User:Egm6321.f09.team5/HW4

 See further comments in google doc for Team 5. Egm6321.f09 21:03, 23 November 2009 (UTC)

Problem Statement
Pg. 19-1. Lengendre Differential Equation (1) Pg. 14-2 w/ n=0, so that the homogenous solution $$ U_{1}(x) = 1 $$

Use Reduction of order method 2 (undetermined factor) to find $$U_{2}(x)$$, 2nd homogeneous solution.

Solution
When n=0, the Legendre equation becomes:

$$ F = (1-x^{2})y^{''} - 2xy^{'} + 0 = 0 \qquad \qquad Equation \quad 1$$

Using the Undetermined Factor as follows:

Assume the full homogeneous solution is of the form:

$$ y(x) = U(x)u_{1}(x) $$

rearranging the differential equation 2 to fit the standard form of equation 2 as follows:

$$ y^{''} + a_{1}(x)y^{'} + a_{0}(x)y = 0 \qquad \qquad Equation \quad 2$$

Equation 1 becomes:

$$ y^{''} - \frac{2x}{(1-x^{2})}y^{'} = 0 \qquad \qquad Equation \quad 3$$

The undetermined factor method allows for the following form:

$$ 0 = U^{'}\left [ a_{1}u_{1} + 2u_{1}^{'} \right ] + U^{''}u_{1} $$

Equation 3 becomes:

$$ 0 = U^{'}\left [ \frac{-2x}{(1-x^{2})} \right ] + U^{''} \qquad \qquad Equation \quad 4 $$

Letting $$ z = U^{'} $$ Equation 4 becomes a L1.ODE.VC which can be solved as follows:

$$ z^{'} + z \left [ \frac{-2x}{(1-x^{2})} \right] = 0 $$

Integrating by parts yields to solve for z yields:

$$ z(x) = c(1-x^{2}) = U^{'}(x) \qquad \qquad where \quad c=\exp(k) $$

Integrating again the result is:

$$ U(x) = c \left [ x - \frac{x^{3}}{3} \right ] + c2 \qquad \qquad. $$                where c and c2 are constants

The expression for the second homogeneous solution is given as:

$$ u_{2} = u_{1} \int_{}^{x} \exp (\ln(1-t^{2})) dt \qquad \qquad Equation \quad 5 $$

Using Equation 5 the second homogeneous solution yields:

$$ u_{2} = \left [ x - \frac{x^{3}}{3} \right ] $$

The Full homogeneous solution can be give as follows:

$$ y(x) = K_{1} + K_{2} \left [ x - \frac{x^{3}}{3} \right ] $$

Problem Statement
King et al. Pg 28. Problem 1.1b

$$ xy^{''} + 2y^{'} + xy = 0 $$          Equation 0

1) Check for exactness, If not exact can it be made exact using integration factor.

2) Use the following trial solutions to find the homogeneous solution

3) Use the undetermined factor method to find the second homogeneous solution, knowing a homogeneous solution

1)
The First step is to check for exactness using the following conditions:

Condition 1: The ODE must be of the form:

$$F=f(x,y,p)y^{''}+g(x,y,p)$$            Equation 1

Where $$ P=y^{'} $$

Condition 2: The following Conditions must be met:

$$    \ f_{xx} + 2pf_{xy} + p^{2}f_{yy} = g_{xp} + pg_{yp} - g_{y}  \qquad $$      Equation 2

$$    \ f_{xp} + pf_{yp} + 2f_{y} = g_{pp}  \qquad $$         Equation 3

$$ f(x)=x $$

$$ g(x)= 2p + xy $$

$$ f_{x}= 1 \qquad \qquad  g_{y}= x $$

$$ f_{xx}= 0  \qquad \qquad g_{yp}= 0 $$

$$ f_{xy}= 0  \qquad \qquad  g_{x}= y  $$

$$ f_{y}= 0        \qquad \qquad \qquad  g_{xp}= 0 $$

$$ f_{yy}= 0    \qquad \qquad  g_{p}= 2 $$

$$ f_{yp}= g_{pp}= f_{xp}=0 $$

Equation 2 yields:

$$ 0 + 2p(0) + p^{2}(0) = 0 + p(0) - x $$

$$ 0 \neq -x $$

This shows that the ODE is not exact.

