User:Egm6321.f09.Team1.sallstrom/HW3

=Problem 1: Solving for H(x,y) to ensure Exactness=

Given
A function is given by
 * {| style="width:100%" border="0" align="left"

F = x^{m}y^{n} \lbrack \sqrt{x}y''+2xy'+3y \rbrack = 0 $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

h(x,y) = x^{m}y^{n} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

Find
Find (m,n) such that Eq 1 is exact.

Solution
The first criteria for exactness for a second order non-liner ODE is that it is of the form
 * {| style="width:100%" border="0" align="left"

F(x,y,p) = f(x,y,p)y'' + g(x,y,p) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

where


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

p = y' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Comparing Eq. 1 with the form of Eq. 3, the functions f and g can be defined as


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

f(x,y,p)=x^{0.5+n}y^{m} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

g(x,y,p)=2x^{n+1}y^{m}p+3x^{n}y^{m+1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

Therefore, the first conidition for exactness is met.

The second condition for exactness is given by the following equations
 * {| style="width:100%" border="0" align="left"

f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

f_{xp} + pf_{yp} + 2f_{y} = g_{pp} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * }

Analyzing Eq. 8 first, it should be noted that most of the terms drop out of the equation because the given f function (Eq. 5) is not a function of p. Therefore all derivatives of the function f wrt. p are zero. In addition to this, $$g_p$$ is constant wrt. $$p$$ and hence $$g_{pp}$$ is zero. Eq. 8 is simplified below


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

2f_y=g_{pp}=0 $$ $$ Taking the derivative of Eq. 5 and substituting into Eq. 9 yields
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

2\left[x^{0.5+n}my^{m-1}\right]=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 10)
 * }
 * }

Eq. 11 must be true for the exactness condition to be met. Therefore, m=0 satisfies this condition.
 * {| style="width:100%" border="0" align="left"


 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

m = 0 $$ $$
 * $$\displaystyle (Eq. 11)
 * }
 * }

Now, the second condition of exactness must be satisfied, specifially Eq. 7. Because m=0, the function f and g can be rewritten yielding
 * {| style="width:100%" border="0" align="left"

f=x^{0.5+n} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 12)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

g=2x^{n=1}p+3x^{n}y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 13)
 * }
 * }

The following derivatives of f and g are calculated in order to plug into Eq. 7.
 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} {f_{xx} = (n+0.5)(n-0.5)x^{-1.5+m}} \\ {f_{xy} = 0} \\ {f_{yy} = 0} \\ {g_{xp} = 2(n+1)x^{n}} \\ {g_{yp} = 0} \\ {g_y=3x^{n}} \\
 * $$\displaystyle
 * $$\displaystyle

\end{array} $$


 * }
 * }

Substiuting into Eq 7 and solving for n to ensure exactness yields
 * {| style="width:100%" border="0" align="left"

(n+0.5)(n-0.5)x^{-1.5+n} = 2(n+1)x^{n}-3x^{n} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 14)
 * }
 * }

The LHS of equation 14 must be zero, since there is nothing to balance a $$x^{-1.5+n}$$ on the right hands side. The LHS is zero if $$n= \pm 0.5$$. If the LHS is zero, so must the RHS. It is zero only if
 * {| style="width:100%" border="0" align="left"


 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

n = \frac{1}{2} $$ $$
 * $$\displaystyle (Eq. 15)
 * }
 * }

=Problem 2: Solving a L1-ODE=

Given
A function is given by
 * {| style="width:100%" border="0" align="left"

\phi(x,y,p) = xp+(2x^{\frac{3}{2}}-1)y - k_1 = k_2 $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

p = y' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }

Find
Solve Eq. 1 for y(x).

Solution
Rewriting Eq. 1 by dividing through by x


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

y'+(2x^{0.5}-x^{-1})y = \frac{k_1 + k_2}{x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

In order to solve this problem, it is desired to form the LHS of the equation in the form of the product rule. The procedure for this is to multiply the entire equation through exp(f(x)) resulting in


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

e^{f(x)}y'+(2x^{0.5}-x^{-1})e^{f(x)}y=\frac{k_1+k_2}{x}e^{f(x)} $$ $$ The LHS needs to be solved such that it is in the form
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

(e^{f(x)}y)'=e^{f(x)}y'+f'(x)e^{f(x)}y $$ $$ By comparison with the LHS Eq. 4, it can be shown that
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f(x)' = 2x^{0.5}-x^{-1} $$ $$ Integrating yields
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f(x) = \frac{4}{3}x^{1.5}-\ln(x) + C = \frac{4}{3}x^{1.5}-\ln(x) $$ $$ Where $$C$$ can be arbitrarily chosen, hence we choose $$C=0$$. Therefore plugging into exp(f(x)) results in
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

e^{f(x)} = \frac{1}{x}e^{1.33x^{1.5}} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

