User:Egm6321.f11.team2/HW6

=R*6.1 - Solve Nonhomogeneous L2-ODE-CC =

Given
As an application of nonhomogeneous L2-ODE-CC:

2nd exactness condition for N2-ODES is:

For problem 2, integrating factor is

Find
1. Find the PDEs that governs the integrating factor $$\displaystyle h(x,y)$$ using 2nd exactness condition for N2-ODEs

2. Solve (6.1.1) using trial solution for integrating factor of $$\displaystyle h(t)={{e}^{\alpha t}}$$

2.1) Find $$ (\bar a_1,\bar a_0)$$ in terms of $$ (a_0,\ a_1,\ a_2) $$.

2.2) Find the quadratic equation for $$\displaystyle \alpha $$.

2.3) Reduced-order equation: lead to

$$\displaystyle e^{\alpha t}[\bar a_1y'+\bar a_0y]=\int e^{\alpha t}f(t)dt$$

$$\displaystyle \bar a_1y'+\bar a_0y=e^{-\alpha t}\int e^{\alpha t}f(t)dt$$

2.4) Use the IFM to solve the equation in part 2.3.

2.5) Show that

$$\displaystyle  \alpha \beta = \frac {a_0}{a_2}$$.

$$\displaystyle  \alpha+\beta=\frac{a_1}{a_2}$$

2.6) Deduce the particular solution $$\displaystyle y_P(t)$$ for general excitation $$\displaystyle f(t)$$

2.7) Verify result with table of particular solutions for

$$\displaystyle f(t) = t exp(bt) $$

2.8) Solve the nonhomogeneous L2-ODE-CC with the following excitation

Hyperbolic function: $$\displaystyle f(t) = tanh t $$

For the coefficients (a0,a1,a2),consider two characteristic equations:

2.8.1) $$\displaystyle (r+1)(r-2) = 0$$

2.8.2) $$\displaystyle (r-4)^2 = 0 $$

2.9) For each case in questions 2.8.1 and 2.8.2, determine the fundamental period of undamped free vibration, and plot the solution for the excitation for about 5 periods, assuming zero initial conditions.

Part 1
Multipling $$\displaystyle H(t,y)$$ in the both sides, (6.1.1) can be rewritten as follows:

In the 2nd exactness condition,

From (6.1.5) and (6.1.6), We can get the following relation:

Assuming that H(t,y) and F(t,y) are only the functions of t in (6.1.2) and (6.1.3), and P=y', we can get the following relations in the 2nd exactness condition:

From (6.1.9) and (6.1.10), we can get the following result:

Part 2.1
where $$ \bar{a}_2 = 0 $$

$$ \frac{d}{dt}\Big[e^{\alpha t}(\bar{a}_1y'+\bar{a}_0y) \Big] = e^{\alpha t}\Big(\bar{a}_1 y + (\alpha \bar{a}_1+\bar{a}_0)y' + \alpha \bar{a}_0 y \Big) = e^{\alpha t}\big[a_2y+a_1y'+a_0y\big] \! $$

Thus, we can the the following relation.

$$ \bar{a}_1 = a_2 \! $$

$$ \bar{a}_0 = a_1 - \alpha a_2 = a_0/\alpha \! $$

Part 2.2
Multiplying $$\displaystyle \alpha $$ in the both second and third sides of (6.1.9) and rewriting it, we can get the following quadratic equation.