The next step is to check whether it can be made exact by using the integration factor method:

The integration factor to be used will be:

$$ h(x,y) = x^{m}y^{n} $$

multiplying the integration factor by the ODE yields the following:

$$ \left [ x^{m}y^{n} \right ] \left[ xy^{''} + 2y^{'} + xy \right ] = 0 $$

$$ x^{m+1}y^{n}y^{''} + 2x^{m}y^{n}y{'} + x^{m+1}y^{n+1} = 0 $$

The next step is to check once again for exactness:

$$ f = x^{m+1}y^{n} $$ $$ g= 2x^{m}y^{n}p + x^{m+1}y^{n+1} $$

$$ f_{x}= y^{n}(m+1)x^{m} \qquad \qquad  g_{y}= 2x^{m}ny^{n-1}p + x^{m+1}(m+1)y^{n} $$

$$ f_{xx}= y^{n}(m+1)mx^{m-1}  \qquad \qquad g_{yp}= 2x^{m}ny^{n-1} $$

$$ f_{xy}= ny^{n-1}(m+1)x^{m}  \qquad \qquad  g_{x}= 2y^{n}pmx^{m-1} + y^{n+1}(m+1)x^{m} $$

$$ f_{y}= x^{m+1}ny^{n-1}        \qquad \qquad \qquad  g_{xp}= 2y^{n}mx^{m-1} $$

$$ f_{yy}= x^{m+1}n(n-1)y^{n-2}    \qquad \qquad  g_{p}= 2x^{m}y^{n} $$

$$ f_{yp}= g_{pp}= f_{xp}=0 $$

Using Equation 3 in order to check for exactness yields the following:

$$ 0 + p(0) + 2(x^{m+1}y^{n-1}n \neq 0 $$

Therefore the new ODE is also concluded not to be exact.

2)
The next step is to find a homogeneous solution by using different trial solutions as follows:

Trial Solution 1

$$ y(x)=\exp^{rx} $$

$$ y^{'}(x) = r\exp^{rx} $$

$$ y^{''}(x) = r^{2}\exp^{rx} $$

using these results back in Equation 0 the following is obtained:

$$ x(r^{2}\exp^{rx}) + 2r\exp^{rx} + x\exp^{rx} = 0 $$

$$ \exp^{rx} (r^{2}(x) + 2r + x) = 0 $$

$$ r^{2}x + 2r + x = 0 $$

This trial solution is abandoned as it does not yield a root for the homogeneous solution

Trial Solution 2

$$ y(x)=x\exp^{rx} $$

$$ y^{'}(x) = (xr\exp^{rx} + x^{'}\exp^{rx}) $$

$$ y^{}(x) = r^{2}x\exp^{rx} + rx^{'}\exp^{rx} + x^{'}r\exp^{rx} + x^{}\exp^{rx} $$

Using these results in Equation 0 the following is obtained:

$$ x^{2}r^{2} + 2rxx^{'} + xx^{''} + 2xr + 2x^{'} = 0 $$

This trial solution is abandoned as it will not yield a root for the homogeneous solution

