User:Egm6321.f09.Team6/HW6

=Problems 1-4=

=Problems 5-7=

=Problems 8=

The Laplace Equation in circular cylinder coordinates $$\displaystyle \Delta \psi = 0 $$

Obtain the separated equations for Laplace equation on circular cylinder coordinates, and identify the Bessel differential equation

$$\displaystyle x^2 y'' + x y' - (x^2 - \nu^2) y = 0 $$ Eq. 1

Solution
Assume the total solution of the Laplace equation is given by

$$\displaystyle f(\xi_1, \xi_2, \xi_3) = X_1(\xi_1) X_2(\xi_2) X_3(\xi_3) $$

Using Laplace operator on the function $$\displaystyle f$$, we can write Laplace equation

$$\displaystyle \Delta f = \frac{1}{\xi_1} \frac{\partial}{\partial \xi_1} \left(\xi_1 \frac{\partial f}{\partial \xi_1} \right) + \frac{1}{\xi_1^2} \frac{\partial^2 f}{\partial \xi_2^2} + \frac{\partial^2 f}{\partial \xi_3^2} = 0 $$

$$\displaystyle X_2 X_3 \frac{1}{\xi_1} \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + X_1 X_3 \frac{1}{\xi_1^2} \frac{\partial^2 X_2}{\partial \xi_2^2} + X_1 X_2 \frac{\partial^2 X_3}{\partial \xi_3^2} = 0 $$

Divide by $$\displaystyle X_1 X_2 X_3$$ $$\displaystyle \frac{1}{\xi_1 X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} + \frac{1}{X_3} \frac{\partial^2 X_3}{\partial \xi_3^2} = 0 $$

The last term is only dependent on $$\displaystyle \xi_3$$, and is the only term dependent on $$\displaystyle \xi_3$$, hence it must be a constant. So we substitute $$\displaystyle \frac{1}{\xi_1 X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} - \beta^2 = 0 $$

where $$\displaystyle \beta^2 = -\frac{1}{X_3} \frac{\partial^2 X_3}{\partial \xi_3^2} $$

Now, multiply by $$\displaystyle \xi_1^2$$ $$\displaystyle \frac{\xi_1}{ X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \frac{1}{X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} - \xi_1^2 \beta^2 = 0 $$

Now, the second term is only dependent on $$\displaystyle \xi_2$$, and it is the only term dependent on $$\displaystyle \xi_2$$. It must also be a constant. We can substitute $$\displaystyle \nu^2 = \frac{1}{X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} $$

So that $$\displaystyle \frac{\xi_1}{ X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \nu^2 - \xi_1^2 \beta^2 = 0 $$

Expand the first term $$\displaystyle \frac{\xi_1^2}{ X_1 } \frac{\partial^2 X_1}{\partial \xi_1^2} + \frac{\xi_1}{ X_1 } \frac{\partial X_1}{\partial \xi_1} + \nu^2 - \xi_1^2 \beta^2 = 0 $$

Multiply by $$\displaystyle X_1$$ $$\displaystyle \xi_1^2 \frac{\partial^2 X_1}{\partial \xi_1^2} + \xi_1 \frac{\partial X_1}{\partial \xi_1} - \left( \xi_1^2 \beta^2 - \nu^2 \right) X_1 = 0 $$

Now, substitute $$\displaystyle x = \xi_1 \beta$$ $$\displaystyle x^2 \frac{\partial^2 X_1}{\partial x^2} + x \frac{\partial X_1}{\partial x} - \left( x^2 - \nu^2 \right) X_1 = 0 $$

Then, with $$\displaystyle X_1 = y(x)$$, we get $$\displaystyle x^2 \frac{\partial^2 y}{\partial x^2} + x \frac{\partial y}{\partial x} - \left( x^2 - \nu^2 \right) y = 0 $$

or

$$\displaystyle x^2 y'' + x y' - \left( x^2 - \nu^2 \right) y = 0 $$

which is the same as Eq. 1, i.e. the Bessel differential equation

=Problem 9 =

Problem Statement
Let $$\displaystyle f = \sum_i g_i $$

Show that,

a) if $$\displaystyle \{g_i\}$$ is odd, then $$\displaystyle f$$ is odd.

b) if $$\displaystyle \{g_i\}$$ is even, then $$\displaystyle f$$ is even.

