User:EGM6321.F10.TEAM1.HW1

Problem 1: Derive (3) and (4) on p1-3
This solution is based on the one previously found by Vasquez et al. The primary improvement is the inclusion of a background section, where discussion of why these equations are important is included.

Background A very common use for the time derivative is in consideration of applied forces. Newton's 2nd law is based in an inertial frame, whereas, whereas most physical systems studied are referenced to a non-inertial coordinate system (the rotating earth). This gives rise to several inertial forces, which go to zero when an inertial reference frame is considered. Amongst these are the Coriolis force and the centrifugal force.

The Coriolis force is calculated as

$${{F}_{c}}=-2m\Omega \times v$$

where m represents the mass of the body, $${\Omega}$$ is the angular velocity of the rotating coordinate system, and v is the linear velocity of the body referenced to the rotating frame. The Coriolis effect of the Earth's rotation can often be ignored, but there are certain engineering applications. Long range ballistic missiles tend to veer off from their intended path if the Coriolis effect is not considered. The Earth's Coriolis effect also causes certain types of waves, the rotation of cyclones, and creates the jet streams and western boundary currents in the oceans. Various other systems may have important Coriolis contributions including rolling jets and rotating missile/gun turrets.

Another important force caused by a rotating frame of reference is the centrifugal force. This force can be defined mathematically as:

$$F=-m\Omega \times (\Omega \times r)$$

where $$ \Omega $$ is defined as above and r is the radius from the center of rotation of the frame. The centrifugal force acts outward from the center of rotation. Some examples of use for this force are: slings, artificial gravity through rotation, spin casting where liquid metal is placed inside of a tube and then the tube spun until the metal is dry thus creating a hollow metal cylinder, and centrifuges to separate materials based on their mass.

Given
The equation

$$f({{y}^{1}}(t),t)$$.

can be used to model the motion of a high speed train.

Find

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$$\frac{\partial }{\partial t}f({{y}^{1}}(t),t)=f{{,}_{s}}{{y}^{1}}{{,}_{t}}+f{{,}_{t}}$$

(p1-3.3)
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$$\frac{d{{t}^{2}}}({{y}^{1}},t)=f{{,}_{s}}{{y}^{1}}{{,}_{tt}}+f{{,}_{ss}}{{\left( {{y}^{1}}{{,}_{t}} \right)}^{2}}+2f{{,}_{st}}{{y}^{1}}{{,}_{t}}+f{{,}_{tt}}$$

(p1-3.4)
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Solution
Equation 3 First, for compactness define:


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$$s:={{y}^{1}}(t)$$

(p1-3.3) so that the given term can be re-written as:
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$$f(s,t)$$

(1.2)
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Next, take a derivative with respect to time of this term. This gives the equation:

$$\frac{\partial }{\partial t}f(s,t)=\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial t}\frac{\partial t}{\partial t}$$


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$$\frac{\partial }{\partial t}f(s,t)=\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial t}$$

(1.3)
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Then substituting back in the definition of s (1):

$$ \frac{d}{dt}f(y^1(t),t)= \frac{\partial f(y^1(t),t)}{\partial s} \frac{\partial y^1}{\partial t}+ \frac{\partial f(y^1(t),t)}{\partial t} $$

Finally, using the notation:

$${{y}^{1}}{{,}_{t}}=\frac{d}{dt}{{y}^{1}}(t)$$

The final equation can be expressed as:


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 * $$\frac{\partial }{\partial t}f({{y}^{1}}(t),t)=f{{,}_{s}}{{y}^{1}}{{,}_{t}}+f{{,}_{t}}$$

(p1-3.3)
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Equation 4 Begin with equation 1.3:

$$\frac{\partial }{\partial t}f(s,t)=\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial t}$$

Then, taking the time derivative term-wise will produce:

$$\frac{\partial }{\partial t}\left( \frac{\partial f}{\partial t} \right)=\frac{{{\partial }^{2}}f}{\partial {{s}^{2}}}\frac{\partial s}{\partial t}\frac{\partial s}{\partial t}+\frac{{{\partial }^{2}}f}{\partial t\partial s}\frac{\partial s}{\partial t}\frac{\partial t}{\partial t}+\frac{\partial f}{\partial s}\frac{\partial {{s}^{2}}}{\partial t\partial s}+\frac{\partial f}{\partial s}\frac{{{\partial }^{2}}s}{\partial {{t}^{2}}}\frac{\partial t}{\partial t}$$

$$\frac{{{\partial }^{2}}f}{\partial {{t}^{2}}}=\frac{{{\partial }^{2}}f}{\partial {{s}^{2}}}\frac{\partial s}{\partial t}\frac{\partial s}{\partial t}+\frac{{{\partial }^{2}}f}{\partial t\partial s}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial s}+\frac{\partial f}{\partial s}\frac{{{\partial }^{2}}s}{\partial {{t}^{2}}}$$

and again substituting in (1) and switching to comma notation will give the final equation:


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$$\frac{d^2f}{dt^2}=f,_s(y^1,t) \frac{d^2y^1}{dt^2}+f,_{ss}\left(\frac{dy^1}{dt} \right)^2+2f,_{st} \frac{dy^1}{dt}+f,_{tt} $$ (p1-3.4)
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Author

User:Egm6321.f10.team1.allison

Given

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$${{c}_{o}}({{y}^{1}},t)=-{{F}^{1}}[1-\overline{R}({{u}^{2}}{{,}_{ss}})({{y}^{1}},t)]-{{F}^{2}}{{u}^{2}}{{,}_{s}}-\frac{T}{R}+M[[1-\overline{R}({{u}^{2}}{{,}_{ss}})][{{u}^{1}}{{,}_{tt}}-\overline{R}({{u}^{2}}{{,}_{stt}})]+{{u}^{2}}{{,}_{s}}{{u}^{2}}{{,}_{tt}}]$$

(p 2-2.3)
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Find
Perform a dimensional analysis on all terms in the equation and describe the physical meaning of each.

Solution
Dimensional analysis Individual variables The abbreviations used herein are: L represents length, M for mass, F for force, and t for time $$\begin{align} & \left[ {{F}^{1}} \right]=F \\ & \left[ {\bar{R}} \right]=L \\ & [u,{}_{ss}]={}^{L}\!\!\diagup\!\!{}_\; \\ & \left[ {{y}^{1}} \right]=L \\ & [t]=t \\ & [{{F}_{2}}]=F \\ & \left[ {{u}^{2}}{{,}_{s}} \right]=L \\ & [M]=m \\ & \left[ u{{,}_{tt}} \right]={}^{L}\!\!\diagup\!\!{}_\; \\ & \left[ u{{,}_{stt}} \right]={}^{1}\!\!\diagup\!\!{}_\; \\ & \left[ {{c}_{o}} \right]=F \\ & [T]=F\times L \\ & [R]=L \\ \end{align}$$<span id="(1)">

Terms Used
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$$[{{F}^{1}}[1-\overline{R}({{u}^{2}}{{,}_{ss}})({{y}^{1}},t)]]=F(1-L{{L}^{-1}})=F$$ (2.1)
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<span id="(1)">
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$$[{{F}^{2}}{{u}^{2}}{{,}_{s}}]=F, [\frac{T}{R}]=T{{L}^{-1}}=F$$ (2.2),(2.3)
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<span id="(1)">
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$$[M1-\overline{R}({{u}^{2}}{{,}_{ss}})][{{u}^{1}}{{,}_{tt}}-\overline{R}({{u}^{2}}{{,}_{stt}})]+{{u}^{2}}{{,}_{s}}{{u}^{2}}{{,}_{tt}} =M((1-L{{L}^{-1}})(L{{S}^{-2}}-LL{{L}^{-1}}{{S}^{-2}})+(L{{L}^{-1}}L{{S}^{-2}}))=ML{{S}^{-2}}=F$$ (2.4)
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$$\therefore[{{c}_{o}}({{y}^{1}},t)]=F$$ (2.5)
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Physical Meaning

Term (1) : $${{u}^{2}}{{,}_{ss}}$$ is the second derivative of axial deformation of guideway with respect to s. $${{u}^{2}}{{,}_{ss}}$$ means curvature of deformed guideway. Physical meaning of Equation (1) is changed transverse force $${{F}^{1}}$$ by guideway deformation.

Term (2) : $${{u}^{2}}{{,}_{s}}$$ is an approximation of slope of deformed guideway. $${{F}^{2}}$$ is a axial force applied to the wheel. Physical meaning of $${{F}^{2}}{{u}^{2}}{{,}_{s}}$$ is transverse force, which is generated by axial force with deformed guideway.

Term (3) : $${T}$$ is a torque applied to the wheel. $${R}$$ is a radius of wheel. Physical meaning of $$\frac{T}{R}$$ is rotating force applied to wheel.

Term (4) : $${{u}^{2}}{{,}_{stt}}$$ is acceleration of the slope. Generated by the guideway is deforming up and down. $${{u}^{1}}{{,}_{tt}}$$ is the second derivative of transverse displacement of the flexible guideway. $${{u}^{2}}{{,}_{tt}}$$ is the second derivative of axial displacement of the flexible guideway. $${M}$$ is a mass of wheel. Physical meaning of Equation (4) is transverse force generated by acceleration of wheel.

