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User talk:Egm6321.f2010.team5.riveros

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title: User talk:Egm6321.f2010.team5.riveros


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User:EGM6321.f10.team6.cook/hw2

= General Form of First Order Differential Equations =

Given
The general form for a first order nonlinear ordinary differential equation is,

Required
Verify the given equation is nonlinear, and give an example of a linear first order ordinary differential equation.

Derivation
Consider a general two-dimensional first order differential equation $$\begin{align}du(x,y)\end{align}$$ which has continuous partial derivatives so that its total differential is

If $$\begin{align} {u=\text{const}} \end{align}$$ then $$\begin{align} du=0 \end{align}$$ (exact equation)

Defining $$M,N$$

So that,

Taking the partial derivatives of $$M,N$$

By assumption of continuity through Schwarz's theorem the mixed derivatives are equal

Which is a necessary and sufficient requirement for eq 1.1.

Demonstrating Nonlinearity
In the derivation only $$\begin{align}du=0 \end{align}$$ and continuous functions were required. Thus, to demonstrate the nonlinearity of the general form only one example must be shown where the general form is nonlinear. It can also be concluded immediately because linearity was not required of $$\begin{align}M\end{align}$$ and $$\begin{align}N\end{align}$$ that nonlinearity is included in the general form. By example, let

Then checking that $$\begin{align}f(cy)=cf(y)\end{align}$$

and thus this function is nonlinear, making the general equation nonlinear in general as one such case has been found.

== Bernoulli Equation ==

Bernoulli's equation is a nonlinear ordinary differential equation which can be transformed into a linear ordinary differential equation.

Equation (1.8) is known as the Bernoulli equation and is nonlinear for $$a$$ greater than or equal to 2 ($$n$$ is an integer). The nonlinearity may be seen clearly by letting $$ \begin{align} {y}={c}{y} \end{align} $$.

It may be written in the form of equation 1.1, like so

Note that the requirement of equation 1.7 is not met and the equation is neither exact nor linear.

Linearization
The equation can be linearized by letting

Differentiating

Which arrives at a new first order linear ordinary differential equation

Note that the partial of $$M$$ with respect to $$y$$ is zero, as is the partial of $$N$$ with respect to $$x$$; thus, equation 1.12 is both exact as well as linear.

Given

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$$ {\phi} \left(x,y\right) = x^{2} y^{3/2} + log (x^{3} y^{2}) = k $$ (4.1)
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Required
Find $$ F(x,y,y') = {\frac {d {\phi}}{d x}} (x,y) $$ and verify that F is exact N1_ODE and invent 3 more

Solution

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$$ F(x,y,y') = {\frac {d{\phi}(x,y)}{d x}} = {\frac} + {\frac} {\frac{dy}{dx}}$$ (4.1)
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$$\begin{align} & M(x,y):={\phi}_{x}(x,y): =\frac{\partial \phi(x,y)}{\partial x} \\ & N(x,y):={\phi}_{y}(x,y): =\frac{\partial \phi(x,y)}{\partial y} \\ \end{align}$$ (4.2)
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So, that is
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$$F(x,y,y') = M\left(x,y\right)+N(x,y)y' $$      (4.3)
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From the given $${{\phi}\left(x,y\right)}$$, we can get M(x,y) and N(x,y)

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$$\begin{align} &M\left(x,y\right) :={\phi}_{x}(x,y)= 2 x y^{\frac{3}{2}} + {\frac{3}{x}}\\ &N(x,y) := {\phi}_{y}(x,y)= {\frac{3}{2}} y^{\frac{1}{2}}x^{2} + {\frac{2}{y}}\\ \end{align} $$      (4.4)
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Follow the $$ 1^{st} $$ condition of exactness,

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$$F(x,y,y') = M\left(x,y\right)+N(x,y)y' = \left( 2 x y^{\frac{3}{2}} + {\frac{3}{x}} \right) + \left( {\frac{3}{2}} y^{\frac{1}{2}}x^{2} + {\frac{2}{y}} \right) y' = 0 $$      (4.5)
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$$ y' = - \frac{M\left(x,y\right)}{N(x,y)} $$      (4.6)
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And then, apply $$ 2^{nd} $$ condition of exactness,

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$$ \underbrace{M_{y}\left(x,y\right)}_{\frac{\partial M(x,y)}{\partial y}}= \underbrace{N_{x}(x,y)}_{\frac{\partial N(x,y)}{\partial x}} $$      (4.7)
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$$ {\frac {{\partial}M(x,y)} {\partial y} } = {\frac {{\partial} \left( 2 x y^{\frac{3}{2}}+{\frac{3}{x}}\right)} {\partial y}} = 3 x y^{\frac{1}{2}} = {\frac {{\partial}\left( {\frac{3}{2}} y^{\frac{1}{2}} x^{2} + {\frac{2}{y}}\right)} {\partial x}} = {\frac {{\partial} N(x,y)}{\partial x}} $$      (4.8)
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So, F is exact N1_ODE.

And 3 more equations,

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$$ {\phi}(x, y) = {\frac{1}{2}} x^{4} sin(y^{2}) = k $$ (4.9)
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$$ {\phi}(x, y) = {\frac{1}{4}}( x^{4} + {6} x^{2} y^{2} + y^{4})= k $$ (4.10)
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$$ {\phi}(x, y) = e^ = k $$ (4.11)
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Required
Find the function of $$f\left( y' \right)$$ such that there is no analytical solution to $$f\left( y' \right)=-\frac{M\left( x,y \right)}{N\left( x,y \right)}$$.

== Solution == The solution of the problem is equivalent to find a N1_ODE that does not satisfy the 1st condition of exactness. In order to a N1_ODE to be exact, it must be in the form of:

The function chosen to test the first condition of exactness is:

From Equation 5.3 we can identify the functions, $$M\left( x,y \right)$$, $$N\left( x,y \right)$$ and $$f\left( y' \right)$$ as:

If we can rearrange Equation 5.3 to fit in the form of the Equation 5.2, than the 1st condition of exactness, if not we found Non-Exact N1_ODE.

Doing some algebra with Equation 5.3 to arrange in the fashion of $$f\left( y' \right)=-\frac{M\left( x,y \right)}{N\left( x,y \right)}$$:

Equation 5.7 does not fit in the same arrangement of Equation 5.2, since we have $$\begin{align} y'+\cos y'\end{align}$$ on the left hand side of the equation instead of only $$\begin{align}y' \end{align} $$.

Therefore there is no analytical solution to the function chosen and also proves that Equation 5.3 is a Non Exact N1_ODE.

Required
Verify that equation 6.1 satisfies the first condition of exactness, equation 6.2.

The first condition of exactness:

where,

Solution
From the equation 6.1,

Total differential of function $$\phi\left(x,y\right)$$,

As$$\phi\left(x,y\right)=k=const$$, then $$ d\mathbf{\phi}=0 $$, We can rewrite the equation 6.5,

For the given problem,

So, we can verify that the equation 6.1 could be converted to the form of the first condition of exactness, equation 6.2.

The mixed derivatives of $$\begin{align}\phi\end{align}$$, $$\begin{align}\phi_{xy},\phi_{yx} \end{align}$$ are both zero, meeting the second condition of exactness. This is clearly seen from equation 6.7 as $$\begin{align}\phi_{x} \end{align}$$ is only a function of $$\begin{align}x\end{align}$$ while $$\begin{align} \phi_{y} \end{align} $$ is just a function of $$\begin{align}y \end{align}$$.

Given
The Euler Integrating Factor Method may be applied to homogeneous differential equations when the equations meet the first condition of exactness, but do not meet the second condition of exactness. That is to say, those equations that can me made to fit the form shown in equation 7.1, but do not meet the criteria shown in equation 7.2 can be made exact using the Euler method. One of the benefits of the Integrating Factor Method is that it can be applied to solve non-homogenous differential equations.

Required
(i) Solve the following non-homogeneous differential equation using the Integrating Factor Method:

(ii) Assume that $${{a}_{1\left( x \right)}}\ne 0$$ for any $$x$$ and find an expression for $${{y}_{\left( x \right)}}$$ in terms of $$\begin{align}{{a}_{0}}\end{align}$$, $$\begin{align}{{a}_{1}}\end{align}$$, $$\begin{align}{b}\end{align}$$ for the following form:

(iii) Apply the integrating factor method to the following equation:

== Solution ==

Explanation of Process
The Integrating Factor Method assumes that some function $${{h}_{\left( x,y \right)}}$$ can be found to make a differential equation exact when it meets the first criterion of exactness (equation 7.1) but not the second (equation 7.2). This factor is multiplied by the original equation to yield equation 7.6.

Applying the second condition of exactness and using the product rule to differentiate each term we obtain:

If we assume that the partial derivative of h with respect to y is zero, then h becomes a function of x only and we are left with the following:

Which can be rearranged to obtain:

Dividing by h and rearranging terms we obtain:

Integrating both sides with respect to x we obtain the following function we can use to obtain h:

If we return to our original non-exact equation given in equation 7.1 we can apply the integrating factor to obtain the following equation:

If we recall that $$h\left( {y}'+{{a}_{0}}y \right)=hb$$, we can rearrange the above to obtain the following:

The reader might recognize that the left side of equation 7.13 is really nothing more than $${{\left( hy \right)}^{\prime }}$$. Substituting into 7.13 we obtain:

After integrating we are left with the solution to y.

Solution to Part I
From equation 7.3 we obtain the following:

Substituting into equation 7.11 we obtain an expression for h:

Combing the result from equation 7.17 with equation 7.15 we obtain the solution to y.

Rearranging the terms:

Solving the first Integral

Using the substitution method

Substituting in the Euqation 7.21 we get

The integral in equation 7.24 cannot be solved using standard techniques. Numerical methods or approximations that employ series expansions must be used.

Solution to Part II
From equation 7.3 we obtain the following:

Substituting into equation 7.11 we obtain an expression for h:

Combing the result from equation 7.26 with equation 7.15 we obtain the solution to y.

Solution to Part III
First we need to rearrange Equation 7.5 to be in the form of:

Rearranging Equation 7.5 we get:

Calculating $${{M}_{y}}$$, $$N$$ and $${{N}_{x}}$$:

Substituting into equation 7.11 we obtain an expression for h:

From equation 7.27 we have a general solution in terms of $${{a}_{0}}$$, $${{a}_{1}}$$, $${b}$$. Substituting the terms given in the problem statement in equation 7.5 we obtain:

Using the substitution method to solve the integral

Substituting Equation 7.33 into 7.15:

Using the substitution method to solve the integral

From Equation 7.34 we get the final answer for $$y(x)$$

Required
1. Show that an integration constant is not necessary in

2. Show the formulation,

agrees with with that in King

which is in the form

3. Find $$y_h(x)$$ independently, i.e, solve $$y'+a_0y=0$$

Part 1
Suppose an integration constant was included

Then recognizing the constant is general, and using exponential rules

The function $$h(x)$$ appears inverted before the integral,

so that the constant can be moved outside of the integration and cancelled with the same constant in the inverted function outside the integration. Thus, the constant is superfluous.

Part 2
The derivation by King is as follows

Multiply by through by

To get

Noting that the left hand side is the product rule carried out for the total derivative

Integrating now that we have a total derivative (note that the integration variable goes from x to s on the right hand side)

Dividing through by the factor on the left hand side arrives at King's form,

Now let $$h(x)$$ be the integrating factor

Noting that integration generates a constant, we can drop $$A$$. Letting $$Q$$ become $$b$$

Thus it is shown that the two forms are equivalent.

Part 3
Finding a solution to

First, assume a solution of the form

taking the derivative

Noting that $$f$$ is the negative integral of $$a_0$$

which is the solution to the homogeneous equation

easily verified by substituting $$y'$$ and $$y$$

Note that the solution is the homogeneous solution, the first term in King's formulation from part 2.

Required
where,
 * Show the given problem is either exact or can be made exact
 * Find $$\phi\left(x,y\right)=k$$

Verify the Exactness
We can get the differential equation from the given conditions.

where,

First condition of exactness
Equation 9.4 satisfies the 1st condition of exactness because it has the form of equation 9.5

Second condition of exactness
Before we proceed, we can rewrite equation 9.4 as $$ {e}^{2y}\neq 0 $$

Checking whether the equation satisfies equation 9.7

It is found equation 9.6 is not exact.

Making the equation exact by multiply Euler integrating factor $$ h\left(x\right) $$.