Solution
The definition of an odd function is that $$\displaystyle f(x) = -f(-x)$$.

a) So, if every $$\displaystyle g_i$$ is odd, then

$$\displaystyle f(x) = \sum_i g_i(x) = \sum_i \left[-g_i(-x)\right] = -\sum_i g_i(-x) = -f(-x) $$

Thus the first case is proven.

The definition of an even function is that $$\displaystyle f(x) = f(-x)$$. So, if every $$\displaystyle g_i$$ is even, then

b) So, if every $$\displaystyle g_i$$ is even, then

$$\displaystyle f(x) = \sum_i g_i(x) = \sum_i g_i(-x) = f(-x) $$

And, thus the second case also hold true.

=Problem 10=

Problem Statement
Show that, for $$\displaystyle k = 0,1,2,...$$,

a) $$\displaystyle P_{2k}(x)$$ is even, and

b) $$\displaystyle P_{2k+1}(x)$$ is odd

Solution
We know that, the Legendre polynomials are given by

$$\displaystyle P_n(x) = \sum_{i=0}^{[n/2]} (-1)^i \frac{(2 n - 2 i)! x^{n - 2i}}{2^n i! (n-i)! (n-2i)!} $$  Eq.(1)

a) The definition of an even function is that $$\displaystyle f(x) = f(-x)$$. Substitute $$\displaystyle n=2k$$ in Eq.(1)

$$\displaystyle P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! x^{2k - 2i}}{2^{2k} i! (2k-1)! (2k-2i)!} $$

$$\displaystyle P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! \left[(x)^2\right]^{k - i}}{2^{2k} i! (2k-1)! (2k-2i)!} $$

Then, $$\displaystyle P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! \left[(-x)^2\right]^{k - i}}{2^{2k} i! (2k-1)! (2k-2i)!} = P_{2k}(-x) $$

Thus $$\displaystyle P_{2k}(x) $$ is even by definition.

b) The definition of an odd function is that $$\displaystyle f(x) = -f(-x)$$. Substitute $$\displaystyle n=2k+1$$ in Eq.(1)

$$\displaystyle P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! x^{2k + 1 - 2i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$

Now, the above equation can be written as

$$\displaystyle P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (x) \left[(x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$   Eq.(2)

$$\displaystyle P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (x) \left[(-x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$   Eq.(3)

Now Eq.(2) at $$\displaystyle -x$$ is

$$\displaystyle P_{2k+1}(-x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (-x) \left[(-x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$   Eq.(4)

Comparing Eqs. (3) and (4) shows that

$$\displaystyle P_{2k+1}(x) = -P_{2k+1}(-x) $$

Thus it is proven that $$\displaystyle P_{2k+1}(x) $$ is odd by definition.

The polynomial
 * {| style="width:100%" border="0" align="left"

q(x) = \sum_{i=0}^4 c_i x^i $$ $$ with the coefficients
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

c_0 = 3, \quad c_1 = 10, \quad c_2 = 15, \quad c_3 = -1, \quad c_4 = 5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

=Problem 11=

Problem Statement
The polynomial

$$\displaystyle q(x) = \sum_{i=0}^4 c_i x^i $$

with

$$\displaystyle c_0 = 3, \quad c_1 = 10, \quad c_2 = 15, \quad c_3 = -1, \quad c_4 = 5 $$

a) Find $$\displaystyle \{ a_i \}$$ S.T. $$\displaystyle q(x) = \sum_{i=0}^4 a_i P_i(x)$$, where $$\displaystyle P_i(x)$$ are Legendre polynomials.

b) Plot $$\displaystyle q = \sum_i c_i x^i = \sum_i a_i P_i(x) $$.