Part 2.3
From Part 2.1,

The above function can be rewriten as follows:

where $$ \beta=\frac{\bar a_0}{\bar a_1}$$

Part 2.4
The integrating factor to use IFM is $$ \bar{h}(t) = exp \Big[ \int \underbrace{\frac{\bar{a}_0}{\bar{a}_1}}_{\beta} dt \Big] = e^{\beta t} \! $$

The IFM would be used for the solution as follows: $$\displaystyle y(t)=\frac{1}{\bar h(t)}\int^t \bar h(s)B(s)ds $$ The result of Part 2.3 will be devided by $$\displaystyle \bar a_1 $$ to obtain B(s). $$\displaystyle y'+\beta y = \frac{e^{-\alpha t}}{\bar a_1} \int e^{\alpha t}F(t)dt = B(s) $$ Then we can combine $$\displaystyle h(t), \ B(s) $$ into $$\displaystyle y(t) $$ to find

Part 2.5
$$ \beta = \frac{\bar{a}_0}{\bar{a}_1} = \frac{a_0/\alpha}{a_2} $$

Multiplying $$\displaystyle \alpha $$ in the both first and last sides of the above equation, we can get the following:

$$ \alpha\beta = \frac{a_0}{a_2} $$

$$ \beta = \frac{\bar{a}_0}{\bar{a}_1} = \frac{a_1 - \alpha a_2}{a_2} = \frac{a_1}{a_2} - \alpha $$

Thus, $$ \alpha + \beta = \frac{a_1}{a_2} $$

Part 2.6
When $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ are the roots of the characteristic equation, the homogeneous solution is

The particular solution $$\displaystyle y_P(t) $$ is

Part 2.7
Since $$\displaystyle F(t) = t e^{bt} $$,

Part 2.8.1
Thus, the L2-ODE is given by

coefficients $$\displaystyle {{a}_{0}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{2}}$$ are

From the root of the above characteristic equations, we can see the following:

$$\displaystyle \alpha=-1, \beta=2$$

The excitation is hyperbolic function as follows: $$\displaystyle f(t)=tanh(t)$$

With the website of "wolframalpha", the general solution y(t) is $$\displaystyle y(t)=\frac{1}{6}\left[1+2e^{-2t}log(e^{2t}+1)-4e^ttan^{-1}(e^{-t})\right]$$

Part 2.8.2
Thus, the L2-ODE is given by

The coefficients $$\displaystyle {{a}_{0}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{2}}$$ are

From the root of the above characteristic equations, we can see the following:

$$\displaystyle \alpha=\beta=4$$

The excitation is hyperbolic function as follows: $$\displaystyle f(t)=tanh(t)$$

With the website of "wolframalpha", the general solution y(t) is $$\displaystyle y(t)=-\frac{Li_2(-e^{2t}\cdot e^{-4t})}{2}-\frac{e^{-2t}}{2}+\frac{1}{16}$$

=R*6.2 - Showing that the form of particular solution in notes agree with the one in book by King and others =

Given
And

Find
with

Solution
From equation Equation 6.2.2 Substituting the value of Wronskian in the above equation Substituting the value of 'h(x)' from Equation 6.2.5 into Equation 6.2.1 Integrating the above Equation by parts and simplifying: Applying dummy integrals: Multiplying the terms inside bracket by $$\displaystyle u_1(x) $$ Taking $$\displaystyle u_1(x) \,\, u_2(x)$$ inside integral. Combining the two integrals:

=R*6.3 - Invalid Roots of a Characteristic Equation=

Given
The L2-ODE-VC: $$\displaystyle (x - 1) y'' - x y' + y = 0 $$(0) Using the Trial Solution $$\displaystyle y = e^{r x}$$ If it is assumed that r is a constant, the characteristic equation is $$\displaystyle (x-1) r^2 - x r + 1 = 0$$

Find
It can be seen that the roots of the characteristic equation are, $$\displaystyle r_1 = 1$$ $$\displaystyle r_2(x) = \frac{1}{x - 1}$$ Explain why $$\displaystyle r_2(x)$$ is not a valid root, i.e., $$\displaystyle u_2(x) = e^{x r_2(x)}$$ is not a valid solution.