Multiply equation 9.11 into equation 9.6

Alternative Approach
It is not necessary to find $$ h\left(x\right) $$ with such verbosity.

$$ N\left(x,y\right) $$ and $$ M\left(x,y\right) $$ are the functions of x only, by making $$ N\left(x\right)=const $$ we can make $$ {M}_{y}\left(x,y\right)={N}_{x}\left(x,y\right)$$

in other words, by just multiplying $${\color{blue}\frac{1}{\mathrm{sin}x}}$$, we have made $$ N\left(x,y\right)=1 $$

then, $$ {M}_{y}\left(x,y\right)={N}_{x}\left(x,y\right)=0$$

This is the same integrating factor that we had found at equation 9.11

Exactness of the new Equation
Let's check the exactness of equation 9.12.

First condition of exactness
Equation 9.12 satisfies the 1st condition of exactness because it has the form of equation 9.5

Second condition of exactness
As equation 9.12 satisfies both the 1st and 2nd condition of exactness, it is an EXACT ODE.

Find $$\phi(x,y)=k$$
As for $$\phi\left(x,y\right)$$

We can find $$\phi\left(x,y\right)=k$$ by integrating $$M\left(x,y\right)$$ for x, or $$N\left(x,y\right)$$ for y.

i.e.

Therefore,

We can find $$k\left(y\right)$$ by $$\frac{\partial \phi(x,y)}{\partial y}$$

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

title: User:EGM6321.f10.team6.cook/hw2


text:
User talk:Rod Corella

 Hello Rod Corella, 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

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User talk:Egm6321.f10.team4.Corella.RL

 Hello Egm6321.f10.team4.Corella.RL, 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

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title: User talk:Egm6321.f10.team4.Corella.RL


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User talk:Egm6321.f10.team5.faraone/hw1

test
Team#5 Page

test mf

title: User talk:Egm6321.f10.team5.faraone/hw1


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User:Egm6321.f10.team5.abc/HW1

Problem 1
Derive the following equations. $$\frac {d} {dt} f(Y^1(t),t) = \frac {\partial f} {\partial S} (Y^1(t),t) \dot {Y}^1 + \frac {\partial f} {\partial t} (Y^1(t),t)$$ $$\frac {d^2 f} {d t^2} = f_{,S}(Y^1,t) \ddot{Y}^1 + f_{,SS} (\dot {Y}^1)^2 + 2 f_{,St} \dot {Y}^1 + f_{,tt}$$

Solution 1
Since $$f(y^1(t),t)$$ is defined as $$f(Y^1(t),t) = \left. f(s,t) \right |_{S=Y^1(t)}$$, we know that $$s$$ is a fuction of $$t$$. Therefore, to take a derivative of $$f$$ with respect to $$t$$, we need to use the chain rule for several variables. $$\frac {df} {dt} = \frac {\partial f} {\partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial t} \frac {\partial t} {\partial t}$$ where $$\frac {\partial t} {\partial t} = 1$$. So we reduce the equation as following. $$\frac {df} {dt} = \frac {\partial f} {\partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial t}$$ Note that $$\frac {\partial s} {\partial t} = \frac {\partial Y^1(t)} {\partial t} = \dot {Y}^1$$. Expressing the first-order derivative formally, we have $$\frac {df(Y^1(t),t)} {dt} = \frac {\partial f(Y^1(t),t)} {\partial s} \dot {Y}^1 + \frac {\partial f(Y^1(t),t)} {\partial t}$$ Therefore P.1-3(3) is derived completely.

Now we take the second derivative of $$f$$. $$\frac {d^2 f} {dt^2} = \frac d {dt} \left ( \frac {df} {dt} \right ) = \frac d {dt} \left (\frac {\partial f} {\partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial t} \right ) = \frac {\partial s} {\partial t} \frac d {dt} \frac {\partial f} {\partial s} + \frac {\partial f} {\partial s} \frac d {dt} \frac {\partial s} {\partial t} + \frac d {dt} \frac {\partial f} {\partial t}$$ Since the chain rule needs to be applied again, we write the equation as the following. $$\frac d {dt} \left ( \frac {df} {dt} \right ) = \frac {\partial s} {\partial t} \left ( \frac \partial {\partial s} \frac {\partial s} {\partial t} + \frac \partial {\partial t} \right ) \frac {\partial f} {\partial s} + \frac {\partial f} {\partial s} \left ( \frac \partial {\partial s} \frac {\partial s} {\partial t} + \frac \partial {\partial t} \right ) \frac {\partial s} {\partial t} + \left ( \frac \partial {\partial s} \frac {\partial s} {\partial t} + \frac \partial {\partial t} \right ) \frac {\partial f} {\partial t}$$ $$= \frac {\partial^2 f} {\partial s^2} \left ( \frac {\partial s} {\partial t} \right )^2 + \frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial s} \frac {\partial s} {\partial t} \frac {\partial^2 s} {\partial s \partial t} + \frac {\partial f} {\partial s} \frac {\partial^2 s} {\partial t^2} + \frac {\partial s} {\partial t} \frac {\partial^2 f} {\partial s \partial t} + \frac {\partial^2 f} {\partial t^2}$$ Because $$\frac {\partial^2 s} {\partial s \partial t} = 0$$ and $$\frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t} + \frac {\partial s} {\partial t} \frac {\partial^2 f} {\partial s \partial t} = 2 \frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t}$$, we can simplify the equation as following. $$\frac {\partial^2 f} {\partial t^2} = \frac {\partial^2 f} {\partial s^2} \left ( \frac {\partial s} {\partial t} \right )^2 + 2 \frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial s} \frac {\partial^2 s} {\partial t^2} + \frac {\partial^2 f} {\partial t^2}$$ Expressing the second-order derivative formally, we have $$\frac {\partial^2 f(Y^1(t),t)} {\partial t^2} = \frac {\partial^2 f(Y^1(t),t)} {\partial s^2} (\dot {Y}^1)^2 + 2 \frac {\partial^2 f(Y^1(t),t)} {\partial t \partial s} \dot {Y}^1 + \frac {\partial f} {\partial s} \ddot {Y}^1 + \frac {\partial^2 f(Y^1(t),t)} {\partial t^2}$$ Therefore P.1-3(4) is derived completely.

Problem 2
Do dimensional analysis of all terms in the following equation, and describe the physical meaning. $$c_0 (y^1,t) = - F^1 [1 - \bar {R} u^2_{,ss} (y^1,t)] - F^2 u^2_{,s} - \frac T R + M \lbrace [1 - \bar R u^2_{,ss}] [u^1_{,tt} - \bar R u^2_{,stt}] + u^2_{,s} u^2_{,tt} \rbrace$$

Solution 2
We first do the dimesional analysis of each variable and state its physical meaning.

$$c_0 = [\frac {ML} {T^2}]$$ : horizontal force acting on wheel

$$F^1 = [\frac {ML} {T^2}]$$ : horizontal component of force

$$\bar R = [L]$$ : rdistance from the center of the guideway to the center of the wheel

$$u^2_{,ss} = [\frac L {L^2}] = [\frac 1 L]$$ : curvature

$$F^2 = [\frac {M L} {T^2}]$$ : vertical component of force

$$u^2_{,s} = [\frac L L] = [1]$$ : slope of transversal deformation(display) of the guideway

$$T = [\frac {M L^2} {T^2}]$$ : torque

$$R = [L]$$ : radius of the wheel

$$M = [M]$$ : mass of the wheel

$$u^1_{,tt} = [\frac L {T^2}]$$ : acceleration of axial deformation (display) of the guideway

$$u^2_{,tt} = [\frac L {T^2}]$$ : acceleration of transversal deformation (display) of the guideway

$$u^2_{,stt} = [\frac L {L T^2}] = [\frac 1 {T^2}]$$ : acceleration of slope of transversal deformation (display) of the guideway

\noindent Now we do the complete dimensional analysis for the equation. $$[\frac {ML} {T^2}] = - [\frac {M L} {T^2}] ([1] - [L \frac 1 L] ) - [\frac {M L} {T^2}] [1] - [\frac {M L^2} {T^2} L] + [M \lbrace ([1] - [L \frac 1 L])([\frac L {T^2}] - [L \frac 1 {T^2}]) + [1 \frac L {T^2}] \rbrace$$ $$= - [\frac {M L} {T^2}] - [\frac {M L} {T^2}] - [\frac {M L} {T^2}] + [M \lbrace ([\frac L {T^2}] - [\frac L {T^2}] - [\frac L {T^2}] + [\frac L {T^2}] + [\frac L {T^2}] \rbrace$$ $$=- [\frac {M L} {T^2}] - [\frac {M L} {T^2}] - [\frac {M L} {T^2}] + [\frac {ML} {T^2}] - [\frac {ML} {T^2}] - [\frac {ML} {T^2}] + [\frac {ML} {T^2}] + [\frac {ML} {T^2}]$$ Is it verified that each term in this equation has the dimension of $$[\frac {ML} {T^2}]$$ which is the unit of force.

Problem 3
Show $$c_3(Y^1,t) \ddot {Y}^1$$ is nonlinear with respect to $$Y^1$$ where $$c_3(Y^1,t) = M[1 - \bar R u^2_{,ss} (Y^1,t)]$$.

Solution 3
Nonlinearity is defined as $$F(\alpha u + \beta v) \ne \alpha F(u) + \beta F(v)$$. $$c_3(Y^1,t) \ddot {Y}^1 = M[1 - \bar R u^2_{,ss} (Y^1,t)] \ddot {Y}^1$$ Given that $$u^2_{,ss} (Y^1,t)$$ is curvature, we define it as $$X$$. $$c_3(Y^1,t) \ddot {Y}^1 = M[1 - \bar R X] \ddot {Y}^1$$ Define a differential operator as below. $$F(\cdot) = M (1 - \bar R X) \frac {\partial^2 (\cdot)} {\partial X^2}$$ This operator must satisfty the condition that $$F(\alpha u + \beta v) = \alpha F(u) + \beta F(v)$$ where $$\alpha$$ and $$\beta$$ are real numbers while $$u$$ and $$v$$ are functions of $$X$$. Now we test the condition of linearity. $$F(\alpha u + \beta v) = M (1 - \bar R X) \frac {\partial^2 (\alpha u + \beta v)} {\partial X^2}$$ $$\alpha F(u) + \beta F(v) = \alpha M (1 - \bar R X) \frac {\partial^2 u} {\partial X^2} + \beta M (1 - \bar R X) \frac {\partial^2 v} {\partial X^2}$$

Problem 4
Boundary Value Problem $$y(a)=\alpha$$, $$y(b)=\beta$$. Find $$c$$,$$d$$ in terms of $$\alpha$$, $$\beta$$.

Solution 4
The function $$y(x)$$ is defined accordingly. $$y(x) = c y_h^1(x) + d y_h^2(x) + y_p(x)$$ Using the boundary value conditions, we get the following relations. $$y(a) = c y_h^1(a) + d y_h^2(a) + y_p(a) = \alpha$$ $$y(b) = c y_h^1(b) + d y_h^2(b) + y_p(b) = \beta$$ Now we need to express $$c$$ and $$d$$ in terms of $$\alpha$$ and $$\beta$$. $$c = \frac {\beta - d y_h^2(b) - y_p(b)} {y_h^1(b)} \qquad d = \frac {\beta - c y_h^1(b) - y_p(b)} {y_h^2(b)}$$ $$y_h^1(a) + \frac {\beta - c y_h^1(b) - y_p(b)} {y_h^2(b)} y_h^2(a) + y_p(a) = c \left (y_h^1(a) - y_h^2(a) \frac {y_h^1(b)} {y_h^2(b)} \right ) + y_h^2(a) \frac {\beta - y_p(b)} {y_h^2(b)} + y_p(a) = \alpha$$ $$c = \frac {\alpha - y_h^2(a) \frac {\beta - y_p(b)} {y_h^2(b)} - y_p(a)} {y_h^1(a) - y_h^2(a) \frac {y_h^1(b)} {y_h^2(b)}} = \frac {\alpha {y_h^2(b)} - y_h^2(a) (\beta - y_p(b)) - y_p(a) {y_h^2(b)}} {y_h^1(a) {y_h^2(b)} - y_h^2(a) {y_h^1(b)}}$$ $$\frac {\beta - d y_h^2(b) - y_p(b)} {y_h^1(b)} y_h^1(a) + d y_h^2(a) + y_p(a) = d \left (y_h^2(a) - \frac {y_h^2(b)} {y_h^1(b)} y_h^1(a)\right ) + \frac {\beta - y_p(b)} {y_h^1(b)} y_h^1(a) + y_p(a) = \alpha$$ $$d = \frac {\alpha - \frac {\beta - y_p(b)} {y_h^1(b)} y_h^1(a) - y_p(a)} {y_h^2(a) - \frac {y_h^2(b)} {y_h^1(b)} y_h^1(a)} = \frac {\alpha y_h^1(b) - (\beta - y_p(b)) y_h^1(a) - y_p(a) y_h^1(b)} {y_h^2(a) y_h^1(b) - y_h^2(b) y_h^1(a)}$$

Problem 5
Verify that $$L_2(y_H^1) = L_2(y_H^2) = 0$$.