Solution
Part a: From Lecture 31-3 Eqs. 1,2,3,4&5


 * $$ \begin{align}

P_{0}\left(x\right) &= 1 \\ P_{1}\left(x\right) &= x \\ P_{2}\left(x\right) &= \frac{1}{2}\left(3x^{2}-1\right) \\ P_{3}\left(x\right) &= \frac{1}{2}\left(5x^{3}-3x\right) \\ P_{4}\left(x\right) &= \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} \end{align} $$


 * Plugging the Legendre polynomials into the expression containing $$\left\{a_{i}\right\}$$, then group the terms by power of $$x$$:


 * $$ \begin{align}

q\left(x\right) &= \sum_{i=0}^{4}\ a_{i} P_{i} \\ &= a_{0} + a_{1} x + a_{2} \frac{1}{2}\left(3x^{2}-1\right) + a_{3}\frac{1}{2}\left(5x^{3}-3x\right) + a_{4}\left(\frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8}\right) \\ &= \left(a_{0}-\frac{1}{2}a_{2}+\frac{3}{8}a_{4}\right) + x\left(a_{1}-3a_{3}\right) + x^{2}\left(\frac{3}{2}a_{2}-\frac{15}{4}a_{4}\right) + x^{3}\left(\frac{5}{2}a_{3}\right) + x^{4}\left(\frac{35}{8}a_{4}\right) \end{align} $$


 * Then, we can set each term of this equation equal to the corresponding term in the definition of $$q\left(x\right)$$,


 * $$ \begin{align}

\sum_{i=0}^{4}\ a_{i} &P_{i} = \sum_{i=0}^{4}\ c_{i} x^{i}& \\ \\ &\text{(1)  }a_{0} - \frac{1}{2}a_{2} + \frac{3}{8}a_{4} &= 3 \\ &\text{(2) }a_{1} - 3a_{3} &= 10 \\ &\text{(3) }\frac{3}{2}a_{2} - \frac{15}{4}a_{4} &= 15 \\ &\text{(4) }\frac{5}{2}a_{3} &= -1 \\ &\text{(5) }\frac{35}{8}a_{4} &= 5 \end{align} $$


 * The last two equations yield the values of $$a_{3}$$ and $$a_{4}$$ directly:


 * $$ \begin{align}

a_{3} &= \frac{-2}{5} \\ a_{4} &= \frac{8}{7} \end{align} $$


 * $$a_{2}$$ is then obtained from equation (3) by plugging in $$a_{4}$$:


 * $$ \begin{align}

\frac{3}{2}a_{2} &= 15 + \frac{15}{4}a_{4} \\ a_{2} &= 10 + \frac{10}{4}\cdot \frac{8}{7} \\ &= \frac{90}{7} \end{align} $$


 * Similarly, $$a_{1}$$ is obtained from equation (2) by plugging in $$a_{3}$$:


 * $$ \begin{align}

a_{1} &= 10 + 3a_{3} \\ &= 10 + \frac{-6}{5} \\ &= \frac{44}{5} \end{align} $$


 * Finally, $$a_{0}$$ is obtained by inserting $$a_{2}$$ and $$a_{4}$$ into equation (1):


 * $$ \begin{align}

a_{0} &= 3 + \frac{1}{2}a_{2} - \frac{3}{8}a_{4} \\ &= 3 + \frac{45}{7} - \frac{3}{7} \\ &= 3 + \frac{42}{7} \\ &= 9 \end{align} $$


 * In summary, the following expressions for $$\left\{a_{i}\right\}$$ were obtained:


 * $$ \begin{align}

a_{0} &= 9,

\quad a_{1} = \frac{44}{5}\\ \\ a_{2} &= \frac{90}{7},

\quad a_{3} = \frac{-2}{5}\\ \\ a_{4} &= \frac{8}{7} \end{align} $$ Part b:

=Problem 12=

Part 1
Problem Statement Without calculating, find the properties of $$A_n$$. Solution First, $$A_n$$ is defined by $$A_n=\frac{2n+1}{2}\int^{+1}_{-1}T_0(1-\mu^2)^2P_n(\mu)d\mu,   n=0,1,2,...$$ where $$\mu=sin(\theta), T_0=const., k=1,2,...$$ One of the properties of Legendre polynomials is that $$P_{2k}$$ is an even function and $$P_{2k}$$ is odd. Since $$(1-\mu^2)^2$$ is even and $$P_{2k}(\mu)$$ is also even, the product must therefore be even. It follows that the integral from -1 to 1 is equal to twice the integral from 0 to 1. $$A_{2k}$$ can therefore be written as $$A_{2k}=(4k+1)\int^{+1}_{0}T_0(1-\mu^2)^2P_{2k}(\mu)d\mu$$ $$A_{2k+1}$$ can be written by $$A_{2k+1}=(4k+1)\int^{1}_{-1}T_0(1-\mu^2)^2P_{2k+1}(\mu)d\mu$$ Since $$(1-\mu^2)^2$$ is even and $$P_{2k+1}(\mu)$$ is odd, the product must be even. The integral of an odd function from -1 to 1 is zero. Therefore, $$A_{2k+1}=0$$