Term (5) : Physical meaning of $${{c}_{o}}({{y}^{1}},t)$$ is sum of transverse force acting on wheel.

Author

User:EGM6321.f10.team1.Kim.TH

Given
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$${{c}_{3}}({{y}^{1}},t){{\ddot{y}}^{1}}+{{c}_{2}}({{y}^{1}},t){{({{\dot{y}}^{1}})}^{2}}+{{c}_{1}}({{y}^{1}},t){{\dot{y}}^{1}}+{{c}_{0}}({{y}^{1}},t)=0$$

(p2-2.3)
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Find
Prove that <span id="(1)">
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$${{c}_{3}}({{y}^{1}},t){{\ddot{y}}^{1}}$$

is nonlinear with respect to $$ y^1 $$.
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Solution
To prove that $${{c}_{3}}({{y}^{1}},t)\frac{d{{t}^{2}}}$$ is a nonlinear equation with respect to $${{y}^{1}}$$, we will consider a specific case for $${{c}_{3}}({{y}^{1}},t)$$. In an effort to make obvious the nonlinear nature of the previous equation, we choose the trigonometric function sine to be included in our definition of $${{c}_{3}}({{y}^{1}},t)$$.

Let

$${{c}_{3}}(y,t)=\sin (y)+{{t}^{2}}$$

Next we need to define a criterion to determine whether or not a given equation is linear or nonlinear. For L to be considered a linear operator, it must have the following properties:

$$L(\phi +\psi )=L(\phi )+L(\psi )$$

$$L(k\phi )=kL(\phi )$$

where k is any scalar.

In order to test our newly formed rules for linearity (and simplify the writing), lets express the equation

$${{c}_{3}}({{y}^{1}},t)\frac{d{{t}^{2}}}=[\sin ({{y}^{1}})+{{t}^{2}}]\frac{d{{t}^{2}}}$$

as an operator L.

Let

$$L(\cdot )=[\sin (\cdot )+{{t}^{2}}]\frac{{{d}^{2}}(\cdot )}{d{{t}^{2}}}$$

We may now apply the prior definitions for linearity as a test for whether L is linear or nonlinear. First we check if the distributive property holds by computing the LHS of the defining equation for linear operators.

$$L(\phi +\psi )=[\sin (\phi +\psi )+{{t}^{2}}]\frac{{{d}^{2}}(\phi +\psi )}{d{{t}^{2}}}$$

and using the trigonometric identity for sine:

$$\sin (\phi +\psi )=\sin (\phi )\cos (\psi )+\cos (\phi )\sin (\psi )$$

we arrive at the following:

$$=[\sin (\phi )\cos (\psi )+\cos (\phi )\sin (\psi )+{{t}^{2}}][\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+\frac{{{d}^{2}}\psi }{d{{t}^{2}}}]$$

which then becomes

$$=[\sin (\phi )\cos (\psi )[\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+\frac{{{d}^{2}}\psi }{d{{t}^{2}}}]+\cos (\phi )\sin (\psi )[\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+\frac{{{d}^{2}}\psi }{d{{t}^{2}}}]+{{t}^{2}}[\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+\frac{{{d}^{2}}\psi }{d{{t}^{2}}}]$$

and expanding into the 6 term expression we have

$$=[\sin (\phi )\cos (\psi )\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+\sin (\phi )\cos (\psi )\frac{{{d}^{2}}\psi }{d{{t}^{2}}}+\cos (\phi )\sin (\psi )\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+\cos (\phi )\sin (\psi )\frac{{{d}^{2}}\psi }{d{{t}^{2}}}+{{t}^{2}}\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+{{t}^{2}}\frac{{{d}^{2}}\psi }{d{{t}^{2}}}$$

We will see in a moment why we picked the trigonometric function sine to demonstrate a test for linearity of an operator. Now we compute the RHS of the defining equation for linear operators.

$$L(\phi )+L(\psi )=[\sin (\phi )+{{t}^{2}}]\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+[\sin (\psi )+{{t}^{2}}]\frac{{{d}^{2}}\psi }{d{{t}^{2}}}$$

distributing the derivatives over the sums we arrive at the following 4 term expression

$$=\sin (\phi )\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+{{t}^{2}}\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+\sin (\psi )\frac{{{d}^{2}}\psi }{d{{t}^{2}}}+{{t}^{2}}\frac{{{d}^{2}}\psi }{d{{t}^{2}}}$$

As shown, the nonlinearities in L with respect to $${{y}^{1}}$$ are obvious in the sine term, and L as defined does not distribute over the given sum. If L did distribute, then we would have seen equivalent expressions between the RHS and LHS of the defining equation for linearity (i.e. the RHS gave 6 terms which were not equivalent to the 4 terms the LHS gave). Therefore, $${{c}_{3}}({{y}^{1}},t)\frac{d{{t}^{2}}}$$ is a nonlinear equation with respect to $${{y}^{1}}$$.