Solution 5
The fuction $$L_2(y)$$ is defined as $$L_2(y)=(1-x^2)y''-2xy'+2y$$ Then we easily verified that $$L_2(y_H^1(x))$$ is 0. $$L_2(y_H^1(x)) = L_2(x) = (1-x^2) 0 - 2 x 1 + 2 x = 0$$ Meanwhile, checking $$L_2(y_H^2(x))$$ is cumbersome. $$L_2(y_H^2(x)) = L_2(\frac x 2 \log \left(\frac {1+x} {1-x} \right)- 1)$$ We calculate $$y_H^2(x)'$$ first. $$y_H^2(x)' = (\frac x 2 \log \left(\frac {1+x} {1-x} \right)- 1)' = \frac 1 2 \log \left(\frac {1+x} {1-x} \right) + \frac x 2 \frac {1-x} {1+x} \left( \frac 1 {1-x} + \frac {1+x} {(1-x)^2} \right)$$ $$ = \frac 1 2 \log \left(\frac {1+x} {1-x} \right) + \frac x 2 \frac {1-x} {1+x} \frac {1-x+1+x} {(1-x)^2} = \frac 1 2 \log \left(\frac {1+x} {1-x} \right) + \frac x {1-x^2}$$ Now we calculate $$y_H^2(x)''$$ using $$y_H^2(x)'$$. $$y_H^2(x)'' = (\frac 1 2 \log \left(\frac {1+x} {1-x} \right) + \frac x {1-x^2})' = \frac 1 2 \frac {1-x} {1+x} \left( \frac 1 {1-x} + \frac {1+x} {(1-x)^2} \right) + \left( \frac 1 {1-x^2} + \frac {2 x^2} {(1-x^2)^2} \right)$$ $$=\frac 1 2 \frac {1-x} {1+x} \frac {1-x+1+x} {(1-x)^2} + \frac {1-x^2+2x^2} {(1-x^2)^2} = \frac 1 {1-x^2} + \frac {1+x^2} {(1-x^2)^2} = \frac 2 {(1-x^2)^2}$$ Substituing the derivatives into the equation of $$L_2(y)$$ $$L_2(y_H^2(x))=(1-x^2) \frac 2 {(1-x^2)^2} - 2 x (\frac 1 2 \log \left(\frac {1+x} {1-x} \right) + \frac x {1-x^2}) + 2 (\frac x 2 \log \left(\frac {1+x} {1-x} \right)- 1)$$ $$= \frac 2 {1-x^2} - \frac {2 x^2} {1-x^2} - 2 = \frac {2(1-x^2)} {1-x^2} - 2 = 2 - 2 =0$$ It is verified that $$L_2(y_H^2(x))$$ is 0.

Therefore we verified that $$L_2(y_H^1(x)) = L_2(y_H^2(x)) = 0$$.

title: User:Egm6321.f10.team5.abc/HW1


text:
User talk:Egm6321.f10.team5.abc

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title: User talk:Egm6321.f10.team5.abc


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User talk:Egm6321.f10.team4.ejm

 Hello Egm6321.f10.team4.ejm, 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

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title: User talk:Egm6321.f10.team4.ejm


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Problem 4
Find $$F(x,y,y') = \frac {d\phi} {dx} (x,y)$$ where $$\phi(x,y) = x^2 y^{\frac 3 2} + \log (x^3 y^2) = k$$ and verify that $$F$$ is exact N1_ODE and invent 3 more.

Solution 4
First, differentiate $$\frac {d\phi} {dx} (x,y)$$ To do so, we use the product rule and the chain rule for the first term in $${phi}(x,y)$$ and a Logarithmic derivative for the second term. So then $$F(x,y,y')$$ comes out to be: $$F(x,y,y') = 2 x y^{\frac 3 2} + \frac 3 {x} + [{\frac 3 2} x^2 y^{\frac 1 2} + \frac 2 y] y'=0$$

where the constant $$k$$ goes to zero after differentiation.

Verification
For $$F(x,y,y')$$ to meet the criteria of an exact N1_ODE it must meet three conditions: 1) Have an integral $$\frac {d\phi} {dx} (x,y)$$ that exists, which it clearly does.

2) Fit the form $$M(x,y) + N(x,y) y' = 0$$, which it does if we make $$M(x,y) = 2 x y^{\frac 3 2} + \frac 3 {x}$$and $$N(x,y) = [{\frac 3 2} x^2 y^{\frac 1 2} + \frac 2 y]$$

3) If $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$, which we can see that it does as well.

Therefore it can be said that $$F(x,y,y')$$ is an exact non-linear first order ordinary differential equation.

Three other examples
Here are three other examples of exact N1_ODE

1) $$F(x,y,y') = 5 x^4 ln(y) + x^5 \frac 1 y y'$$

2) $$F(x,y,y') = 5 x^{\frac 3 2} y^2 + 4 x^{\frac 5 2} y y' $$

3) $$F(x,y,y') = 3 \frac 1 x y^3 + 9 ln(x) y^2 y' $$

Problem 1 - Derive EOM (SC-N1-ODE)
From Meeting 13, p. 13-1 ~ p. 13-2

Given
A projectile of mass (m) is being shot through the air at an angle (α) to the horizontal with velocity (v). The projectile experiences a drag force due to air resistance, $$F_{air} = k v^n \,$$ Where $$k \,$$ is the coefficient of air resistance and is a constant in this problem. The force equations are therefore given as such, $$m(\frac {dv_x} {dt}) = -k v^{n} {\cos {\alpha}}$$ $$\displaystyle (Eq. 1.1) $$ $$m(\frac {dv_y} {dt}) = -k v^{n} {\sin {\alpha}} - mg$$ $$\displaystyle (Eq. 1.2) $$

Find
For the given situation: $$m(\frac {dv_y} {dt}) = -k v^{n} {\sin {\alpha}} - mg$$
 * 1) Derive the equations of motion for the projectile
 * 2) For the particular case $$k = 0 \,$$ verify that $$y(x)\,$$ is a parabola
 * 3) Consider $$k \ne 0 \,$$ and $$v_{xo}=0 \,$$, therefore

becomes

$$m(\frac {dv_y} {dt}) = -k v_y^{n} {\sin {\alpha}} - mg$$

 * 3.1 Find $$v_y (t)\,$$ and $$y(t) \,$$ for $$m = constant \,$$
 * 3.2 Find $$v_y (t)\,$$ and $$y(t) \,$$ for $$m = m(t) \,$$

 * 4. Is $$m(\frac {dv_y} {dt}) = -k v_y^{n} {\sin {\alpha}} - mg$$ exact or can it be made exact?

Part 1
The projectile is given an initial velocity $$ v $$ in a direction that is at an angle $$ \alpha $$ to the horizontal. The Equations of motion (Eq. 1.1 and Eq. 1.2) given in the problem statement are acquired through summing the forces on the projectile and decomposing those forces into their vector quantities acting on the projectile. We sum the forces by looking at which ones are having an effect on the motion of the object.

1) inertia of the projectile after being fired from its initial position:

$$F_{inertia} = m(\frac {dv} {dt}) $$

which must be decomposed into its x and y components:

$$F_{x-inertia} = m(\frac {dv_x} {dt}) $$

$$F_{y-inertia} = m(\frac {dv_y} {dt}) $$

2) The effects of gravity on the y-component of the projectile's inertia:

$$F_{g} = -m g \,$$

$$g =  \,$$ acceleration  due  to  gravity

3) The force of air resistance on the projectile's motion decomposed into its x and y directional components:

$$F_{x-air} = -k v^{n} {\cos {\alpha}} \,$$

$$F_{y-air} = -k v^{n} {\sin {\alpha}} \,$$

Now if we sum the forces in the x and y directions and consider the force equilibrium in those directions we obtain the Equations of Motion given in the problem statement:

$$\sum {F_x} = m(\frac {dv_x} {dt}) = -k v^{n} {\cos {\alpha}}$$

$$\sum {F_y} = m(\frac {dv_y} {dt}) = -k v^{n} {\sin {\alpha}} - mg$$

Part 2
For the particular case where $$ k = 0 $$ equations 1.1 and 1.2 reduce to:

$$ m(\frac {dv_x} {dt}) = 0$$

which reduces to

$$ (\frac {dv_x} {dt}) = 0$$$$\displaystyle (Eq. 1.3) $$

and

$$m(\frac {dv_y} {dt}) = -mg $$

which reduces further to

$$(\frac {dv_y} {dt}) = -g $$$$\displaystyle (Eq. 1.4) $$

Now we can integrate Eqs. 1.3 and 1.4 with respect to time in order to get expressions for velocity and position in the x and y directions. We will start with the x-direction equations:

x-component:

$$ \frac {dv_x} {dt} = 0$$

Integrating with respect to time we obtain,

$$ v_x(t) = c_1 \,$$$$\displaystyle (Eq. 1.5) $$

Using the initial condition $$ v_x(t=0) = v_{xo} \,$$ the constant of integration $$ c_1 \,$$ can be found,

$$ c_1 = v_{xo} \,$$

So Eq. 1.5 becomes,

$$ v_x(t) = v_{xo} \,$$ $$\displaystyle (Eq. 1.6) $$

Now we repeat the same procedure for Eq. 1.4 to obtain the y component of the velocity vector. Once we integrate Eq. 1.4 with respect to time we get,

$$ v_y(t) = -gt + c_2 \,$$$$\displaystyle (Eq. 1.7) $$

Then using the initial condition $$ v_y(t=0) = v_{yo} \,$$ we can determine that $$ c_2 = v_{yo} \,$$ so then Eq. 1.7 becomes,

$$ v_y(t) = -gt + v_{yo} \,$$$$\displaystyle (Eq. 1.8) $$

Now we integrate Eq. 1.7 and 1.8 with respect to time once more to obtain the equations for position as a function of time in the x and y directions. After integrating with respect to time Eq. 1.7 and 1.8 become,

$$ x(t) = v_{xo}t + c_3 \,$$$$\displaystyle (Eq. 1.9) $$

and

$$ y(t) = - \frac 1 2 gt^2 + v_{yo}t + c_4 \,$$$$\displaystyle (Eq. 1.10) $$

Using initial conditions for position $$ x(t=0) = x_{o}\,$$ and $$ y(t=0) = y_{o}\,$$ then $$c_3\,$$ and $$c_4\,$$ can be solved for,

$$ c_3 = x_{o}\,$$

and

$$ c_4 = y_{o}\,$$

Plugging these values back into Eq. 1.9 and 1.10 yields the position equations for the x and y directions,

$$ x(t) = v_{xo}t + x_o \,$$$$\displaystyle (Eq. 1.11) $$

$$ y(t) = - \frac 1 2 gt^2 + v_{yo}t + y_o \,$$$$\displaystyle (Eq. 1.12) $$

To express y as a function of x we will manipulate eq. 1.11 to get t in terms of x. This yeilds the expression,

$$ t = \frac {x-x_o} {v_{xo}} \,$$$$\displaystyle (Eq. 1.13) $$

We then plug this value in for t in Eq. 1.12 to get,

$$ y(x) = - \frac 1 2 g(\frac {x-x_o} {v_{xo}})^2 + v_{yo}(\frac {x-x_o} {v_{xo}}) + y_o \,$$$$\displaystyle (Eq. 1.14) $$

to confirm that Eq. 1.14 is a parabola we can simply compare the general form of a parabolic function to Eq. 1.14,

$$ y(x) = ax^2 + bx + c \,$$

$$ y(x) = - \frac 1 2 g(\frac {x-x_o} {v_{xo}})^2 + v_{yo}(\frac {x-x_o} {v_{xo}}) + y_o \,$$

where $$a$$, $$b$$ and $$c$$ are all constants. It is apparent that Eq. 1.14 fits the form of a parabolic function so it can be concluded that our expression for $$y(x)$$ is a parabola.