Trial Solution 3

$$ y(x)=x^{-1}\exp^{rx} $$

$$ y^{'}(x) = (rx^{-1}\exp^{rx} - x^{-2}\exp^{rx}) $$

$$ y^{''}(x) = \exp^{rx}(-2rx^{-2} + r^{2}x^{-1} + 2x^{-3}) $$

Using these results in Equation 0 the following is obtained:

$$\exp^{rx} \left [ r^{2} + 1 \right ] = 0 $$

From the previous equation the following result is found:

$$ r = \sqrt[]{-1} = i $$

$$ y(x) = x^{-1}\exp^{ix} $$

Using the de Moivre relationships the following form is obtained for the homogeneous solution:

$$ y(x)= x^{-1}(cosx + isinx) $$

3)
Given a homogeneous solution the full solution can be found using the undetermined factor method. The full solution is assumed to be of the form:

$$ y(x) = U(x)u_{1}(x) $$

Where $$ u_{1}(x) $$ is the known homogeneous solution.

$$ u_{1}(x)= \frac{sin(x)}{x} $$

Equation 0 is organized as follows:

$$ y^{''} + \frac{2}{x}y^{'} + y = 0 $$

where:

$$ a_{0}=1 \qquad \qquad a_{1} = 2x^{-1} $$

Because of the undetermined factor method the following can be written:

$$ 0 = U^{'} \left [ x^{-2}\sin(x) + 2x^{-1}\cos(x) \right ] + U^{''}\frac{\sin(x)}{x} $$  Equation 3

A variable is defined as follows:

$$ z=U^{'} $$

Equation 3 can be rewritten as follows:

$$ z\left [ x^{-2}\sin(x) + 2\cos(x)x^{-1} \right] + z^{'} \frac{\sin(x)}{x} $$

Solving the previous L1.ODE.VC using integration by parts and then solving for $$ U $$ the following is identified:

$$ u_{2}(x) = u_{1} \int{}^{x} \frac{1}{(u_{1}(t))^{2}} \exp^{-\int{}^{t} a_{1}(s)ds} dt $$

$$ \int{}^{t} a_{1}(s)ds = 2ln(x) $$

$$ \int{}^{x} \frac{1}{(u_{1}(t))^{2}} \exp^{-\int{}^{t} a_{1}(s)ds} dt = \int{}^{x}\frac{t^{2}}{\sin(t)^{2}}t^{-2} = -cot(x)  $$

$$ u_{2}(x) = \frac{\sin(x)}{x}\frac{-\cos(x)}{\sin(x)} = -\frac{\cos(x)}{x}  $$

This result confirms the results found in part 2 for the full homogeneous solution.

Problem Statement
Pg. 21-3

$$ U_{1}(x)= \exp^{x} $$

$$ \qquad  (x-1)y^{''}-xy^{'} + y =0   \qquad \qquad Equation \quad 1$$

Explain how do you know the homogeneous solution given.

Use the following trial solution:

$$ y(x)= \exp^{rx} \qquad \qquad r \quad is \quad constant$$

Use the Reduction of order method 2 (undetermined factor) to find 2nd equation.

NOTE:

-Find r1,r2. How many valid homogeneous solutions?

If only one valid homogeneous solution, $$\quad U_{1}(x)=\exp^{r_{1}x} \quad$$, find $$\quad U_{2}(x) \quad$$ using undetermined factor.

Solution
The first step is to try to find a homogeneous solution by the trial solution technique:

The trial solution used will be $$ y(x) = \exp^{rx} $$

Using the trial solution the following is obtained:

$$ y(x) = \exp^{rx} $$

$$ y^{'}(x) = r\exp^{rx} $$

$$ y^{''}(x) = r^{2}\exp^{rx} $$

Using the previous values and substituting them to Equation 1, the following is found:

$$ (x-1)r^{2}\exp^{rx} - x(r\exp^{rx}) + \exp^{rx} = 0 $$

$$ \exp^{rx} (r^{2}x - r^{2} - xr + 1) = 0 $$

$$ r^{2}x - r^{2} - xr + 1 $$

It is observed that when r = 1 a solution is found. Therefore the following is obtained:

$$ u_{1}(x) = \exp^{x} $$

With the homogeneous solution found the next step is to find second homogeneous solution. The first step is to rearrange Equation 1 as follows:

$$ y^{''} - (\frac{x}{x-1}y^{'} + (\frac{1}{x-1})y = 0 $$

Using the undetermined factor method the full solution is said to be of the following form:

$$ y(x) = U(x)u_{1}(x) $$

with this method the following can be written:

$$ 0= U^{'} \left[ a_{1}u_{1} + 2u_{1}^{'} \right ] + U^{''}u_{1} $$

A variable Z is now defined as follows:

$$ z=U^{'} $$

In adddition the following is also identified:

$$ a_{1} = (\frac{-x}{(x-1)}) $$

$$ a_{0} = (\frac{1}{(x-1)} $$

With this form an expression can be written as follows:

$$ u_{2} = u_{1} \int_{}^{x} \frac{1}{(u_{1}(t))^{2}} \exp (- a*_{1}(t)dt) \qquad \qquad Equation \quad 2$$

The value for a* is found as follows:

$$ a* = \int{}^{t} a_{1}(s)ds = - \left [x + ln(x-1) \right ] $$

Using this value into Equation 2 yields and analyzing the integral as a first step yields:

$$ \int{}^{x} \frac{1}{\exp^{2t}} \left[ \exp^{t} + (t-1) \right ] dt $$

$$ = - \exp^{-x} + \frac{\exp^{-2x}}{-4} \left [ 2x-1 \right ] $$

$$ u_{2} = \exp^{x} \left [ \exp^{-x} + \frac{\exp^{-2x}}{-4} \left [ 2x-1 \right ] \right ] $$

$$ u_{2} = 1 + \frac{\exp^{-x}\left[2x-1 \right ]}{-4} \qquad \qquad Equation \quad 3$$

It is then concluded that the full homogeneous solution to Equation 1 is as follows:

$$ y(x) = K_{1}\exp^{x} + K_{2} (1 + \frac{\exp^{-x}\left[2x-1 \right ]}{-4}) $$

$$ K_{1} \quad and \quad K_{2} $$ are constants to be found using boundary conditions or initial conditions.

Problem Statement
Pg. 22-1 King et al. Pg. 28, Pb 1.3a

Find the General Solution to the following Non-homogeneous L2.ODE.VC:

$$ y^{''} - 2y^{'} + y = x^{1.5}\exp^{x} $$ Equation 0

Solution
Because of linearity and superposition the following can be stated about the full solution:

$$ y(x) = y_{h} + y_{p} $$

where:

$$y_{h} = homogeneous \quad solution $$

$$y_{p} = particular\quad solution $$

The first task is to find the Homogeneous solution. This will be done by using the trial solution method as follows:

$$ y(x) = \exp{rx} $$

$$ y^{'}(x) = r\exp{rx} $$

$$ y^{''}(x) = r^{2}\exp{rx} $$

The Homogeneous ODE would be of the following form:

$$ y^{''} - 2y^{'} + y = 0 $$

Plugging the trial solution values to this equation the following is obtained:

$$ r^{2}\exp{rx} - 2r\exp{rx} + \exp{rx} = 0 $$

This simplifies to:

$$ (r-1)^{2} = 0 $$

$$ r = 1 $$

From this the following is identified:

$$ u_{1}(x) = \exp{x} $$

The next step is then to use the method of undetermined factor to identify the full solution:

$$ f(x) = U^{'}(a_{1}u_{1} + 2u_{1}^{'}) + U^{''}u_{1} $$

where:

$$ a_{1} = -2 \qquad \qquad a_{0} = 1 \qquad \qquad f(x)= x^{1.5}\exp^{x} $$

note: $$ a*_1 = \int{}^{t} a_{1}(s)ds = -2t $$

$$u_{2} = u_{1} \int_{}^{x} \frac{1}{(u_{1}(t))^{2}} \exp (- a*_{1}(t)dt)$$

$$u_{2} = exp(x) \int_{}^{x} \frac{1}{(exp(2t))} \exp (2t)dt$$

$$u_{2} = \int_{}^{x}\ 1dt \exp(x)$$

$$u_{2} = x\exp(x)$$

$$y_{p} = \int_{}^{x} \ s^{1.5} \ exp(s) \frac{(exp(s) \ x \ exp(x) - exp(x) \ s \ exp(s)}{exp(2s)} ds$$