Part 2
Problem Statement Compute three non-zero coefficients of $$A_n$$ analytically, using either $$\theta$$ or $$\mu$$ as integrating variables. Solution First, $$P_n(x)$$ is defined by $$P_n(x)=\sum_{r=0}^{m}(-1)^r\frac{(2n-2r)!x^{n-2r}}{2^nr!(n-r)!(n-2r)!}$$ Therefore, the following Legendre Polynomials are given $$P_0(x)=1$$ $$P_1(x)=x$$ $$P_2(x)=\frac{1}{2}(3x^2-1)$$ $$P_3(x)=\frac{1}{2}(5x^3-3x)$$ $$P_4(x)=\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8}$$ The first three non-zero coefficients are $$A_0$$, $$A_2$$ and $$A_4$$. First, we solve for $$A_0$$. $$A_0=\int^{1}_{0}T_0(1-\mu^2)^2d\mu=T_0\left[\mu-\frac{2}{3}\mu^3+\frac{1}{5}\mu^5\right]^{1}_{0}=\frac{8}{15}T_0$$ Next, $$A_2=5\int^{1}_{0}T_0(1-\mu^2)^2\left[\frac{1}{2}(3\mu^2-1)^2\right]d\mu=\frac{5}{2}T_0\left[\mu^9-\frac{24}{7}\mu^7+\frac{22}{5}\mu^5-\frac{8}{3}\mu^3+\mu\right]^{1}_{0}=-\frac{16}{21}T_0$$ Finally, $$A_4=9\int^{1}_{0}T_0(1-\mu^2)^2\left(\frac{35}{8}\mu^4-\frac{15}{4}\mu^2+\frac{3}{8}\right)d\mu=9T_0\left[\frac{35}{72}\mu^9-\frac{25}{14}\mu^7+\frac{19}{8}\mu^5-\frac{3}{2}\mu^3+\frac{3}{8}\mu\right]^{1}_{0}=\frac{8}{35}T_0$$ =Problem 13=

Problem Statement
Verify the values for $${x_i}$$, $${w_i}$$ and $$P_n(x)$$ in the table for Gauss-Legendre quadrature on the Gaussian Quadrature Wikipedia page.