Part 3
For the case where $$v_{xo}=0 \,$$, and $$k \ne 0$$ we can see from Eq. 1.1 that as long as $$v^n \ne 0 \,$$ then $$\alpha = 90^\circ \,$$ so that there are no net forces acting on the projectile in the x direction which leaves Eq. 1.2 as the net force acting on the projectile. However, since $$\alpha = 90^\circ \,$$ Eq. 1.2 can be re-written as,

$$m(\frac {dv_y} {dt}) = -k v_y^{n}- mg \,$$ $$\displaystyle (Eq. 1.15) $$

Part 3.1
Solving Eq. 1.15 for $$v_y(t)$$ and $$y(t)$$ when $$m =$$ constant. Start with the equation of motion (Eq. 1.15),

$$m(\frac {dv_y} {dt}) = -k (v_y)^{n}- mg \,$$

In order to solve this we must first show exactness. So we first put Eq 1.15 into a form that shows the first condition of exactness is met

$$m(\frac {dv_y} {dt})+ k v_y^{n}+ mg = 0\,$$
$$\displaystyle (Eq. 1.16) $$

So that we can see that the two parts, $$M(v_y,t)$$ and $$N(v_y,t)$$ are,

$$M(v_y,t)=k v_y^{n}+ mg $$

$$N(v_y,t)=m \,$$

Now that the first condition of exactness has been shown we must see if the second condition of exactness is met,

$$ M(v_y,t)_{v_y} = N(v_y,t)_t$$
$$\displaystyle (Eq. 1.17) $$

Which is clearly not met if $$m$$ is a constant

$$M(v_y,t)_{v_y} = knv_y^{n-1} \ne N(v_y,t)_{t} = \frac{dm(t)}{dt}$$
$$\displaystyle (Eq. 1.18) $$

Therefore, the equation is non-exact.

However, the equation can be made exact through the integrating factor method

Then expressing as a total derivative,

Keeping in mind that Eq. 1.20 is a sum of the forces so we can introduce the $$\phi$$ notation that will simplify the equation a bit.

$$ \sum F = \frac {\partial \phi} {\partial t} dt + \frac {\partial \phi} {\partial v_y} dv_y =0 $$

Therefore the $$\phi$$ derivatives are as follows,

$$\phi_t = h(v_y,t)(kv_y^n+mg)$$

$$\phi_{v_y} = h(v_y,t)m)$$

$$\phi_{t,v_y} = \phi_{v_y,t}$$

Now, we test for exactness. In order for Eq. 1.20 to be exact it must meet the condition,

$$h_{v_y}N+hN_{v_y}=h_tM+hM_t \,$$
$$\displaystyle (Eq. 1.21) $$

So we substitute the problem parameters into Eq. 1.21 and get,

$$\frac{\partial h(v_y,t)}{\partial v_y}(kv_y^n+mg)+h(v_y,t)\frac{\partial(kv_y^n+mg)}{\partial v_y} = \frac{\partial h(v_y,t)}{\partial t}m+h(v_y,t)\frac{\partial m}{\partial t}$$
$$\displaystyle (Eq. 1.22) $$

Letting $$\partial h / \partial v_y = 0$$ and since $$m$$ is a constant $$\partial m / \partial t = 0$$. So Eq. 1.22 reduces to,

$$h(v_y,t)\frac{\partial(kv_y^n+mg)}{\partial v_y} = \frac{\partial h(v_y,t)}{\partial t}m$$
$$\displaystyle (Eq. 1.23) $$

Now, we manipulate Eq 1.23 in order to obtain an expression for $$h(v_y,t)$$,

$$h(v_y,t)\frac{\partial(kv_y^n+mg)}{\partial v_y} = \frac{\partial h(v_y,t)}{\partial t}m$$
$$\displaystyle (Eq. 1.24) $$

$$h(v_y,t)\frac{nkv_y^n}{m} = \frac{\partial h(v_y,t)}{\partial t}$$
$$\displaystyle (Eq. 1.25) $$

$$h(v_y,t) = \exp(\frac{kv_y^n}{m}t)$$
$$\displaystyle (Eq. 1.26) $$

Therefore the $$\phi$$ derivative equations can be solved to get $$\phi$$,

$$ \phi_t = \exp(\frac{kv_y^n}{m}t)(kv_y^n+mg) $$
$$\displaystyle (Eq. 1.27) $$

$$ \phi = \int{\phi_t dt} = \int{ \exp(\frac{kv_y^n}{m}t)(kv_y^n+mg)dt} $$
$$\displaystyle (Eq. 1.28) $$

$$ \phi =(kv_y^n+mg)\frac{m}{kv_y^n}\exp(\frac{kv_y^n}{m}t)+f(v_y) = c_5 $$
$$\displaystyle (Eq. 1.29) $$

$$ \phi_v = m\exp(\frac{kv_y^n}{m}t) $$
$$\displaystyle (Eq. 1.30) $$

$$ \phi = \int{\phi_{v_y} dv_y} = \int{ m\exp(\frac{kv_y^n}{m}t)dv_y} = c_5 $$
$$\displaystyle (Eq. 1.31) $$

$$ \phi = m\int{\exp(\frac{kv_y^n}{m}t)dv_y} + f(t) = c_5 $$
$$\displaystyle (Eq. 1.32) $$

We can now combine these $$\phi$$ equations to come up with an expression of the sum of the forces that is exact,

$$ \sum F =\frac {\partial \phi} {\partial t} dt + \frac {\partial \phi} {\partial v_y} dv_y =\phi = m\int{\exp(\frac{kv_y^n}{m}t)dv_y} + (kv_y^n+mg)\frac{m}{kv_y^n}\exp(\frac{kv_y^n}{m}t) = c_5 $$
$$\displaystyle (Eq. 1.33) $$

This resulting equation is not integrable and therefore no explicit general form of $$v_y$$ was found. However, through the integrating factor method an exact non-linear expression was found. Numerical solutions can be found given a value of n, and initial conditions.

Part 3.2
For the condition where $$m=m(t) \ne 0$$, the integrating factor becomes even more complicated because in Eq 1.22 the initial simplification that occurs when $$ \frac {\partial m} {\partial t} = 0$$ does not happen. This will result in another implicit solution for $$v_y$$ when the integrating factor method is applied.

Part 3.2
If $$m(t)\ne0$$ the integrating factor method is complicated in equation 1.22 as the partial of $$m$$ with respect to time remains, complicating the expression for $$h$$

Author and Proof-reader
[Author]

[Proof-reader]

=HW 5=

Given
The Euler L2-ODE-VC: $$ a_2x^2y''+a_1xy'+a_0y=0 \,$$
$$\displaystyle (Eq. 3.1) $$

The Euler L2-ODE-CC: $$ b_2y''+b_1y'+b_0y=0 \,$$
$$\displaystyle (Eq. 3.2) $$

The characteristic equation: $$ (r-\lambda)^2 = r^2-2\lambda r + \lambda^2 = 0 \,$$
$$\displaystyle (Eq. 3.3) $$

Find
1.1) Find $$a_2$$, $$a_1$$ , $$a_0$$ such that Eqn. 3.3 is the characteristic equation of Eqn. 3.1

1.2) Find the first homogeneous solution, $$y_1$$

1.3) Find the second homogeneous solution, $$y_2$$

1.4) Find the general homogeneous solution

2.1) Find $$b_2$$, $$b_1$$ , $$b_0$$ such that Eqn. 3.3 is the characteristic equation of Eqn. 3.2

2.2) Find the first homogeneous solution, $$y_1$$

2.3) Find the second homogeneous solution, $$y_2$$

2.4) Find the general homogeneous solution

Part 1.1
The coefficients of Eqn. 3.1 (a terms) can be found if we use the trial solution method. The trial solution for an Euler L2-ODE-VC is: $$ y=x^r \,$$
$$\displaystyle (Eq. 3.1.1) $$

Therefore we can calculate the first and second derivatives of the trial solution:

$$ y'=rx^(r-1) \,$$
$$\displaystyle (Eq. 3.1.2) $$

$$ y''=r(r-1)x^(r-2) \,$$
$$\displaystyle (Eq. 3.1.3) $$

Then substitute back into Egn. 3.1 so that it becomes,

$$ a_2x^2r(r-1)x^(r-2)+a_1xrx^(r-1)+a_0x^r=0 \,$$
$$\displaystyle (Eq. 3.1.4) $$

Then simplify,

$$ x^r[a_2r(r-1)+a_1r+a_0]=0 \,$$
$$\displaystyle (Eq. 3.1.5) $$

$$ a_2r^2-a_2r+a_1r+a_0=0 \,$$
$$\displaystyle (Eq. 3.1.6) $$

$$ a_2r^2+(a_1-a_2)r+a_0=0 \,$$
$$\displaystyle (Eq. 3.1.7) $$

We then set Eqn. 3.1.7 equal to the expanded out version of the characteristic equation (Eqn. 3.3) to obtain the values of the coefficients,

$$ a_2r^2+(a_1-a_2)r+a_0=r^2-2\lambda r+\lambda^2 \,$$
$$\displaystyle (Eq. 3.1.8) $$

It is then clear that,

$$ a_2 = 1 \,$$
$$\displaystyle (Eq. 3.1.9) $$

$$ (a_1-a_2)= -2\lambda ; a_1 = a_2-2\lambda = 1-2\lambda \,$$
$$\displaystyle (Eq. 3.1.10) $$

$$ a_0=\lambda^2 \,$$
$$\displaystyle (Eq. 3.1.11) $$

Part 1.2
To find the first homogeneous solution use the trial solution (Eqn. 3.1.1) and the characteristic equation (Eqn. 3.3). We find the roots of Eqn. 3.3 and substitute the first (real) one in for $$r$$ in the trial solution (Egn. 3.1.1). In this particular case the two roots are the same so $$r_1=r_2=\lambda$$. Substituting this into Eqn. 3.1.1 we get,

$$ y_1=x^\lambda \,$$
$$\displaystyle (Eq. 3.1.12) $$

Part 1.3
Since $$\lambda$$ has two identical roots we can find the second homogeneous solution using the form,

$$ y_2(x)=U(x)y_1(x) \,$$
$$\displaystyle (Eq. 3.1.13) $$

We then calculate the first and second derivatives of Eqn 3.1.13. The $$(x)$$ argument will be ignored to simplify the equations.

$$ y_2'=U'y_1+Uy_1' \,$$
$$\displaystyle (Eq. 3.1.14) $$

$$ y_2=Uy_1+2U'y_1'+Uy_1'' \,$$
$$\displaystyle (Eq. 3.1.15) $$

Plug Eqns. 3.1.13, 3.1.14, 3.1.15 into Eqn 3.1 and simplify,

$$ a_2x^2(Uy_1+2U'y_1'+Uy_1)+a_1x(U'y_1+Uy_1')+a_0Uy_1=0 \,$$
$$\displaystyle (Eq. 3.1.16) $$

$$ U(a_2x^2y_1+a_1xy_1'+a_0y_1)+U'(2a_2x^2y_1'+a_1xy_1)+Ua_2x^2y_1=0 \,$$
$$\displaystyle (Eq. 3.1.17) $$