Egm6321.fall10.team1.lang 07:23, 14 September 2010 (UTC)

Given
given the boundary values:

$$\begin{align} & y(a)=\alpha \\ & y(b)=\beta \\ \end{align}$$.

Find
Solve for variables c and d in the following equation:

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$$y(x)=cy_{H}^{1}(x)+dy_{H}^{2}(x)+{{y}_{p}}(x)$$

(p5-3)
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Solution
Substituting the boundary point data into the given equation yields:

<span id="(1)">
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$$y(a)= \alpha = cy_{H}^{1}(a)+dy_{H}^{2}(a)+{{y}_{p}}(a)$$ (4.1),(4.2)
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$$y(b)= \beta = cy_{H}^{1}(b)+dy_{H}^{2}(b)+{{y}_{p}}(b)$$ (4.3),(4.4)
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Using algebra to solve equations (4.2) and (4.4) for variable c yields:

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$$c = \frac{1}{y_{H}^{1}(a)}[\alpha - dy_{H}^{2}(a)-{{y}_{p}}(a)]$$ (4.5)
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<span id="(1)">
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$$c = \frac{1}{y_{H}^{1}(b)}[\beta - dy_{H}^{2}(b)-{{y}_{p}}(b)]$$ (4.6)
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Setting equations (4.5) and (4.6) equal to one another by recognizing that c=c yields:

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$$\frac{1}{y_{H}^{1}(a)}[\alpha - dy_{H}^{2}(a)-{{y}_{p}}(a)]= \frac{1}{y_{H}^{1}(b)}[\beta - dy_{H}^{2}(b)-{{y}_{p}}(b)]$$ (4.7)
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Solving equation (4.7) for variable d using algebra:

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$$\frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}[\alpha - dy_{H}^{2}(a)-{{y}_{p}}(a)]= \beta - dy_{H}^{2}(b)-{{y}_{p}}(b)$$ (4.8)
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$$\frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}[\alpha -{{y}_{p}}(a)]+{{y}_{p}}(b)-\beta = \frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}dy_{H}^{2}(a) - dy_{H}^{2}(b) = d(\frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}y_{H}^{2}(a) - y_{H}^{2}(b))$$ (4.9),(4.10)
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<span id="(1)">
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$$d = \frac{\frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}[\alpha -{{y}_{p}}(a)]+{{y}_{p}}(b)-\beta}{\frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}y_{H}^{2}(a) - y_{H}^{2}(b)}$$ (4.11)
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Equation (4.11) is the final answer for variable d.

Now, variable c may be found by substituting equation (4.11) into equation (4.5):

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$$c = \frac{1}{y_{H}^{1}(a)}[\alpha - \frac{\frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}[\alpha -{{y}_{p}}(a)]+{{y}_{p}}(b)-\beta}{\frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}y_{H}^{2}(a) - y_{H}^{2}(b)}y_{H}^{2}(a)-{{y}_{p}}(a)]$$ (4.12)
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Equation (4.12) is the final answer for variable c, and so the solution is now complete.

EGM6321.f10.team1.russo 22:23, 13 September 2010 (UTC)

Given
Legendre's Equation with n=1 and two solutions <span id="(1)">
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$$\begin{align} & {(1-{{x}^{2}})}{y_H}''-{2x}{y_H}'+{2}{y_H}=0 \\ \end{align}$$

(5.1)
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$$ y_H^1=x $$ and $$ y_H^2=\frac{x}{2} \log \left( \frac{1+x}{1-x} \right ) - 1 $$

Find
Show that:

$$ y_{H}^{1} $$ and $$ y_{H}^{2} $$ are both valid homogeneous solutions to the given equation.