Solution
For $$P_1=x$$, letting $$x=0$$ yields $$P_1=0$$. Similarly, if we substitute $$x=/pm\sqrt{\frac{1}{3}}$$ into the equation for $$P_2$$, we have $$P_2=\frac{1}{2}(3\frac{1}{3}-1)=0$$. We then perform the same substitution for the remainder of the Legendre Polynomials. $$P_3=\frac{1}{2}=(5(0)^3-3(0))=0$$ and $$P_3=\frac{1}{2}\left(5\left(\frac{3}{5}\right)^{3/2}-3\sqrt{\frac{3}{5}}\right)=0$$ $$P_4=\frac{35}{8}\left(\sqrt{(3+2\sqrt{6/5})/7}\right)^4-\frac{15}{4}\left(\sqrt{(3+2\sqrt{6/5})/7}\right)^2+\frac{3}{8}=0$$ Additionally, $$P_4=\frac{35}{8}\left(\sqrt{(3-2\sqrt{6/5})/7}\right)^4-\frac{15}{4}\left(\sqrt{(3-2\sqrt{6/5})/7}\right)^2+\frac{3}{8}=0$$ Finally, we substitute the roots for the 5th degree Legendre Polynomial, $$P_5$$. $$P_5=\frac{1}{8}\left(63\left(\frac{1}{3}\sqrt{5-2\sqrt{10/7}}\right)^5-70\left(\frac{1}{3}\sqrt{5-2\sqrt{10/7}}\right)^3+15\left(\frac{1}{3}\sqrt{5-2\sqrt{10/7}}\right)\right)=0$$ and $$P_5=\frac{1}{8}\left(63\left(\frac{1}{3}\sqrt{5+2\sqrt{10/7}}\right)^5-70\left(\frac{1}{3}\sqrt{5+2\sqrt{10/7}}\right)^3+15\left(\frac{1}{3}\sqrt{5+2\sqrt{10/7}}\right)\right)=0$$ Next, the weight $${w_i}$$ is given as $$ w_i = \frac{2}{\left( 1-x_i^2 \right) (P'_n(x_i))^2}$$ The first derivatives of the Legendre Polynomials are: $$P'_1=1$$ $$P'_2=3x$$ $$P'_3=\frac{1}{2}(15x^2-3)$$ $$P'_4=\frac{35}{2}x^3-\frac{15}{2}x$$ $$P'_5=\frac{1}{8}\left(\frac{63}{5}x^4-\frac{70}{3}x^2+15\right)$$ Substituting the roots and Legendre Polynomial derivatives into the weight equation yields the following results: $$w_1=2$$ $$w_2=1$$ $$w_{3,1}=\frac{8}{9}$$ $$w_{3,2}=\frac{5}{9}$$ $$w_{4,1}=\frac{8+\sqrt{30}}{36}$$ $$w_{4,2}=\frac{8-\sqrt{30}}{36}$$ $$w_{5,1}=\frac{128}{225}$$ $$w_{5,2}=\frac{322+13\sqrt{70}}{900}$$ $$w_{5,3}=\frac{322-13\sqrt{70}}{900}$$ =Problem 14=

Part 1
Problem From Lecture p.22-2, King p.28 problem 1.1.b: Verify the exactness of the following differential equation using the Integrating Factor Method $$xy''+2y'+xy=0$$ Solution In order to satisfy the first condition of exactness, the differential equation must be able to be expressed using the following functions $$F=f(x,y,p)y''+g(x,y,p)$$ The equation meets the first condition for exactness where $$f(x,y,p)=x$$ $$g(x,y,p)=2p+xy$$ Now, we check to see if the equation satisfies the second condition for exactness where $$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$ $$f_{xp}+pf_{yp}+2f_{y}=g_{pp}$$ Since $$f_{xx}=f_{xy}=f_{yy}=g_{xp}=g_{yp}=0$$ and $$g_y=x$$ the first equation is not satisfied and the differential equation is not exact.

Part 2
Problem Solve the differential equation using the following trial solutions:
 * 2.1) $$y(x)=e^{rx}$$
 * 2.2) $$y(x)=xe^{rx}$$
 * 2.3) $$y(x)=\frac{1}{x}e^{rx}$$

Solution 2.1) By substitution, we find that $$xr^2e^{rx}+2re^{rx}+xe^{rx}=0$$  dividing by $$e^{rx}$$, we have  $$xr^2+2r+x=0$$  Letting $$x=1$$ , we have  $$r^2+2r+x=0$$  Therefore,  $$r_{1,2}=1$$  and the particular solution is  $$y_p=c_1e^{-t}+c_2te^{-t}$$

2.2) By substitution, we have $$x(2re^{rx}+r^2xe^{rx})+2(e^{rx}+rxe^{rx})+x^2e^{rx}=0$$  Dividing by $$e^{rx}$$ and letting $$x=0$$  $$r^2+4r+3=0$$  This yields the following particular solution  $$y_p=c_1e^{-t}+c_2e^{-3t}$$

2.3) By substitution, we have $$x\left(\frac{2}{x^3}e^{rx}-\frac{2r}{x^2}e^{rx}+\frac{r^2}{x}e^{rx}\right)+r\left(-\frac{1}{x^2}e^{rx}+\frac{r}{x}e^{rx}\right)+x\left(\frac{1}{x}e^{rx}\right)=0$$  Dividing by $$e^{rx}$$ and letting $$x=0$$  $$r^2+1=0$$  This yields the following particular solution  $$y_p=c_1cos(t)+c_2sin(t)$$

Part 3
Problem Use the Undetermined Coefficient Method to find $$u_2(x)$$, knowing $$u_1(x)=\frac{sin(x)}{x}$$. Compare to above results.