This can be simplified further noting that $$U$$ is multiplied by Eqn 3.1, which equals $$0$$. So that completely eliminates one term in Egn. 3.1.17. We then substitute in the calculated values of the $$a$$ coefficients and the values for $$y_1$$ from part 1.1 and simplify,

$$ U'(2a_2x^2y_1'+a_1xy_1)+U''a_2x^2y_1=0 \,$$
$$\displaystyle (Eq. 3.1.18) $$

$$ U'[2x^2\lambda x^{\lambda -1}+(1-2\lambda )xx^\lambda ]+U''x^2x^\lambda =0 \,$$
$$\displaystyle (Eq. 3.1.19) $$

$$ U'[2x^2\lambda x^{\lambda -1}+(1-2\lambda )xx^\lambda ]+U''x^2x^\lambda =0 \,$$
$$\displaystyle (Eq. 3.1.19) $$

$$ U'[2\lambda x^{\lambda +1}+x^{\lambda +1}-2\lambda x^{\lambda +1}]+U''x^{\lambda +2} =0 \,$$
$$\displaystyle (Eq. 3.1.20) $$

$$ U'x^{\lambda +1}+U''x^{\lambda +2} =0 \,$$
$$\displaystyle (Eq. 3.1.21) $$

$$ (U'+ U''x)x^{\lambda +1} =0 \,$$
$$\displaystyle (Eq. 3.1.22) $$

$$ U'+ U''x =0 \,$$
$$\displaystyle (Eq. 3.1.23) $$

We then use variation of parameters to reduce the order of Eqn. 3.1.23,

$$ m=U' \,$$
$$\displaystyle (Eq. 3.1.24) $$

So we substitute in the change of parameters into Egn. 3.1.24 and use separation of variables to solve for $$m$$,

$$ m+ m'x =0 \,$$
$$\displaystyle (Eq. 3.1.25) $$

$$ m + x \frac {dm} {dx}=0 \,$$
$$\displaystyle (Eq. 3.1.26) $$

$$ m = -x \frac {dm} {dx} \,$$
$$\displaystyle (Eq. 3.1.27) $$

$$ \frac {m} {dm} = - \frac {x} {dx} \,$$
$$\displaystyle (Eq. 3.1.28) $$

Then invert to get,

$$ \frac {dm} {m} = - \frac {dx} {x} \,$$
$$\displaystyle (Eq. 3.1.29) $$

Then integrate each side and simplify,

$$ ln (m) = - (ln (x) + k_1) \,$$
$$\displaystyle (Eq. 3.1.30) $$

$$ m = \frac 1 x e^{- k_1} \,$$
$$\displaystyle (Eq. 3.1.31) $$

$$ k_2 = e^{-k_1} \,$$
$$\displaystyle (Eq. 3.1.32) $$

$$ m = \frac 1 x k_2 \,$$
$$\displaystyle (Eq. 3.1.33) $$

Then substitute the identity $$m = U'$$ into Eqn. 3.1.33 and integrate to solve for $$U$$,

$$ U = \int U' dx = \int \frac 1 x k_2 dx \,$$
$$\displaystyle (Eq. 3.1.34) $$

$$ U = k_2 [ln(x) + k_3] \,$$
$$\displaystyle (Eq. 3.1.35) $$

Substitute $$U$$ back into Eqn. 3.1.13 along with the known value of $$y_1$$ to obtain the 2nd homogeneous solution,

$$ y_2 = k_2 [ln(x) + k_3]x^{\lambda} \,$$
$$\displaystyle (Eq. 3.1.36) $$

Part 1.4
The general homogeneous solution of the Euler L2-ODE-VC will have the form,

$$ y(x) = C_1 y_1(x)+C_2 y_2(x) \,$$
$$\displaystyle (Eq. 3.1.37) $$

Plugging Eqns. 3.1.36 and 3.1.12 into Eqn. 3.1.37 and simplifying results in the general homogeneous solution,

$$ y(x) = C_1 x^{\lambda} + C_2 [k_2 [ln(x) + k_3]x^{\lambda}] \,$$
$$\displaystyle (Eq. 3.1.38) $$

$$ y(x) = x^{\lambda}[C_1 + C_2 k_2 ln(x) + C_2 k_3] \,$$
$$\displaystyle (Eq. 3.1.39) $$

$$ C_3 = C_1 + C_2 k_3 \,$$
$$\displaystyle (Eq. 3.1.40) $$

$$ C_4 = C_2 k_2 \,$$
$$\displaystyle (Eq. 3.1.41) $$

Therefore the general homogeneous solution simplifies to,

$$ y(x) = x^{\lambda}[C_4 + C_5 ln(x)] \,$$
$$\displaystyle (Eq. 3.1.42) $$

Part 2.1
The coefficients of Eqn. 3.2 (b terms) can be found if we use the trial solution method. The trial solution for an Euler L2-ODE-CC is: $$ y=e^{rx} \,$$
$$\displaystyle (Eq. 3.2.1) $$

Therefore we can calculate the first and second derivatives of the trial solution:

$$ y'=re^{rx} \,$$
$$\displaystyle (Eq. 3.2.2) $$

$$ y''=r^2e^{rx} \,$$
$$\displaystyle (Eq. 3.2.3) $$

Then substitute back into Egn. 3.2 so that it becomes,

$$ b_2r^2e^{rx} + b_1re^{rx} + b_0e^{rx}=0 \,$$
$$\displaystyle (Eq. 3.2.4) $$

Then simplify,

$$ e^{rx}[b_2r^2+b_1r+b_0]=0 \,$$
$$\displaystyle (Eq. 3.2.5) $$

$$ b_2r^2+b_1r+b_0=0 \,$$
$$\displaystyle (Eq. 3.2.6) $$

We then set Eqn. 3.2.6 equal to the expanded out version of the characteristic equation (Eqn. 3.3) to obtain the values of the coefficients,

$$ b_2r^2+b_1r+b_0=r^2-2\lambda r+\lambda^2 \,$$
$$\displaystyle (Eq. 3.2.7) $$

It is then clear that,

$$ b_2 = 1 \,$$
$$\displaystyle (Eq. 3.2.8) $$

$$ b_1= -2\lambda \,$$
$$\displaystyle (Eq. 3.2.9) $$

$$ b_0=\lambda^2 \,$$
$$\displaystyle (Eq. 3.2.10) $$

Part 2.2
To find the first homogeneous solution use the trial solution (Eqn. 3.2.1) and the characteristic equation (Eqn. 3.3). We find the roots of Eqn. 3.3 and substitute the first (real) one in for $$r$$ in the trial solution (Egn. 3.2.1). In this particular case the two roots are the same so $$r_1=r_2=\lambda$$. Substituting this into Eqn. 3.2.1 we get,

$$ y_1=e^{\lambda x} \,$$
$$\displaystyle (Eq. 3.2.11) $$

Part 2.3
Since $$\lambda$$ has two identical roots we can find the second homogeneous solution using the form,

$$ y_2(x)=U(x)y_1(x) \,$$
$$\displaystyle (Eq. 3.2.12) $$

We then calculate the first and second derivatives of Eqn 3.2.12. The $$(x)$$ argument will be ignored to simplify the equations.

$$ y_2'=U'y_1+Uy_1' \,$$
$$\displaystyle (Eq. 3.2.13) $$

$$ y_2=Uy_1+2U'y_1'+Uy_1'' \,$$
$$\displaystyle (Eq. 3.2.14) $$

Plug Eqns. 3.2.12, 3.2.13, 3.2.14 into Eqn 3.2 and simplify,

$$ b_2(Uy_1+2U'y_1'+Uy_1)+b_1(U'y_1+Uy_1')+b_0Uy_1=0 \,$$
$$\displaystyle (Eq. 3.2.15) $$

$$ U(b_2y_1+b_1y_1'+b_0y_1)+U'(2b_2y_1'+b_1y_1)+Ub_2y_1=0 \,$$
$$\displaystyle (Eq. 3.2.16) $$

This can be simplified further noting that $$U$$ is multiplied by Eqn 3.2, which equals $$0$$. So that completely eliminates one term in Egn. 3.2.16. We then substitute in the calculated values of the $$a$$ coefficients and the values for $$y_1$$ from part 2.1 and simplify,

$$ U'(2b_2y_1'+b_1y_1)+U''b_2y_1=0 \,$$
$$\displaystyle (Eq. 3.2.17) $$

$$ U'[2\lambda e^{\lambda x}+(-2\lambda )e^{\lambda x}]+U''e^{\lambda x} =0 \,$$
$$\displaystyle (Eq. 3.2.18) $$

$$ U''e^{\lambda x} =0 \,$$
$$\displaystyle (Eq. 3.2.19) $$

Therefore

$$ U'' =0 \,$$
$$\displaystyle (Eq. 3.2.20) $$

Then Integrate Eqn. 3.2.20 twice with respect to $$x$$

$$ U' = k_1 \,$$
$$\displaystyle (Eq. 3.2.21) $$

$$ U = x k_1 + k_2 \,$$
$$\displaystyle (Eq. 3.2.22) $$

Then plug Eqn. 3.2.22 and 3.2.1 into Eqn. 3.2.12 to get,

$$ y_2(x) = e^{\lambda x} (x k_1 + k_2) \,$$
$$\displaystyle (Eq. 3.2.23) $$

Part 2.4
The general homogeneous solution of the Euler L2-ODE-CC will have the form,

$$ y(x) = C_1 y_1(x)+C_2 y_2(x) \,$$
$$\displaystyle (Eq. 3.2.24) $$

Plugging Eqns. 3.2.23 and 3.2.12 into Eqn. 3.2.24 and simplifying results in the general homogeneous solution,

$$ y(x) = C_1 e^{\lambda x} + C_2 e^{\lambda x} (x k_1 + k_2) \,$$
$$\displaystyle (Eq. 3.2.25) $$

$$ y(x) = e^{\lambda x}[C_1 + C_2 + k_2] + C_2 k_1 x e^{\lambda x} \,$$
$$\displaystyle (Eq. 3.2.26) $$

$$ C_3 = C_1 + C_2 + k_3 \,$$
$$\displaystyle (Eq. 3.2.27) $$

$$ C_4 = C_2 k_1 \,$$
$$\displaystyle (Eq. 3.2.28) $$

Therefore the general homogeneous solution simplifies to,

$$ y(x) = C_3 e^{\lambda x} + C_4 x e^{\lambda x} \,$$
$$\displaystyle (Eq. 3.2.29) $$

=HW 6=

Given
$$ P_2(x) = \frac 1 2 (3x^2-1) \,$$
$$\displaystyle (Eq. 1.1) $$

$$ Q_2(x) = \frac 1 4 (3x^2-1) log(\frac {1+x} {1-x}) - \frac 3 2 x \,$$
$$\displaystyle (Eq. 1.2) $$

Find
Assuming Eq. 1.1 is the first homogeneous solution to the Legendre function show that Eq. 1.2 is a second homogeneous solution using the variation of parameters method.

Solution
First we must identify the Equation that Eq. 1.1 and 1.2 are solutions to. The Legendre function is written as

$$ 0 = (1-x^2)y'' - 2xy' + n(n+1)y \,$$
$$\displaystyle (Eq. 1.3) $$

For a second order equation $$n=2$$ so the Legendre function becomes

$$ 0 = (1-x^2)y'' - 2xy' + 6y \,$$
$$\displaystyle (Eq. 1.4) $$

To obtain the second homogeneous solution the general form of a homogeneous L2-ODE-VC will be needed,

$$ 0 = a_2y'' + a_1 y' + a_0 y \,$$
$$\displaystyle (Eq. 1.5) $$

So we use the general solution form,

$$ y(x)=U(x) u_1(x) \,$$
$$\displaystyle (Eq. 1.6) $$

Where $$U(x)$$ is unknown, $$u_1(x) = P_2(x)$$ and $$u_2(x)$$ is the second homogeneous solution in this case. So we now solve for $$u_2(x)$$ in a general form,

$$ u_2(x)=U(x)u_1(x) \,$$
$$\displaystyle (Eq. 1.7) $$

$$ u_2'=U'u_1+Uu_1' \,$$
$$\displaystyle (Eq. 1.8) $$

$$ u_2=Uu_1+2U'u_1'+Uu_1'' \,$$
$$\displaystyle (Eq. 1.9) $$

Plug Eqs. 1.7, 1.8, 1.9 into Eq. 1.5 and simplify,

$$ a_2(Uu_1+2U'u_1'+Uu_1)+a_1(U'u_1+Uu_1')+a_0Uu_1=0 \,$$
$$\displaystyle (Eq. 1.10) $$

$$ U(a_2u_1+a_1u_1'+a_0u_1)+U'(2a_2u_1'+a_1u_1)+Ua_2u_1=0 \,$$
$$\displaystyle (Eq. 1.11) $$

This can be simplified further noting that $$U$$ is multiplied by the original general form of the L2-ODE-VC, which equals $$0$$. So that completely eliminates one term in Eg. 1.11. We then get the resulting equation,

$$ U'(2a_2u_1'+a_1u_1)+U''a_2u_1=0 \,$$
$$\displaystyle (Eq. 1.12) $$

Then we force $$a_2 = 1$$, use the identity $$Z=U'$$ to reduce the order of Eq. 1.12, and divide through by $$u_1$$ to get,

$$ Z(\frac{2u_1'} {u_1} + a_1)+Z'=0 \,$$
$$\displaystyle (Eq. 1.13) $$

This now makes the equation of first order. We can then use the integrating factor method to make the equation exact and thus easier to solve for $$U(x)$$. According to Eq. (1) on p.10-3 of the class notes the integrating factor for a linear first order ODE is of the form,

$$ h(x) = exp[\int a_0(s) ds] \,$$
$$\displaystyle (Eq. 1.14) $$

In this case $$a_0 = (\frac{2u_1'} {u_1} + a_1)$$ so Eq. 1.14 becomes,

$$ h(x) = exp[\int (\frac{2u_1'} {u_1} + a_1) dx] \,$$
$$\displaystyle (Eq. 1.15) $$