Solution
$$\displaystyle r_2(x)$$ is not a valid root because it is assumed for the trial solution r is a constant, thus r cannot be a function of x. To see this more clearly we can assume that r is a function of x and see what the result would be. This would make the trial solution $$\displaystyle y = e^{x r(x)}$$ Using this trial solution it can be seen that a different characteristic equation is obtained. $$\displaystyle y' = (r(x) x )' e^{x r(x)}$$ $$\displaystyle y = (r(x) x ) e^{x r(x)} + ((r(x) x )')^2 e^{x r(x)}$$ $$\displaystyle (x - 1) ((r(x) x )'' + ((r(x) x )')^2) - x ((r(x) x )') + 1 = 0$$ It can be seen that this characteristic equation is not the same as the characteristic equation obtained by assuming r is a constant. This shows that $$r_2(x)$$ is not a valid root.

R*6.4 – Using the Method of Variation of Parameters to Find a Linearly Independent Solution to a Homogeneous L2-ODE When One Solution is Known

Given
$$\displaystyle u_1 := e^x$$

$$\displaystyle F(u_1) = 0$$

Find
Using u1 and the method of Variation of Parameters, determine another, linearly independent solution of 6.4.1.

Solution
Let

$$\displaystyle u = m(x)\,u_1 = m(x)\,e^x$$

Then

$$ \begin{align} u' &= m'\,e^x + m\,e^x \\ u &= m\,e^x + 2m'\,e^x + m\,e^x \end{align} $$

Substituting these values into 6.4.1, we find

$$ \begin{align} F(u) &= (x-1)\,( m''e^x + 2m'e^x + m\,e^x ) - x\,( m'e^x + m\,e^x ) + m\,e^x \\ &= \left[ (x-1) \right] m''e^x + \left[2(x-1) - x \right] m'e^x + \cancel{\left[ (x-1) - (x-1) \right]}\, m\,e^x \\ &= \left[ (x-1)\,m'' + (x-2)\,m' \right]\,e^x \\ &= 0 \end{align} $$

For this to be true for all x, the expression in brackets must be equal to zero. This gives us a differential equation for m:

$$\displaystyle (x-2)\,m' + (x-1)\,m'' = 0$$

Since m does not appear in this equation, we may reduce its order by defining p := m'.

As noted above, 6.4.2 is in the particular form for an N1-ODE, but it is not exact since

$$\displaystyle M_p = x-2 \quad\neq\quad 1 = N_x$$

Using the IFM and assuming hp = 0 yields

$$ \begin{align} \frac{h_x}{h} &= \frac{1}{N}\,(M_p - N_x) \\ \frac{h_x}{h} &= \left(\frac{1}{x-1}\right)(x-3) \\ \int \frac{dh}{h} &= \int \frac{x-3}{x-1}\,dx \\ \log h &= x - 2\log (x-1) \\ \log h + \log (x-1)^2 &= x \\ \log (h\,(x-1)^2) &= x \\ h &= \frac{e^x}{(x-1)^2} \\ \end{align} $$

We see that our assumption about hp holds. Multiplying 6.4.2 through by h gives

$$ \begin{align} \frac{e^x\,p'}{x-1} + \frac{x-2}{(x-1)^2}\,e^x\,p &= 0 \\ \frac{d}{dx} \left( \frac{e^x\,p}{x-1}\right) &= 0 \\ \frac{e^x\,p}{x-1} &= k_1 \\ p &= k_1\,(x-1)e^{-x} \end{align} $$

Since p := m', we may integrate the expression above to determine m:

$$\displaystyle m(x) = -k_1\, x e^{-x} + k_2$$

Finally, substituting this expression for m into the definition of u := m ex we arrive at

$$\displaystyle u(x) = k_2\, e^x + k_3\, x \qquad k_3 := -k_1$$

Since u is a solution of 6.4.1, and it is a linear combination of u1 and x, we finally conclude that the other homogeneous solution of 6.4.1 is

$$\displaystyle u_2 = x $$

=R*6.5 - Find two homogeneous solutions =

Given
$$\displaystyle (x+1)y''-(2x+3)y' + 2y = 0 $$(6.5.1) $$\displaystyle y = e^{rx} $$(6.5.2)