Solution Part 1: Showing that $$u_1$$ is a solution to the differential equation.

$$y = x^{-1}sinx$$

$$\frac{dy}{dx}=\frac{d}{dx}\left(x^{-1}sinx\right)=-x^{-2}sinx+x^{-1}cosx$$

$$\frac{d^2y}{dx^2}=\frac{d^2}{dx^2}\left(x^{-1}sinx\right)=2x^{-3}sinx-2x^{-2}cosx-x^{-1}sinx$$

Substituting into the original equation, we have:

$$x\left(2x^{-3}sinx-2x^{-2}cosx-x^{-1}sinx\right)+2\left(-x^{-2}sinx+x^{-1}cosx\right)+x\left(x^{-1}sinx\right)=0$$

The left-hand side of the equation above reduces to zero, proving that $$u_1$$ is a solution to the differential equation.

Solution Part 2: Solving for $$u_2$$. We will look for a solution of the form $$y(x)=U(x)u_1(x)$$. First, we represent the first and second derivative as follows:

$$\frac{dy}{dx}=\frac{dU}{dx}u_1+U\frac{du_1}{dx}$$

$$\frac{d^2y}{dx^2}=\frac{d^2U}{dx^2}u_1+2\frac{dU}{dx}\frac{du_1}{dx}+U\frac{d^2u_1}{dx^2}$$

Next, we substitute the above equations into the differential equation.

$$x\left(\frac{d^2U}{dx^2}u_1+2\frac{dU}{dx}\frac{du_1}{dx}+U\frac{d^2u_1}{dx^2}\right)+2\left(\frac{dU}{dx}u_1+U\frac{du_1}{dx}\right)+x\left(U(x)u_1(x)\right)=0$$

We can now collect the terms.

$$U\left(\frac{d^2U}{dx^2}+2\frac{du_1}{dx}+xu_1\right)+xu_1\frac{d^2U}{dx^2}+\frac{dU}{dx}\left(2x\frac{du_1}{dx}+2u_1\right)=0$$

Since $$u_1$$ is a solution of the differential equation, the term multiplying by $$U$$ is zero. Next, we define $$Z:=\frac{dU}{dx},$$ and $$\frac{dZ}{dx}=\frac{d^2U}{dx^2}$$. By removing the $$u_1$$ terms, which are equal to zero and substituting the new definitions, we arrive at the following first-order, linear differential equation:

$$xu_1\frac{dZ}{dx}+Z\left(2x\frac{du_1}{dx}+2u_1\right)=0$$

Dividing by $$Zxu_1$$, we have

$$\frac{1}{Z}\frac{dZ}{dx}+\frac{2}{u_1}\frac{du_1}{dx}+\frac{2}{x}=0$$

By integration, we have

$$log|Z|+2log|u_1|+\int^x\frac{2}{s}ds=C$$

This can be used to solve for $$Z$$.

$$Z=\frac{k}{u_{1}^{2}}exp\left(-\int^x\frac{2}{s}ds\right)=\frac{dU}{dx}$$, where $$k=e^C$$

Integrating the above equation gives us an equation for $$U(x)$$.

$$U(x)=\int^x\frac{c}{u_{1}^{2}(t)}exp\left(-\int^{t}\frac{2}{s}ds\right)dt+\tilde{c}$$

This gives us an equation for $$y(x)$$.

$$y(x)=u_1(x)\int^x\frac{c}{u_{1}^{2}(t)}exp\left(-\int^{t}\frac{2}{s}ds\right)dt+\tilde{c}u_1(x)$$

Finally, we arrive at the equation for $$u_2(x)$$.

$$u_2(x)=u_1(x)\int^{x}\frac{1}{u_{1}^{2}(t)}exp\left(-\int^{t}\frac{2}{s}ds\right)dt$$

By solving the above equation, we arrive at the solution for $$u_2$$.

$$u_2(x)=x^{-1}cosx$$

=Contributing Members= Egm6321.f09.Team6.PaulMoore 10:21, 18 November 2009 (UTC)

Egm6321.f09.team6.breaux 17:19, 18 November 2009 (UTC)

Egm6321.f09.Team6.dianafoster 18:08, 18 November 2009 (UTC)

Egm6321.f09.Team06.sada 18:44, 18 November 2009 (UTC)