Simplifying Eq. 1.15 results in,

$$ h(x) = u_1^2 exp[\int a_1 dx] \,$$
$$\displaystyle (Eq. 1.16) $$

Then multiply Eq. 1.13 through by Eq. 1.16 and simplify we find that

$$ hZ' + h'Z = (hZ)' = 0 \,$$
$$\displaystyle (Eq. 1.17) $$

Solving for $$Z$$ results in

$$ Z = \frac {1} {h(x)} \int 0 dx = \frac {k_2} {h(x)} \,$$
$$\displaystyle (Eq. 1.18) $$

Invoking the identity $$Z = U'$$ to solve for $$ U $$.

$$ U = \int \frac {k_2} {h} dx +k_1 \,$$
$$\displaystyle (Eq. 1.19) $$

Then plug Eq. 1.19 back into Eq. 1.6 to get the expression,

$$ y(x) = u_1 [\int \frac {k_2} {h} dx + k_1] \,$$
$$\displaystyle (Eq. 1.20) $$

Then multiply through by the $$u_1$$ term,

$$ y(x) = u_1 \int \frac {k_2} {h} dx + k_1u_1 \,$$
$$\displaystyle (Eq. 1.21) $$

Knowing that the complete homogeneous solution to the 2nd order equation is a linear combination of all homogeneous solutions means,

$$ u_2 = u_1 \int \frac {k_2} {h} dx \,$$
$$\displaystyle (Eq. 1.22) $$

and $$ k_1 = 1 $$

Now we make the Legendre function, Eq. 1.4, fit the homogeneous form of our general L2-ODE-VC so that $$a_2 = 1$$ so that we can find $$a_1$$ in order to sole for $$h(x)$$

$$ 0 = y'' - \frac {2x} {(1-x^2)} y' + \frac {6} {(1-x^2)}y \,$$
$$\displaystyle (Eq. 1.23) $$

Therefore,

$$ a_1 = \frac {-2x} {(1-x^2)} \,$$
$$\displaystyle (Eq. 1.24) $$

$$ a_0 = \frac {6} {(1-x^2)} \,$$
$$\displaystyle (Eq. 1.25) $$

Now plug Eq. 1.22 and Eq. 1.1 back into Eq. 1.16 and simplify the expression for $$h(x)$$ to obtain

$$ h(x) = (\frac 3 2 x^2 - \frac 1 2)^2 (x^2 -1) \,$$
$$\displaystyle (Eq. 1.26) $$

Then plug Eq. 1.26 and Eq. 1.1 into Eq. 1.22 to get the expression for the 2nd homogeneous solution. We will also assume $$k_2=-1$$ for reasons that will become clear later. So now the expression for $$u_2$$ becomes,

$$ u_2(x) = [\frac 1 2 (3 x^2 - 1)] (\int \frac{-1} {(\frac 3 2 x^2 - \frac 1 2)^2 (x^2 - 1)}) dx \,$$
$$\displaystyle (Eq. 1.27) $$

This simplifies to

$$ u_2(x) = \frac 1 2 (3 x^2 - 1) \frac 1 2 [\frac{-6x} {3x^2 - 1} - log (1-x) + log (x+1)] \,$$
$$\displaystyle (Eq. 1.28) $$

This equation then simplifies to the form of Eq. 1.2

$$ u_2(x) = Q_2(x) = \frac 1 4 (3x^2-1) log(\frac {1+x} {1-x}) - \frac 3 2 x \,$$
$$\displaystyle (Eq. 1.29) $$

Given
From p.28 problem 1.1 of King we are given

Part 1 A) $$ (x-1)y'' - xy'+y = f(x) = 0 \,$$
$$\displaystyle (Eq. 2.1.1) $$

with solution

$$ u_1(x) = e^x \,$$
$$\displaystyle (Eq. 2.1.2) $$

and

B) $$ xy'' + 2y'+ xy = f(x) = 0 \,$$
$$\displaystyle (Eq. 2.2.1) $$

with solution

$$ u_1(x) = x^{-1}sin(x) \,$$
$$\displaystyle (Eq. 2.2.2) $$

Part 2

Instead of homogeneous functions, make $$f(x) = sin(x)$$ for Eqs. 2.1.1 and 2.2.1 and find the general solutions.

Find
Show that $$u_1$$ in parts 1 and 2 are homogeneous solutions to their respective ODE's. Then use variation of parameters to find $$u_2$$ for Eq. 2.1.1 and Eq. 2.2.1

Part 1
Part A

First we must show that Eq. 2.1.2 is a solution to Eq. 2.1.1. To do this we simply plug Eq. 2.1.2 in for $$y$$ in Eq. 2.1.1. To do this we first must find the 1st and 2nd derivatives of Eq. 2.1.2

$$ u_1' = e^x \,$$
$$\displaystyle (Eq. 2.1.3) $$

$$ u_1'' = e^x \,$$
$$\displaystyle (Eq. 2.1.4) $$

Then plug Eq. 2.1.2, 2.1.3 and 2.1.4 into Eq. 2.1.1

$$ (x-1)e^x - x e^x + e^x = 0 \,$$
$$\displaystyle (Eq. 2.1.5) $$

It is then clear that the RHS reduces to zero, which matches the LHS, so we have confirmed that Eq. 2.1.2 is a solution of Eq. 2.1.1.

Next, we wil find the 2nd homogeneous solution to Eq. 2.1.1 using variation of parameters. to do this we use the general form of the homogeneous L2-ODE-VC,

$$ a_2y'' + a_1y' + a_0y = 0 \,$$
$$\displaystyle (Eq. 2.1.6) $$

To solve the general form we use the general form of the solution,

$$ y(x)=U(x) u_1(x) \,$$
$$\displaystyle (Eq. 2.1.7) $$

Where $$U(x)$$ is unknown, $$u_1(x)$$ is Eq. 2.1.2 and $$u_2(x)$$ is the second homogeneous solution in this case. So we now solve Eq. 2.1.6 for $$y(x)$$ in a general form. To do this we must have the 1st and 2nd derivatives of $$y(x)$$,

$$ y =U(x)u_1(x) \,$$
$$\displaystyle (Eq. 2.1.7) $$

$$ y'=U'u_1+Uu_1' \,$$
$$\displaystyle (Eq. 2.1.8) $$

$$ y=Uu_1+2U'u_1'+Uu_1'' \,$$
$$\displaystyle (Eq. 2.1.9) $$

Plug Eqs. 2.1.7, 2.1.8, 2.1.9 into Eq. 2.1.6 and simplify,

$$ a_2(Uu_1+2U'u_1'+Uu_1)+a_1(U'u_1+Uu_1')+a_0Uu_1=0 \,$$
$$\displaystyle (Eq. 2.1.10) $$

$$ U(a_2u_1+a_1u_1'+a_0u_1)+U'(2a_2u_1'+a_1u_1)+Ua_2u_1=0 \,$$
$$\displaystyle (Eq. 2.1.11) $$

This can be simplified further noting that $$U$$ is multiplied by the original general form of the L2-ODE-VC, which equals $$0$$. So that completely eliminates one term in Eg. 2.1.11. We then get the resulting equation,

$$ U'(2a_2u_1'+a_1u_1)+U''a_2u_1=0 \,$$
$$\displaystyle (Eq. 2.1.12) $$

Then we force $$a_2 = 1$$, use the identity $$Z=U'$$ to reduce the order of Eq. 2.1.12, and divide through by $$u_1$$ to get,

$$ Z(\frac{2u_1'} {u_1} + a_1)+Z'=0 \,$$
$$\displaystyle (Eq. 2.1.13) $$

This now makes the equation of first order. We can then use the integrating factor method to make the equation exact and thus easier to solve for $$U(x)$$. According to Eq. (1) on p.10-3 of the class notes the integrating factor for a linear first order ODE is of the form,

$$ h(x) = exp[\int a_0(s) ds] \,$$
$$\displaystyle (Eq. 2.1.14) $$

In this case $$a_0 = (\frac{2u_1'} {u_1} + a_1)$$ so Eq. 2.1.14 becomes,

$$ h(x) = exp[\int (\frac{2u_1'} {u_1} + a_1) dx] \,$$
$$\displaystyle (Eq. 2.1.15) $$

Simplifying Eq. 2.1.15 results in,

$$ h(x) = u_1^2 exp[\int a_1 dx] \,$$
$$\displaystyle (Eq. 2.1.16) $$

Then multiply Eq. 2.1.13 through by Eq. 2.1.16 and simplify we find that

$$ hZ' + h'Z = (hZ)' = 0 \,$$
$$\displaystyle (Eq. 2.1.17) $$

Solving for $$Z$$ results in

$$ Z = \frac {1} {h(x)} \int 0 dx = \frac {k_2} {h(x)} \,$$
$$\displaystyle (Eq. 2.1.18) $$

Invoking the identity $$Z = U'$$ to solve for $$ U $$.

$$ U = \int \frac {k_2} {h} dx +k_1 \,$$
$$\displaystyle (Eq. 2.1.19) $$

Then plug Eq. 2.1.19 back into Eq. 2.1.7 to get the expression,

$$ y(x) = u_1 [\int \frac {k_2} {h} dx + k_1] \,$$
$$\displaystyle (Eq. 2.1.20) $$

Then multiply through by the $$u_1$$ term,

$$ y(x) = u_1 \int \frac {k_2} {h} dx + k_1u_1 \,$$
$$\displaystyle (Eq. 2.1.21) $$

Knowing that the complete homogeneous solution to the 2nd order equation is a linear combination of all homogeneous solutions means,

$$ u_2 = u_1 \int \frac {k_2} {h} dx \,$$
$$\displaystyle (Eq. 2.1.22) $$

and $$ k_1 = 1 $$

Now we make the Eq. 2.1.1 fit the homogeneous form of our general L2-ODE-VC so that $$a_2 = 1$$ so that we can find $$a_1$$ in order to solve for $$h(x)$$

$$ 0 = y'' - \frac {x} {(x-1)} y' + \frac {1} {(x-1)}y \,$$
$$\displaystyle (Eq. 2.1.23) $$

Therefore,

$$ a_1 = \frac {-x} {(x-1)} \,$$
$$\displaystyle (Eq. 2.1.24) $$

$$ a_0 = \frac {1} {(x-1)} \,$$
$$\displaystyle (Eq. 2.1.25) $$

Now plug Eq. 2.1.22 and Eq. 2.1.2 back into Eq. 2.1.16 and simplify the expression for $$h(x)$$ to obtain

$$ h(x) = (\frac {e^x} {x-1} \,$$
$$\displaystyle (Eq. 2.1.26) $$

Then plug Eq. 2.1.26 and Eq. 2.1.2 into Eq. 2.1.22 to get the expression for the 2nd homogeneous solution. So now the expression for $$u_2$$ becomes,

$$ u_2(x) = e^x (k_2 \int \frac {(x-1)} {e^x}) dx \,$$
$$\displaystyle (Eq. 2.1.27) $$

This simplifies to

$$ u_2(x) = e^x \frac {-k_2x} {e^x} \,$$
$$\displaystyle (Eq. 2.1.28) $$

After the negative sign is absorbed into the constant $$k_2$$, the 2nd homogeneous solution simplifies to the form

$$ u_2(x) = k_2 x \,$$
$$\displaystyle (Eq. 2.1.29) $$

Part B

First we must show that Eq. 2.2.2 is a solution to Eq. 2.2.1. To do this we simply plug Eq. 2.2.2 in for $$y$$ in Eq. 2.2.1. To do this we first must find the 1st and 2nd derivatives of Eq. 2.2.2

$$ u_1' = -x^{-2} sin(x) + x^{-1} cos(x) \,$$
$$\displaystyle (Eq. 2.2.3) $$

$$ u_1'' = 2x^{-3} sin(x) - 2x^{-2} cos(x) - x^{-1} sin(x) \,$$
$$\displaystyle (Eq. 2.2.4) $$

Then plug Eq. 2.2.2, 2.2.3 and 2.2.4 into Eq. 2.2.1

$$ [2x^{-2} sin(x) - 2x^{-1} cos(x) - sin(x)] + [2x^{-1} cos(x) - 2x^{-2} sin(x)] + sin(x) = 0 \,$$
$$\displaystyle (Eq. 2.2.5) $$

It is then clear that the RHS reduces to zero, which matches the LHS, so we have confirmed that Eq. 2.2.2 is a solution of Eq. 2.2.1.