Find
Find $$\displaystyle u_1(x), u_2(x) $$ of eq 6.5.1 using the trial solution eq 6.5.2

Solution
$$\displaystyle y = e^{rx} \,\,\,\, y' = re^{rx} \,\,\, y'' = r^2e^{rx} \,\,\, $$(6.5.3) plugging eq 6.5.3 into eq 6.5.1 $$\displaystyle (x+1)r^2e^{rx}-(2x+3)re^{rx}+2e^{rx} = 0 $$ $$\displaystyle (x+1)r^2 - (2x + 3)r + 2 = 0 $$ $$\displaystyle r = \frac {(2x+3) \pm \sqrt{(2x+3)^2-8(x+1)}}{2(x+1)} $$ $$\displaystyle r = \frac{(2x+3) \pm (2x+1)}{2(x+1)} $$ $$\displaystyle r_1 = 2, \,\,\,\,r_2 = \frac{(2x+3)-2x - 1 }{2(x+1)} $$ Now we can find u(x), that is $$\displaystyle u_1(x)= e^{2x} $$ From lecture note 34-5, $$\displaystyle u_2(x)= u_1(x) \cdot \int \frac{1}{u_1^2(x)}exp[-\int a_1(x)dx]dx $$ $$\displaystyle a_1(x) = \frac {-(2x+3)}{x+1}$$ $$\displaystyle u_2(x) = e^{2x} \cdot \int{ e^{-4x}\cdot e^{\int \frac{2x+3}{x+1}dx}dx} $$ $$\displaystyle u_2(x) = e^{2x}\cdot \int{ e^{-4x}\cdot e^{(2x + log (x+1))}dx}$$ $$\displaystyle u_2(x) = e^{2x}\cdot \int{ e^{-4x}\cdot e^{2x} \cdot (x+1)dx}$$ $$\displaystyle u_2(x) = e^{2x}\cdot \int{ e^{-2x}\cdot (x+1)dx}$$ $$\displaystyle u_2(x) = e^{2x}\cdot e^{-2x}\cdot \frac {-(2x+3)}{4}$$ $$\displaystyle u_2(x) = \frac {-(2x+3)}{4} $$ Here -1/4 is constant. Thus, homogeneous solution is that $$\displaystyle y(x) = c_1\cdot e^{2x} + c_2\cdot (2x+3) $$

=R*6.6 - Finding L2-ODE-VC =

Given
Trial solution

$$\displaystyle y(x) = \frac {e^{rx}}{\sin x}$$ (6.6.1)

and characteristic equation.

$$\displaystyle (r-2)(r-\frac {1}{1+x})=0$$ (6.6.2)

Find
Find an L2-ODE-VC whose characteristic equation is 6.6.2

Solution
From 6.6.1, we have

$$\displaystyle y'= \frac{r\,{e}^{r\,x}}{\mathrm{sin}\left( x\right) }-\frac{{e}^{r\,x}\,\mathrm{cos}\left( x\right) }{{\mathrm{sin}\left( x\right) }^{2}}$$ (6.6.3)

$$\displaystyle y''= \frac{{r}^{2}\,{e}^{r\,x}}{\mathrm{sin}\left( x\right) }+\frac{{e}^{r\,x}}{\mathrm{sin}\left( x\right) }-\frac{2\,r\,{e}^{r\,x}\,\mathrm{cos}\left( x\right) }{{\mathrm{sin}\left( x\right) }^{2}}+\frac{2\,{e}^{r\,x}\,{\mathrm{cos}\left( x\right) }^{2}}{{\mathrm{sin}\left( x\right) }^{3}}$$ (6.6.4) then

$$\displaystyle y'=\left(r-\frac{\,\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}\right)\frac{{e}^{r\, x}}{\mathrm{sin}\left(x\right)}$$ (6.6.5)