$$y_{p} = 1.5 \ x^{0.5} \ exp(x)-2.5 \ x^{1.5} \ exp(x)$$

$$y = y_h+y_p$$

$$y(x) = A\ exp(x) + B\ x\ exp(x) -1.5\ x^{0.5}\ exp(x) - 2.5\ x^{1.5}\ exp(x)$$

Problem Statement
Pg. 22-3

Describe in words, a step by step method of attacking similar problems when given only homogeneous or non-homogeneous L2.ODE.VC

Solution
In order to solve the homogeneous or non-homogeneous L2.ODE.VC the following method may be employed.

For homogeneous ODE's simply apply the following method

1) Verify exactness of the ODE.

Test for exactness. Use the Reduction of Order Method 0 if missing dependent variable. If that works, then there is an exact homogeneous solution. If the ODE is not exact apply the Integrating Factor Method to obtain an exact equation.

2) Apply trial solutions.

This step involves guessing a value for y(x) and applying values of y(x) and it's derivatives to the ODE. Once the correct form of y(x) is assumed, the roots of the characteristic are determined. If one of these roots is a constant, we have solved for our first homogeneous solution. The first homogeneous solution takes the form of the y(x) which yielded the characteristic equation. The first homogeneous solution contains the constant root of the characteristic equation as its coefficient.

3) Apply Undetermined Factor Method to find the second homogeneous solution.

4) Use Variation of Parameter Method to find the complete solution.

For a non-homogeneous L2.ODE.VC

Need to find homogeneous solution and particular solution. Find homogeneous solution using the previous steps. Find particular solution by using Trial Solution, Varying Parameters, or Alternative Method. Alternative Method can be used in Reduction of Order Method 2 if we know only one homogeneous solution obtained from trial solution.

Problem Statement
Pg. 23-1

Non-homogeneous L2.ODE.VC

a.) $$ (x-1)y^{''} - xy^{'} + y = x $$

b.) $$ xy^{''} + 2y^{'} + xy = x $$

Two Methods:

-Variation of Parameters (after knowing $$ U_{1}(x) and U_{2}(x) $$

-Alternative Method (bypass $$ U_{2} $$

Solution a.):
For a.) $$ (x-1)y^{''} - xy^{'} + y = x $$

Variation of Parameters:
For $$ (x-1)y^{''} - xy^{'} + y = x $$

From problem 3 we know u1 and u2:

$$ u_{1}(x) = \exp(x) $$

$$ u_{2} = 1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4} $$

So the homogenous part would be:

$$ y(x) = K_{1}\exp(x) + K_{2} (1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4}) $$

Next we need to find the particular solution:

$$y_p=P_1(x)u_1(x)+P_2(x)u_2(x)$$

We need to use the Wronskian matrix:

$$\underline{W}=\begin{bmatrix} u_1 & u_2\\ u'_1 & u'_2 \end{bmatrix}$$

$$W=det(\underline{W})$$

$$W=u_1u'_2-u'_1u_2$$

$$W=exp(x)*\frac{\partial }{\partial x}[ 1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4} ]-exp(x)*[ 1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4}] $$

$$W=exp(x)*(\frac{xexp(-x)}{2}-\frac{exp(-x)}{2}-\frac{exp(-x)}{4})-exp(x)-\frac{x}{2}+\frac{1}{4}$$

$$W=\frac{x}{2}-\frac{1}{2}-\frac{1}{4}-exp(x)-\frac{x}{2}+\frac{1}{4}$$

$$W=\frac{1}{2}-exp(x)$$

from 19-3,

$$P_1(x)=-\int \frac{u_2(s)f(s)}{W}ds+A$$

$$P_2(x)=\int \frac{u_1(s)f(s)}{W}ds+B$$

The constants are arbitrary, and can be set to 0

$$P_1(x)=)=-\int \frac{x[1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4}]}{\frac{1}{2}-exp(x)}dx$$