Next, we wil find the 2nd homogeneous solution to Eq. 2.2.1 using variation of parameters. to do this we use the same techniques to find the general solutions to $$h(x)$$, $$Z$$ , and $$U$$ as we did in Part 1. So we can use Eqs. 2.1.16, 2.1.18 , 2.1.19 , and 2.1.21 shown below.

$$ h(x) = u_1^2 exp[\int a_1 dx] \,$$
$$\displaystyle (Eq. 2.1.16) $$

$$ Z = \frac {1} {h(x)} \int 0 dx = \frac {k_2} {h(x)} \,$$
$$\displaystyle (Eq. 2.1.18) $$

$$ U = \int \frac {k_2} {h} dx +k_1 \,$$
$$\displaystyle (Eq. 2.1.19) $$

$$ y(x) = u_1 \int \frac {k_2} {h} dx + k_1u_1 \,$$
$$\displaystyle (Eq. 2.1.21) $$

Again, knowing that the complete homogeneous solution to the 2nd order equation is a linear combination of all homogeneous solutions means,

$$ u_2 = u_1 \int \frac {k_2} {h} dx \,$$
$$\displaystyle (Eq. 2.1.22) $$

and $$ k_1 = 1 $$

Now we make the Eq. 2.2.1 fit the homogeneous form of our general L2-ODE-VC so that $$a_2 = 1$$ so that we can find $$a_1$$ in order to solve for $$h(x)$$

$$ 0 = y'' - \frac {2} {x} y' + y \,$$
$$\displaystyle (Eq. 2.2.6) $$

Therefore,

$$ a_1 = \frac {2} {x} \,$$
$$\displaystyle (Eq. 2.2.7) $$

$$ a_0 = 1 \,$$
$$\displaystyle (Eq. 2.2.8) $$

Now plug Eq. 2.2.7 and Eq. 2.2.2 into Eq. 2.1.16 and simplify the expression for $$h(x)$$ to obtain

$$ h(x) = sin^2(x) \,$$
$$\displaystyle (Eq. 2.2.9) $$

Then plug Eq. 2.2.9 and Eq. 2.2.2 into Eq. 2.1.22 to get the expression for the 2nd homogeneous solution. So now the expression for $$u_2$$ becomes,

$$ u_2(x) = [\frac {sin(x)} {x}] [k_2 \int \frac {1} {sin^2(x)}] dx \,$$
$$\displaystyle (Eq. 2.2.10) $$

This simplifies to

$$ u_2(x) = [\frac {sin(x)} {x}] [-k_2cot(x)] \,$$
$$\displaystyle (Eq. 2.2.11) $$

The 2nd homogeneous solution then simplifies to the form

$$ u_2(x) = k_2 [\frac {-cos(x)} {x}] \,$$
$$\displaystyle (Eq. 2.2.12) $$

Part 2
Part A

To find the general solution of

$$ sin(x) = (x-1)y'' - xy' + y \,$$
$$\displaystyle (Eq. 2.1.30) $$

we assume the general solution is of the form

$$ y(x) = A(x) u_1 + B(x) u_2 \,$$
$$\displaystyle (Eq. 2.1.31) $$

Since we already know $$u_1$$ and $$u_2$$ from part 1 A we need to find $$A(x)$$ and $$B(x)$$. To do this we utilize the equations,

$$ A(x) = - \int \frac 1 W u_2(x) f(x) dx \,$$
$$\displaystyle (Eq. 2.1.32) $$

and

$$ B(x) = \int \frac 1 W u_1(x) f(x) dx \,$$
$$\displaystyle (Eq. 2.1.33) $$

where W is the Wronskian. The wronskian is the determinant of a simple 2x2 matrix in this case. It will simplify to,

$$ W =u_1 u_2' - u_2 u_1' = e^x -xe^x = e^x (1-x) \,$$
$$\displaystyle (Eq. 2.1.34) $$

Now we plug Eq. 2.1.34 and 2.1.29 into Eq. 2.1.32, where the constant will be neglected in this case for $$u_2$$ and $$f(x) = sin(x)$$ so we get,

$$ A(x) = - \int \frac {x sin(x)} {e^x (1-x)} dx \,$$
$$\displaystyle (Eq. 2.1.35) $$

For simplicity we will leave Eq. 2.1.35 in its integral form. Now we calculate $$B(x)$$. Plug Eq. 2.1.2 and Eq. 2.1.34 into Eq. 2.1.33 to get

$$ B(x) = \int \frac {sin(x)} {(1-x)} dx \,$$
$$\displaystyle (Eq. 2.1.36) $$

Again, for simplicity we will leave Eq. 2.1.36 in its integral form. So we then plug Eqs. 2.1.35, 2.1.36 into Eq. 2.1.31 to get our general solution for Part A,

$$ y(x) = -e^x \int \frac {x sin(x)} {e^x (1-x)} dx + x \int \frac {sin(x)} {(1-x)} dx \,$$
$$\displaystyle (Eq. 2.1.37) $$

Part B We will now use the same procedure to find the general solution to the nonhomogeneous equation,

$$ sin(x) = xy'' + 2y' + xy \,$$
$$\displaystyle (Eq. 2.2.13) $$

Our Wronskian is now,

$$ W = u_1 u_2' - u_2 u_1' = (x^{-1} sin(x)) (x^{-2}cos(x) + x^{-1}sin(x)) - (-x^{-1}cos(x)) (-x^{-2}sin(x) + x^{-1}cos(x)) \,$$
$$\displaystyle (Eq. 2.2.14) $$

which simplifies all the way down to,

$$ W = x^{-2} \,$$
$$\displaystyle (Eq. 2.2.15) $$

So the coefficients of our general solution $$A(x)$$ and $$B(x)$$ can be abtained by pluging Eq. 2.2.15, 2.2.12, and 2.2.2 into Eq. 2.1.32 and 2.1.33 as they are needed,

$$ A(x) = - \int x^2 \frac {-cos(x)} {x} sin(x) dx = \frac 1 8 [sin(2x)-2xcos(2x)] \,$$
$$\displaystyle (Eq. 2.2.16) $$

$$ B(x) = \int x^2 \frac {sin(x)} {x} sin(x) dx = \frac {x^2} {4} - \frac 1 4 x sin(2x)- \frac 1 8 cos(2x) \,$$
$$\displaystyle (Eq. 2.2.17) $$

Now we plug Eq. 2.2.17, 2.2.16, 2.2.2 and 2.2.12 into Eq. 2.1.31 to obtain the general solution for Eq. 2.2.13.

$$ y(x) = \frac {sin(x)} {8x} [sin(2x)-2xcos(2x)] - \frac {cos(x)} {x} [\frac {x^2} {4} - \frac 1 4 x sin(2x)- \frac 1 8 cos(2x)] \,$$
$$\displaystyle (Eq. 2.2.18) $$

=HW 7=

Given
$$ Q_n(x) = P_n(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {2n-2j+1} {(2n-j+1)j} P_{n-j}(x) \,$$
$$\displaystyle (Eq. 7.1) $$

$$ J = 1+2[\frac {n-1} {2}] \,$$
$$\displaystyle (Eq. 7.2) $$

Find
Use Eq. 7.1 to show $$Q_n(x)$$ is even or odd depending on $$n$$

Solution
In order to get a preliminary solution we must review when a function is even or odd. A function is said to be odd if $$-f(x) = f(-x)$$ and even if $$f(x) = f(-x)$$.

Since Eq. 7.1 is a combination of functions we must also review the properties of what happens when even/odd functions are combined either through multiplication or summation.

Even/Odd function Basic Properties

1) The sum of a number of odd functions is odd

2) The sum of a number of even functions is even

3) The sum of an even and an odd function is not even or odd, unless one of the functions is equal to zero

4) The product of an even (or odd) function and a constant multiple is even (odd)

5) The product of two even functions is even

6) The product of two odd functions is even

7) the product of an even and odd function is odd

Reference: Even and Odd functions

Now, to use these properties in eq. 7.1 we must determine whether $$P_n$$ is even or odd based on $$n$$. To do that we simply recall that $$P_n$$ is odd if $$n$$ is odd and $$P_n$$ is even if $$n$$ is even.

Reference: Legendre Polynomials

So, we can now look at Eq. 7.1 in terms of evenness and oddness. For the first case we assume $$ n = even $$

$$ Q_n(x) = (even) tanh^{-1}(x) - 2 \sum_{j=1,3,5,..}^J \frac {2n-2j+1} {(2n-j+1)j} (odd) \,$$
$$\displaystyle (Eq. 7.3) $$

Eq. 7.3 must be further simplified in terms of it evenness/oddness. By simple visual analysis of a graph of $$tanh^{-1}(x)$$ or by plugging in random numbers and then their negative counterparts we can see that

$$ tanh^{-1}(x) = odd \,$$
$$\displaystyle (Eq. 7.4) $$

Also, we can apply Basic Properties (1) and (4) above to see that the summation term is the sum of a number of odd functions after each has been multiplied by a constant. Therefore the entire second term of Eq. 7.3 is an odd function. This allows us to reduce Eq. 7.3 to

$$ Q_n(x) = (even) (odd) - (odd) \,$$
$$\displaystyle (Eq. 7.5) $$

Then using Basic Properties (1) and (7) we see that when

$$ n = even \,$$

$$ Q_n(x) = odd \,$$
$$\displaystyle (Eq. 7.6) $$

Now we use the same process to determine if $$Q_n(x)$$ is even or odd if $$n$$ is odd

If $$n = odd$$, then $$P_n(x) = odd$$ and $$P_{n-j}(x) = even$$. Also keeping in mind that $$tanh^{-1}(x)$$ is always odd. So Eq. 7.1 reduces to

$$ Q_n(x) = (odd) (odd) - 2 \sum_{j=1,3,5,..}^J (constant) (even) \,$$
$$\displaystyle (Eq. 7.7) $$

Using Basic Properties (2), (4) and (5) it is clear that when

$$ n = odd \,$$

Eq. 7.7 reduces to

$$ Q_n(x) = even \,$$
$$\displaystyle (Eq. 7.8) $$

To prove that $$Q_n(x)$$ is even when $$n$$ is odd and vice versa we will look at the specific examples when $$n = 1, 2, 3, 4$$

To do this we must first get the Legendre Polynomials for $$n = 0, 1, 2, 3, 4$$. The quickest way to do this is to use the Rodriquez formula.

$$ P_n(x) = \frac {1} {2^n n!} \frac {d^n} {dx^n} (x^2-1)^n \,$$
$$\displaystyle (Eq. 7.9) $$

Reference: Rodrigues Formula

From Eq. 7.9 we can get

$$ P_0(x) = 1 \,$$
$$\displaystyle (Eq. 7.10) $$

$$ P_1(x) = x \,$$
$$\displaystyle (Eq. 7.11) $$

$$ P_2(x) = \frac {1} {2} (3x^2-1) \,$$
$$\displaystyle (Eq. 7.12) $$

$$ P_3(x) = \frac {1} {2} (5x^3-3x) \,$$
$$\displaystyle (Eq. 7.13) $$

$$ P_4(x) = \frac {1} {8} (35x^4 - 30x^2 + 3) \,$$
$$\displaystyle (Eq. 7.14) $$

For $$n=1$$

First, we see for $$n=1$$ Eq. 7.2 reduces to

$$ J = 1+2[\frac {1-1} {2}] = 1 \,$$
$$\displaystyle (Eq. 7.15) $$

Then plug Eq. 7.15, 7.11 into Eq 7.1 to get

$$ Q_1(x) = (x) tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^1 \frac {2(1)-2(j)+1} {(2(1)-(j)+1)(j)} P_{1-j}(x) \,$$
$$\displaystyle (Eq. 7.16) $$