$$\displaystyle y''=\left({r}^{2}+1-\frac{2\, r\,\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}+\frac{2\,{\mathrm{cos}\left(x\right)}^{2}}{{\mathrm{sin}\left(x\right)}^{2}}\right)\frac{{e}^{r\, x}}{\mathrm{sin}\left(x\right)}$$ (6.6.6) So we can assume that

$$\displaystyle a_{2}(x)\left({r}^{2}+1-\frac{2\, r\,\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}+\frac{2\,{\mathrm{cos}\left(x\right)}^{2}}{{\mathrm{sin}\left(x\right)}^{2}}\right)\frac{{e}^{r\, x}}{\mathrm{sin}\left(x\right)}+a_{1}(x)\left(r-\frac{\,\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}\right)\frac{{e}^{r\, x}}{\mathrm{sin}\left(x\right)}+a_{0}(x)\frac{{e}^{r\, x}}{\mathrm{sin}\left(x\right)}=0$$ (6.6.7)

$$\displaystyle a_{2}(x){r}^{2}+\left(a_{1}(x)-a_{2}(x)\frac{2\,\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}\right)r-a_{1}(x)\frac{cos\left(x\right)}{\mathrm{sin}\left(x\right)}+a_{2}(x)+a_{2}(x)\frac{2\,{\mathrm{cos}\left(x\right)}^{2}}{{\mathrm{sin}\left(x\right)}^{2}}+a_{0}(x)=0$$ (6.6.8) comparing 6.6.8 with the 6.6.2 we have

$$\displaystyle a_{2}=x+1$$ (6.6.9) $$\displaystyle a_{1}=2(x+1)\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}-(2x-3)$$ (6.6.10) $$\displaystyle a_{0}=1-x-(2x-3)\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}$$ (6.6.11) Thus

$$\displaystyle \left(x+1\right)y''+\left(2(x+1)\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}-(2x-3)\right)y'+\left(1-x-(2x-3)\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}\right)y=0$$ (6.6.12)

=R*6.7 - Homogeneous Solution of Legendre Function Using Variation of parameters=

Given
The general form of the Legendre Equation is

$$\displaystyle P_2(x)$$ should be

for the following result of $$\displaystyle Q_2(x)$$

Find
Show

using the Method of Variation of Parameters.

Solution
When n=2, the given Legendre Equation is as

(6.7.4) cam be rewritten as follows:

Let $$\displaystyle y(x) = U(x)\cdot {u}_{1}(x)$$ and $$\displaystyle P_2(x) = {u}_{1}(x)$$.

where $$\displaystyle \begin{align} &y = U {u}_{1} \\ &y' = U {u}_{1}'+U' {u}_{1} \\ &y= U {u}_{1}+2U' {u}_{1}'+U'' {u}_{1} \\ \end{align} $$

To reduce the order of (6.7.6), $$\displaystyle Z:=U'$$ willb be used.

To use integrating factor method, we first need to find the integrating factor, h(x), as follows:

The soloution of Z is

From the definition of Z(x)= U',

Solving y(x) with U(x),

2nd homogeneous solution is

Thus,

=R*6.8 - Solving L2-ODE-VC using variation of parameters. =

Given
and

Find
For both equations: 1) Find the first homogenous solution by trial solution. 2) Find the complete solution by variation of parameters with excitation $$ \displaystyle f(x)= exp(\alpha x) $$

Solution
Solving for homogenous solution of Equation 6.8.1 Consider the general trial solution $$ \displaystyle y=x^c e^{rx} $$ Taking derivative to obtain $$ \displaystyle y' $$ and $$ \displaystyle y'' $$ Substituting the derivatives back into equation 6.8.1
 * {| style="width:100%" border="0" align="left"