$$P_2(x)=\int \frac{xexp(x)}{\frac{1}{2}-exp(x)}dx$$

The full solution is:

$$ y(x) = y_h+y_p = K_1u_1+K_2u_2 + P_1u_1 + P_2u_2 $$

$$ y(x)= K_{1}\exp(x) + K_{2} (1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4})-exp(x)\int \frac{x[1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4}]}{\frac{1}{2}-exp(x)}dx+(1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4})\int \frac{xexp(x)}{\frac{1}{2}-exp(x)}dx$$

Alternative Method
This is the form of the equation we are using:

$$y''+a_1(x)y'+a_0(x)y=f(x)$$

$$y''+\frac{x}{(x-1)}y'+\frac{1}{(x-1)}y=\frac{x}{(x-1)}$$

We know the value of $$u_1$$

$$y(x)=U(x)u_1(x)$$

We don't know the value of U(x)

$$U'(a_1u_1+2u_1')+U''u_1=f(x)$$

Missing dependent variable U, so use Reduction of Order Method 0

let Z(x) = U'(x)

$$Z(a_1u_1+2u_1')+Z'u_1=f(x)$$

Solve for Z(x):

$$Z(x)=\frac{}{}\int^xh(s)f(s)ds$$

$$h(x)=exp(\int^x a_0(s)ds)$$

$$h(s)=h(x)=exp(\int^x(a_1u_1+2u_1')dx)=exp(\int^x(\frac{x}{(x-1)}exp(x)+2exp(x))dx$$

$$h(x)=exp(x)-exp(x)(x-1)+2exp(x)$$

$$h(s)=exp(s)-exp(s)(s-1)+2exp(s)$$

$$Z(x)=\frac{}{}\int^x(exp(s)-exp(s)(s-1)+2exp(s))(\frac{s}{(s-1)})ds=\int^x(exp(s)-sexp(s)+exp(s)+2exp(s)-sexp(s)+s^2exp(s)-sexp(s)-2sexp(s))ds$$

$$Z(x)=\int^x(4exp(s)-5sexp(s)+s^2exp(s))ds$$

$$Z(x)=4exp(x)-5exp(x)(x-1)+exp(x)(x^2-2x+2)$$

$$U(x)=\int^x(Z(s))ds$$

$$U(x)=\int^x(4exp(s)-5exp(s)(s-1)+exp(s)(s^2-2s+2))ds$$

$$U(x)=4exp(x)-5exp(x)(x-2)+exp(x)(x^2-4x+6)$$

The solution is:

$$y(x)=U(x)u_1$$

$$y(x)=(4exp(x)-5exp(x)(x-2)+exp(x)(x^2-4x+6))exp(x)$$

Solution b.):
For b.) $$ xy^{''} + 2y^{'} + xy = x $$

Variation of Parameters:
For $$ xy^{''} + 2y^{'} + xy = x$$

From problem 2 we know u1 and u2:

$$ u_{1}(x)= \frac{sin(x)}{x} $$

$$ u_{2}(x)= -\frac{cos(x)}{x} $$

So the homogenous part would be:

$$ y(x) = K_{1}(\frac{sin(x)}{x}) - K_{2}(\frac{cos(x)}{x}) $$

Next we need to find the particular solution:

$$y_p=P_1(x)u_1(x)+P_2(x)u_2(x)$$

We need to use the Wronskian matrix:

$$\underline{W}=\begin{bmatrix} u_1 & u_2\\ u'_1 & u'_2 \end{bmatrix}$$

$$W=det(\underline{W})$$

$$W=u_1u'_2-u'_1u_2$$

$$W=(\frac{sin(x)}{x})\frac{\partial }{\partial x}(-\frac{cos(x)}{x})-\frac{\partial }{\partial x}(\frac{sin(x)}{x})(-\frac{cos(x)}{x}) $$