It is then clear that $$j=1$$ so once we plug that in and Eq. 7.10 for $$P_0(x)$$ Eq. 7.16 reduces to

$$ Q_1(x) = (x) tanh^{-1}(x) - (1) \,$$
$$\displaystyle (Eq. 7.17) $$

To test whether it is even or odd we can plug in $$x= \frac 1 2$$ and $$x= \frac {-1} {2}$$

$$ Q_1(\frac 1 2) = -0.725347... \,$$
$$\displaystyle (Eq. 7.18) $$

and

$$ Q_1(\frac {-1} {2}) = -0.725347... \,$$
$$\displaystyle (Eq. 7.19) $$

Therefore we can see that $$Q_1(x)$$ is an even function

Used Wolfram Alpha for numerical computation

For $$n=2$$

First, we see for $$n=2$$ Eq. 7.2 reduces to

$$ J = 1+2[\frac {2-1} {2}] = 2 \,$$
$$\displaystyle (Eq. 7.20) $$

Then plug Eq. 7.12, and 7.20 into Eq 7.1 to get

$$ Q_2(x) = (\frac 1 2 (3x^2-1)) tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^2 \frac {2(2)-2(j)+1} {(2(2)-(j)+1)(j)} P_{2-j}(x) \,$$
$$\displaystyle (Eq. 7.21) $$

It is then clear that $$j=1$$ so once we plug that in and Eq. 7.11 for $$P_1(x)$$ Eq. 7.21 reduces to

$$ Q_2(x) = (\frac 1 2 (3x^2-1)) tanh^{-1}(x) - (\frac 3 2) (x) \,$$
$$\displaystyle (Eq. 7.22) $$

To test whether it is even or odd we can plug in $$x= \frac 1 2$$ and $$x= \frac {-1} {2}$$

$$ Q_2(\frac 1 2) = -0.818663... \,$$
$$\displaystyle (Eq. 7.23) $$

and

$$ Q_2(\frac {-1} {2}) = 0.818663... \,$$
$$\displaystyle (Eq. 7.24) $$

Therefore we can see that $$Q_2(x)$$ is an odd function

Used Wolfram Alpha for numerical computation

For $$n=3$$

First, we see for $$n=3$$ Eq. 7.2 reduces to

$$ J = 1+2[\frac {3-1} {2}] = 3 \,$$
$$\displaystyle (Eq. 7.25) $$

Then plug Eq. 7.13, and 7.25 into Eq 7.1 to get

$$ Q_3(x) = (\frac 1 2 (5x^3-3x)) tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^3 \frac {2(3)-2(j)+1} {(2(3)-(j)+1)(j)} P_{3-j}(x) \,$$
$$\displaystyle (Eq. 7.26) $$

It is then clear that $$j=1, 3$$ so once we plug that in we see that Eq. 7.26 reduces to

$$ Q_3(x) = (\frac 1 2 (5x^3-3x)) tanh^{-1}(x) - 2 [\frac 5 6 (\frac 1 2 (3x^2-1)) + \frac 1 12 (1)] \,$$
$$\displaystyle (Eq. 7.27) $$

Eq. 7.27 can then be further simplified to get

$$ Q_3(x) = (\frac 1 2 (5x^3-3x)) tanh^{-1}(x) - \frac 5 6 (3x^2-1) - \frac 1 6 \,$$
$$\displaystyle (Eq. 7.28) $$

To test whether it is even or odd we can plug in $$x= \frac 1 2$$ and $$x= \frac {-1} {2}$$

$$ Q_3(\frac 1 2) = -0.198655... \,$$
$$\displaystyle (Eq. 7.29) $$

and

$$ Q_3(\frac {-1} {2}) = -0.198655... \,$$
$$\displaystyle (Eq. 7.30) $$

Therefore we can see that $$Q_3(x)$$ is an even function

Used Wolfram Alpha for numerical computation

For $$n=4$$

First, we see for $$n=4$$ Eq. 7.2 reduces to

$$ J = 1+2[\frac {4-1} {2}] = 4 \,$$
$$\displaystyle (Eq. 7.31) $$

Then plug Eq. 7.14, and 7.31 into Eq 7.1 to get

$$ Q_4(x) = [\frac 1 8 (35x^4 - 30x^2 + 3)] tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^4 \frac {2(4)-2(j)+1} {(2(4)-(j)+1)(j)} P_{4-j}(x) \,$$
$$\displaystyle (Eq. 7.32) $$

It is then clear that $$j=1, 3$$ so once we plug that in we see that Eq. 7.32 reduces to

$$ Q_4(x) = [\frac 1 8 (35x^4 - 30x^2 + 3)] tanh^{-1}(x) - 2 [\frac 7 8 (\frac 1 2 (5x^3-3x)) + \frac 3 18 (x)] \,$$
$$\displaystyle (Eq. 7.33) $$

Eq. 7.33 can then be further simplified to get

$$ Q_4(x) = [\frac 1 8 (35x^4 - 30x^2 + 3)] tanh^{-1}(x) - \frac 7 8 (5x^3-3x)) - \frac 1 3 (x)] \,$$
$$\displaystyle (Eq. 7.34) $$

To test whether it is even or odd we can plug in $$x= \frac 1 2$$ and $$x= \frac {-1} {2}$$

$$ Q_4 (frac 1 2) = 0.440175... \,$$
$$\displaystyle (Eq. 7.35) $$

and

$$ Q_4(\frac {-1} {2}) = -0.440175... \,$$
$$\displaystyle (Eq. 7.36) $$

Therefore we can see that $$Q_4(x)$$ is an odd function

Used Wolfram Alpha for numerical computation

From the generalization of evenness and oddness made in the first part of the problem along with the examples provided it has been proven that Eq. 7.1 is an odd function when "n" is an even integer and even when "n" is an odd integer.

title: User talk:Egm6341.f10.team5.robcarroll


text:
Talk:Motivation and emotion/Book/2010/Hunger motivation

Feedback
This is by far the best developed chapter so far - congratulations. I used this chapter as an example in my Motivation and Emotion lecture last week. Sincerely, James. -- Jtneill - Talk - c 21:14, 15 September 2010 (UTC)
 * I'm glad it provided a an example. Please feel free to edit the chapter and make it suited for the types of students that you have in mind. This is a topic that I teach to first year medical students. --JWSchmidt 21:57, 15 September 2010 (UTC)

Diagram
Just noticed that User:KmpH created this diagram which might be useful for this chapter: -- Jtneill - Talk - c 11:25, 15 December 2010 (UTC)


text:
User:Jiangpx/6321 hw2

=prob1 - 7-1=

=prob 2 - 7-1: verify (7) is a N1-ODE=

Given
$$\begin{align} & M(x,y)=4{{x}^{7}}+\sin y \\ & N(x,y)={{x}^{2}}{{y}^{3}} \\ \end{align}$$

Find
$$(4{{x}^{7}}+\sin y)+({{x}^{2}}{{y}^{3}})y'=0$$ is a N1-ODE

Solution
In general, N1-ODE has the form of $$M(x,y)+N(x,y)y'=0$$,

in which, $$y'$$ tells that it is of order 1 and the coefficients M and N are nonlinear - they violate the definition of linear operator, defined as follows:

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

$$\displaystyle L(cx)=cL(x)$$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$\displaystyle L(x+y)=L(x)+L(y)$$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Hence, the given equation is a N1-ODE.

=prob 3 - 8-1=

=prob 4 - 8-2=

=prob 5 - 9-2=

=prob 6 - 9-3=

Given
$$y(x)={{\sin }^{-1}}(k-15{{x}^{5}})$$

$$\begin{align} & M(x,y)=75{{x}^{4}} \\ & N(x,y)=\cos y \\ \end{align}$$

Verify
$$y(x)={{\sin }^{-1}}(k-15{{x}^{5}})$$ satisfies $$M(x,y)+N(x,y)y'=0$$

Solution
The ODE to be solved is

$$75{{x}^{4}}+(\cos y)y'=0$$

and from $$y(x)={{\sin }^{-1}}(k-15{{x}^{5}})$$

we get $$\sin y=k-15{{x}^{5}}$$

Differentiate both sides, $$\cos y\cdot y'=-75{{x}^{4}}$$, which is the ODE to be solved.

Hence, verified.

=prob 7 - 10-3 & 11-1= =prob 8 - 11-2= =prob 9 - 12-2=

title: User:Jiangpx/6321 hw2


text:
Motivation and emotion/Book/Unselected topics

This is an older list of some possible topic areas that may not have been sufficiently well covered in the book chapters.

Motivation

 * 1) Biological motives (needs)
 * 2) Pleasure
 * 3) Satiety
 * 4) Thermoregulation
 * 5) Thirst
 * 6) Brain (Neurophysiological structures and needs
 * 7) Freud's drive theory
 * 8) Instinct
 * 9) Historical development of motivation
 * 10) Incentive
 * 11) Learning theory
 * 12) Attention and learning
 * 13) Classical conditioning
 * 14) Economics of motivation
 * 15) Instrumental learning
 * 16) Social incentive theory
 * 17) Mini-theories
 * 18) Self-determination theory
 * 19) Theory of planned behaviour
 * 20) Theory of reasoned action
 * 21) More motivational theories
 * 22) Motivation and behaviour
 * 23) Motivation and cognition
 * 24) Cognitive consistency
 * 25) Distraction
 * 26) Feedback
 * 27) Motivational interviewing
 * 28) Motivation and culture
 * 29) Motivation and education
 * 30) History of motivation and education
 * 31) Influencers of educational motivation
 * 32) Cultural
 * 33) Educational setting
 * 34) Familial
 * 35) Peers
 * 36) University student motivation
 * 37) Motivation and psychological disorders
 * 38) Neurobiology of motivation
 * 39) Personality
 * 40) Sensation seeking
 * 41) Motivation and self
 * 42) Needs
 * 43) Psycho-analytic perspectives of motivation
 * 44) Psychological reactance
 * 45) Thematic Apperception Test
 * 46) Psychopathologies
 * 47) Obessive-compulsive disorder
 * 48) Types of motivation
 * 49) Amotivation / Demotivation
 * 50) Intimacy
 * 51) Power
 * 52) Social
 * 53) Will power

Emotion

 * 1) Aspects of emotion
 * 2) Emotional regulation
 * 3) Emotional valence
 * 4) Negative emotions
 * 5) Positive emotions
 * 6) Children
 * 7) Emotional intelligence
 * 8) Assessment of emotion
 * 9) Self-report
 * 10) PANAS
 * 11) Physiological measurements
 * 12) Behavioural observations
 * 13) Basic/Core emotions
 * 14) Plutchik
 * 15) For more models and an overview, see
 * 16) Emotion and behaviour
 * 17) Brain design and emotion
 * 18) Emotion and cognition
 * 19) Appraisal
 * 20) Attention
 * 21) Attributions and expectations
 * 22) Decision-making
 * 23) Learning
 * 24) Sadness
 * 25) Emotion and environment
 * 26) Emotion and personality
 * 27) Emotion and psychological disorders
 * 28) Dementia
 * 29) Emotion and specific groups
 * 30) Infants
 * 31) Adults
 * 32) Elderly
 * 33) Emotion and specific topics
 * 34) Apologising
 * 35) Health
 * 36) Learning
 * 37) Memory
 * 38) Personality
 * 39) Historical development of emotion
 * 40) Specific emotions
 * 41) Core emotions not yet covered:
 * 42) Disgust
 * 43) Sadness
 * 44) Detailed list of specific emotions
 * 45) Stress, arousal and coping
 * 46) Theories
 * 47) James-Lange Theory
 * 48) Cannon-Bard Theory
 * 49) Schacter-Singer Theory
 * 50) Somatic
 * 51) Neurobiological
 * 52) Cognitive theories
 * 53) Well-being


text:
User talk:Creasyhuhu

Kipling's Puck of Pout's Hill
Kipling's Puck of Pout's Hill The W ind in the Willows shares with contemporary children's books are a faith in the rural—and in rural England, especially—and an uncomfortable awareness of threats to the status quo. (The book that best sums up these preoccupations is Kipling's Puck of Pout's Hill, published in 1906.) http://www.webdaytona.com Nor is this a book about animals, despite the ink that has been spilt attempting to link it to animal fictions. For virtually all of the time the characters are, as Margaret Blount puts it, 'Olympians, middle-aged men. . . doing nothing as becomes animals, http://www.solecolor.com yet very much involved with the real world .. . For animals, read chaps.'6 Fred Inglis regards the heroes as 'the model men of private means w hom its readers once hoped to become... The four friends translate readily into the heroes of John Buchan and Sapper [and] P. G. Wodchousc.'7 Of course, as Roger Sale noted, the characters are more than chaps: 'It will not do to say that they are human beings, because G rahame's fantasy depends on his being able to. . . http://www.easylulu.com not give them an age, a biography, a past'.* Thus they are partly creatures of fable, unencumbered by at least some of the complexities attached to human life (such as servants and except in Otter's case relatives), and this enables them to be simultaneously univcrsals and many-layered individuals.

title: User talk:Creasyhuhu