Solving the above Equation for constants 'r' and 'c' :
 * $$\Rightarrow x^{c+1}\left[r^2-r \right]+x^c\left[ 2rc-r^2-c+1 \right]+x^{c-1}\left[c^2-c-2rc \right]+x^{c-2}\left[c-c^2 \right]=0$$
 * }
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} i) & c=0 & r=1\\  ii) & c=1 & r=0\\ \end{array} $$ Therefore First Homogeneous solution is
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

and second homogeneous solution is

Solution is a linear combination of two homogenous solutions: Now solving for complete solution if the forcing function $$ \displaystyle f(x)= e^{{\alpha} x }$$ Above equation can be rewritten as:

Now Therefore: Now Integrating factor is defined as: Therefore: Now particular solution is defined as : Now complete solution is defined as:

Solving for homogenous solution for Equation 6.8.2 Assuming the general trial solution $$ \displaystyle y=x^c e^{rx} $$ Taking derivative to obtain $$ \displaystyle y' $$ and $$ \displaystyle y'' $$ and substituting the values back into equation 6.8.2, we obtain: Solving the above Equation for constants 'r' and 'c' :
 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} i) & c=-1 & r=i\\  ii) & c=-1 & r=-i\\ \end{array} $$ Therefore First Homogeneous solution is
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

and second homogeneous solution is

Solution is a linear combination of two homogenous solutions: Now solving for complete solution if the forcing function $$ \displaystyle f(x)= e^{{\alpha} x }$$ Above equation can be rewritten as:

Now

Therefore: Now Integrating factor is defined as:

Therefore: Now particular solution is defined as :

Now complete solution is defined as:

.

=R6.9 - Using Variation of Parameters to Determine a Homogeneous Solution=

Given
$$\displaystyle ( 1 - x^2 ) y'' - 2 x y' + 2 y = f(x)$$(1) Where, $$\displaystyle f(x) = 1 $$(2) $$\displaystyle u_{1}(x) = P_1(x) = x$$(3)

Find
Given that equation (3) is the 1st homogeneous solution for the Non-homogeneous Legendre equation (L2-ODE-VC) in equation (1) find the solution $$\displaystyle y(x)$$ by variation of parameters.