$$W=(\frac{sin(x)}{x})(\frac{cos(x)}{x^2}+\frac{sin(x)}{x})-(\frac{cos(x)}{x}-\frac{sin(x)}{x^2})(-\frac{cos(x)}{x})$$

$$W=(\frac{sin(x)cos(x)}{x^3}+\frac{sin^2(x)}{x^2})+(\frac{cos^2(x)}{x^2}-\frac{sin(x)cos(x)}{x^3})$$

from 19-3,

$$P_1(x)=-\int \frac{u_2(s)f(s)}{W}ds+A$$

$$P_2(x)=\int \frac{u_1(s)f(s)}{W}ds+B$$

The constants are arbitrary, and can be set to 0

$$P_1(x)=-\int \frac{-cos(x)}{(\frac{sin(x)cos(x)}{x^3}+\frac{sin^2(x)}{x^2})+(\frac{cos^2(x)}{x^2}-\frac{sin(x)cos(x)}{x^3})}dx$$

$$P_2(x)=\int \frac{sin(x)}{(\frac{sin(x)cos(x)}{x^3}+\frac{sin^2(x)}{x^2})+(\frac{cos^2(x)}{x^2}-\frac{sin(x)cos(x)}{x^3})}dx$$

The full solution is:

$$ y_p=(-\int \frac{-cos(x)}{(\frac{sin(x)cos(x)}{x^3}+\frac{sin^2(x)}{x^2})+(\frac{cos^2(x)}{x^2}-\frac{sin(x)cos(x)}{x^3})}dx)(\frac{sin(x)}{x})+(\int \frac{sin(x)}{(\frac{sin(x)cos(x)}{x^3}+\frac{sin^2(x)}{x^2})+(\frac{cos^2(x)}{x^2}-\frac{sin(x)cos(x)}{x^3})}dx)(\frac{-cos(x)}{x})$$

Alternative Method:
This is the form of the equation we are using:

$$y''+a_1(x)y'+a_0(x)y=f(x)$$

$$y''+\frac{2}{x}y'+y=1$$

We know the value of $$u_1$$

$$y(x)=U(x)u_1(x)$$

We don't know the value of U(x)

$$U'(a_1u_1+2u_1')+U''u_1=f(x)$$

Missing dependent variable U, so use Reduction of Order Method 0

let Z(x) = U'(x)

$$Z(a_1u_1+2u_1')+Z'u_1=f(x)$$

Solve for Z(x):

$$Z(x)=\frac{}{}\int^xh(s)f(s)ds$$

$$h(x)=exp(\int^x a_0(s)ds)$$

$$h(s)=h(x)=exp(\int^x(a_1u_1+2u_1')dx)=exp(\int^x(\frac{2sin(x)}{x^2}+\frac{2cos(x)}{x}-\frac{2sin(x)}{x^2})dx$$

$$h(x)=\int^x(\frac{2cos(x)}{x})dx$$

$$h(s)=\int^x(\frac{2cos(s)}{s})ds$$

$$Z(x)=\int^x(\int^s(\frac{2cos(s)}{s})ds)dx$$

$$U(x)=\int^x(Z(s))ds$$

$$U(x)=\int^x(\int^s(\int^s(\frac{2cos(s)}{s})ds)ds)dx$$

The solution is:

$$y(x)=U(x)u_1$$

$$y(x)=(\int^x(\int^s(\int^s(\frac{2cos(s)}{s})ds)ds)dx)(\frac{sin(x)}{x})$$

Contributing Authors
--Egm6321.f09.team5.GV 20:24, 21 October 2009 (UTC)

--Egm6321.f09.team5.risher 20:55, 21 October 2009 (UTC)

--Egm6321.f09.team5.bear 20:55, 21 October 2009 (UTC)