{x}{\frac{a_0(x)}{a_1(x)}ds+k_1} }}\left[\int ^{x}{\frac{e^{\int^{s}{\frac{a_0(s)}{a_1(s)}ds+k_1} }*b(s)}{a_1(s)}ds+k_2}\right] $$.     (2.11.5)
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$$ 3)$$ Since $$ a_1(x)=x^2+1, a_0(x)=x^3, b(x)=x^4 $$

Then we can express the equation $$ a_1(x)y'+a_0(x)y=b(x) $$ as
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$$ \underbrace{\frac{x^3}{x^2+1}y}_{\displaystyle M(x,y)}+\underbrace{1}_{\displaystyle N(x,y)}*y'=\frac{x^4}{x^2+1} $$     (2.11.6)
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The integrating factor is
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$$ \begin{align} h(x)&=e^{\int^{x}{\frac{a_0(x)}{a_1(x)}ds+k_1} }\\ &=e^{\int^{s}{\frac{s^3}{s^2+1}ds+k_1} }\\ &=e^{\int^{x}{s-\frac{s}{s^2+1}ds+k_1} }\\ &=e^{\frac{x^2}{2}-\frac{1}{2}log(x^2+1)+k_1}\\ &=e^{\frac{x^2}{2}+k_1}/\sqrt{x^2+1} \end{align} $$     (2.11.7)
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Using eq. $$ y(x)=\frac{1}{h(x)}\left[\int ^{x}{\frac{h(s)b(s)}{a_1(s)}ds+k_2}\right] $$ and replacing the integrating factor and right side function with the general terms, yields


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$$ y(x)=\frac{(x^2+1)^\frac{1}{2}}{e^{\frac{1}{2}x^2+k_1}}[\int ^{x}\frac{e^{\frac{1}{2}s^2+k_1}}{(s^2+1)^\frac{1}{2}}\frac{s^4}{s^2+1}+k_{2}]=\frac{(x^2+1)^\frac{1}{2}}{e^{\frac{1}{2}x^2+k_1}}[\int ^{x}\frac{s^4 e^{\frac{1}{2}s^2+k_1}}{(s^2+1)^\frac{3}{2}}+k_{2}] $$     (2.11.8)
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Comments
- reformatted formulas and correct solutions for 1) and 3) --Egm6321.f12.team5.kim (talk) 04:57, 19 September 2012 (UTC)

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