Solution
Let us assume that the solution takes the form, $$\displaystyle y(x) = U(x) u_1(x)$$ Where $$\displaystyle U(x) $$ is an unknown factor to be determined. Then, $$\displaystyle \left [ y = U u_1 \right ] \times a_0(x)$$ $$\displaystyle \left [ y' = U u'_1 + U' u_1 \right ] \times a_1(x)$$ $$\displaystyle \left [ y = U u_1 + 2 U' u'_1 + U'' u_1 \right ] \times a_2(x)$$ $$\displaystyle a_0 y + a_1 y' + a_2 y = U \left [ a_0 u_1 + a_1 u'_1 + a_2 u_1 \right ] + U' \left [ a_1 u_1 + 2 a_2 u'_1 \right ] + a_2 U'' u_1 = f(x) $$ Since $$\displaystyle u_1(x) $$ is a homogeneous solution, $$\displaystyle a_0 u_1 + a_1 u'_1 + a_2 u''_1 = 0$$ Using a substitution of variables, $$\displaystyle Z:= U'$$ We obtain the equation, $$\displaystyle \tilde a_1(x) Z' + \tilde a_0(x) Z = \frac {f(x)}{a_2(x) u_1(x)}$$ Where, $$\displaystyle \tilde a_1(x) = 1$$ $$\displaystyle \tilde a_0(x) = \frac {a_1(x) u_1(x) + 2 a_2(x) u'_1(x)}{a_2(x) u_1(x)}$$ In order to solve for $$\displaystyle Z(x) $$ the integrating factor method can be used. Because all variables are a function of x we can assume that the integration factor is a function of x, thus $$\displaystyle h(x) = \exp \left [ \int \tilde a_0(x) dx \right ]$$ $$\displaystyle h(x) = \exp \left [ \int \frac {a_1(x)}{a_2(x)} + \frac {2 u'_1(x)}{u_1(x)} dx \right ]$$ $$\displaystyle h(x) = u^2_1(x) \exp \left [ \int \frac {a_1(x)}{a_2(x)} dx \right ]$$ $$\displaystyle Z(x) = \frac {1}{h(x)} \left [ k_2 + \int h(x) \frac {f(x)}{u_1(x)} dx \right ]$$ $$\displaystyle U(x) = k_1 + \int Z(x) dx$$ Substituting in all of the variables we obtain the equation for the solution, $$\displaystyle y(x) = k_1 u_1(x) + u_1(x) \int \frac{1}{h(x)} \left [ k_2 + \int h(x) \frac{f(x)}{u_1(x)} dx \right ] dx$$ $$\displaystyle y(x) = k_1 u_1(x) + k_2 u_1(x) \int \frac{1}{h(x)} dx + u_1(x) \int \frac{1}{h(x)} \left [ \int h(x) \frac{f(x)}{u_1(x)} dx \right ] dx$$ $$\displaystyle y(x) = k_1 x + k_2 x \int \left [ \frac{1}{x^2} \exp \left [ - \int \frac{-2x}{1-x^2} dx \right ] \right ] dx + x \int \left [ \frac {1}{x^2} \exp \left [ - \int \frac{-2x}{1-x^2} dx \right ] \int \left [ x \exp \left [ \int \frac{-2x}{1-x^2} dx \right ] \right ] dx \right ] dx$$ $$\displaystyle y(x) = k_1 x + k_2 x \int \left [ \frac{1}{x^2 (1 - x^2 )} \right ] dx + x \int \left [ \frac {1}{x^2 ( 1 - x^2 ) } \int ( x (1 - x^2) ) dx \right ] dx$$ $$\displaystyle y(x) = k_1 x + k_2 \left [ \frac{x}{2} \log \left ( \frac {1+x}{1-x} \right ) - 1 \right ] + x \int \left [ \frac {1}{x^2 ( 1 - x^2)} (\frac{1}{2} x^2 - \frac{1}{4} x^4) \right ] dx$$ $$\displaystyle y(x) = k_1 x + k_2 \left [ \frac{x}{2} \log \left ( \frac {1+x}{1-x} \right ) - 1 \right ] + x \int \left [ \frac {2 - x^2}{4(1 - x^2)} \right ] dx$$ $$\displaystyle y(x) = k_1 x + k_2 \left [ \frac{x}{2} \log \left ( \frac {1+x}{1-x} \right ) - 1 \right ] + \frac {x}{8} \left [ 2 x + \log \left ( \frac {x+1}{1-x} \right ) \right ] $$ Where the second homogeneous solution is, $$\displaystyle y(x) = \frac{x}{2} \log \left ( \frac {1+x}{1-x} \right ) - 1$$ And the particular solution is, $$\displaystyle y(x) = \frac {x}{8} \left [ 2 x + \log \left ( \frac {x+1}{1-x} \right ) \right ] $$

=Contributing Members= Egm6321.f11.team2.kim I solved R*6.5 and proofread R*6.3, R*6.7, R*6.8 and R*6.9 Egm6321.f11.team2.johri.y 19:15, 16 November 2011 (UTC) I solved problem R*6.2 and R*6.8. Egm6321.f11.team2.Xia I solved R*6.6 and proofread R*6.4 [User:Egm6321.f11.team2.kim|Egm6321.f11.team2.kim]] I solved R*6.5 and proofread R*6.3, R*6.7, R*6.8 and R*6.9 Egm6321.f11.team2.epps 20:21, 16 November 2011 (UTC) I solved R*6.4 and proofread R6.2, R6.3, R6.5, R6.6, and R6.7.

Egm6321.f11.team2.rho21:49, 16 November 2011 (UTC) I solved R*6.1 and R*6.7 and proofread R6.4, R6.5, R6.8, and R6.9. Egm6321.f11.team2.wise 21:57, 16 November 2011 (UTC) I solved R*6.3 and R6.9 and proofread R6.7, and R6.8