Solution
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$${{y}_{H}^{1}}=x$$

(5.2) <span id="(1)">
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$${{y}_{H}^{1}}'=1$$

(5.3)
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$${{y}_{H}^{1}}''=0$$

(5.4) plugging equations (5.2)-(5.4) into equation (5.1):
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results in:

$$\begin{align} & (1-{{x}^{2}})[0]-2x[1]+2[x]=0 \\ & -2x+2x=0 \\ & 0=0 \\ \end{align}$$

$$\begin{align} & \therefore {{y}_{H}^{1}}=x \end{align}$$ is a valid solution.

similarly, the second solution

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$${y_{H}^{2}}=\frac{x}{2}\ln \left( \frac{1+x}{1-x} \right)-1$$

(5.5)
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can be shown to be a homogeneous solution to the solution by:

$$\begin{array}{*{35}{l}} {} & y{{_{H}^{2}}^{\prime }}=\frac{1}{2}\ln \left( \frac{1+x}{1-x} \right)+\frac{x}{2}{{(\ln \left( 1+x \right)-\ln \left( 1-x \right))}^{\prime }} \\ \end{array}$$ <span id="(1)">
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$$=\frac{1}{2}\ln \left( \frac{1+x}{1-x} \right)+\frac{x}{2}\left( \frac{1}{1+x}+\frac{1}{1-x} \right)$$

(5.6)
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$$\begin{array}{*{35}{l}} {} & y{{_{H}^{2}}^{\prime \prime }}=\frac{1}{2}\left( \frac{1}{1+x}+\frac{1}{1-x} \right)+\frac{1}{2}\left( \frac{1}{1+x}+\frac{1}{1-x} \right)+\frac{x}{2}\left( \frac{-1}+\frac{1} \right) \\ \end{array}$$ <span id="(1)">
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$$=\left( \frac{1}{1+x}+\frac{1}{1-x} \right)+\frac{x}{2}\left( \frac{-1}+\frac{1} \right)$$

(5.7)
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$$\begin{array}{*{35}{l}} {} & (1-{{x}^{2}})y{{_{H}^{2}}^{\prime \prime }}=(1-{{x}^{2}})\left( \frac{1}{1+x}+\frac{1}{1-x} \right)+\frac{x(1-{{x}^{2}})}{2}\left( \frac{-1}+\frac{1} \right) \\ {} & =(1-x+1+x)+\frac{1}{2}\left( \frac{-x(1-x)}{(1+x)}+\frac{x(1+x)}{(1-x)} \right) \\ \end{array}$$ <span id="(1)">
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$$=2+\frac{x}{2}\left( \frac{4x}{1-{{x}^{2}}} \right)=2+\left( \frac{2{{x}^{2}}}{1-{{x}^{2}}} \right)$$

(5.8)
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$$\begin{array}{*{35}{l}} {} & -2xy{{_{H}^{2}}^{\prime }}=-2x\left[ \frac{1}{2}\ln \left( \frac{1+x}{1-x} \right)+\frac{x}{2}\left( \frac{1}{1+x}+\frac{1}{1-x} \right) \right] \\ {} & =-2x\left[ \frac{1}{2}\ln \left( \frac{1+x}{1-x} \right)+\left( \frac{x}{1-{{x}^{2}}} \right) \right] \\ \end{array}$$ <span id="(1)">
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$$=-x\ln \left( \frac{1+x}{1-x} \right)-\frac{2{{x}^{2}}}{1-{{x}^{2}}}$$

(5.9)
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$$\begin{array}{*{35}{l}} {} & 2y_{H}^{2}=2\left[ \frac{x}{2}\ln \left( \frac{1+x}{1-x} \right)-1 \right] \\ \end{array}$$ <span id="(1)">
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$$=x\ln \left( \frac{1+x}{1-x}\right)-2$$

(5.10)
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Equations (5.5)-(5.10) can then be substituted into equation (5.1) $$\begin{align} {(1-{{x}^{2}})}{y_{H}^{2}}''-{2x}{y_{H}^{2}}'+{2}{y_{H}^{2}} \\ \end{align}$$ $$={2}+{\frac{{2x}^{2}}{1-{{x}^{2}}}}-{x}\ln \left( \frac{1+x}{1-x} \right)-\frac{{2x}^{2}}{1-{{x}^{2}}}+{{x}\ln \left( \frac{1+x}{1-x} \right)-{2}}=0 $$

$$\begin{align} \therefore {y_{H}^{2}}=\frac{x}{2}\ln \left( \frac{1+x}{1-x} \right)-1

\end{align}$$ is a valid solution.

Author

User:EGM6321,F10.TEAM1.WILKS

Signatures
Solved problem 1 --Bryan Allison (BA)

Solved problem 2 --Taeho Kim (TK)

Solved problem 3 --Paul Lang (PL)

Solved problem 4 --Charles Russo (CR)

Solved problem 5 --192.91.147.35 17:06, 15 September 2010 (UTC)User:EGM6321,F10.TEAM1.WILKS|David Wilks (DW)]]