Going back to Eq. 5 and integrating both sides results in
 * {| style="width:100%" border="0" align="left"

e^{f(x)}y=(k_1+k_2)\int^x\frac{1}{x}e^{f(x)}dx $$ $$ A change of variable will now be used to reduce confusion within the integral
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

e^{f(x)}y=(k_1+k_2)\int^x\frac{1}{a}e^{f(a)}dx $$ $$ Finally, the function y is solved for by dividing by exp(f(x)) and substituing for f(x) and f(a).
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }
 * {| style="width:100%" border="0" align="left"


 * style="width:30%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

y=xe^{-1.33x^{1.5}}(k_1+k_2)\int^x\frac{1}{\xi} e^{\frac{4}{3}\xi^{1.5}}d\xi $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

=Problem 3: Determining Mathamatical Structure yield an exact L2_ODE_VC=

Given
The function is a class of exact Linear Second Order ODE with varying coefficents.

Find
In order to invent new exact L2-ODE_VC, find the mathamatical structure of phi that yields the above class.

Solution
The form of a L2_ODE_VC is given by

The form of a 2N-ODE is given by
 * {| style="width:100%" border="0" align="left"

F(x,y,p,y) = P(x) y + Q(x) p + R(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

where


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

p = y' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }

The first exactness condition for a L2_ODE_VC is


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

F(x,y,p,y) = \phi_x + \phi_y p + \phi_p y = \frac{d\phi(x,y,p)}{dx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

By comparing equation 3 and 5, we see that


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

\phi_p = P(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Integrating yields


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

\phi = P(x) p + C(x,y) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

Then


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

\phi_x = P'(x) p + \frac{\partial C(x,y)}{\partial x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }


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

\phi_y = \frac{\partial C(x,y)}{\partial y} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

Substituting these into Eq. 3 yields


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

F(x,y,p,y) = P(x) y + \underbrace{\left(P'(x) + \frac{\partial C}{\partial y}\right)}_{Q(x)} p + \underbrace{\frac{\partial C}{\partial x}}_{R(x) y} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

We notice by comparing Eq. 8 with Eq. 1 that


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

\frac{\partial C}{\partial x} = R(x) y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

Integrating the above yields


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

C = y\underbrace{\int_0^x R(\xi) d\xi}_{T(x)} + K(y) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

Substitute this into Eq. 5


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

\phi(x,y,p) = P(x) p + T(x) y + K(y) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

The second exactness condition is
 * {| style="width:100%" border="0" align="left"

f_0 - \frac{df_1}{dx} + \frac{d^2 f_2}{dx} = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 12)
 * }
 * }

where


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

f_i = \frac{\partial F}{\partial y^{i}} = \frac{\partial}{\partial y^{i}}\left(\frac{d F}{dx}\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 13)
 * }
 * }

We have


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

F=\frac{d\phi}{dx} = P(x) y'' + \left( P'(x) + T(x) + K'(y)\right)p + T'(x) y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * }

Then


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

f_0 = \frac{\partial F}{\partial y} = K'(y) + T'(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\frac{df_1}{dx} = \frac{d}{dx}\left(\frac{\partial F}{\partial p}\right) = \frac{d}{dx}\left( P'(x) + T(x) + K'(y)\right) = P(x) + T'(x) + K(y) p $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\frac{d^2 f_2}{dx^2} = \frac{d^2 }{dx^2} \left(\frac{\partial F}{\partial y}\right) = \frac{d^2 P(x)}{dx^2} = P(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * }

Putting these together in Eq. 12 yields
 * {| style="width:100%" border="0" align="left"

K'(y) + T'(x) - P(x) - T'(x) - K(y) p + P(x) = K'(y) - K(y) p = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * }

=Problem 4: Solving for N=1 form of Exactness Conditions=

Given
A non linear 1st Order ODE is defined by
 * {| style="width:100%" border="0" align="left""

F(x,y,y') = 0= \frac{d\phi}{dx}(x,y) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

Find
Show:
 * {| style="width:100%" border="0" align="left""

1) f_o ( \phi) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left""
 * {| style="width:100%" border="0" align="left""

2) f_1 = \phi_y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * {| style="width:100%" border="0" align="left""
 * {| style="width:100%" border="0" align="left""

3) f_o - \frac{df_1}{dx} = 0 \Leftrightarrow \phi_{xy}=\phi_{yx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Solution
Starting with the general definition
 * {| style="width:100%" border="0" align="left"

f_n = \frac{\partial F(x,y,p)}{\partial y^{n}} $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

p=y' $$ $$ Solving Part 1:
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

Pluging n=0 into Eq. 5 yields
 * {| style="width:100%" border="0" align="left"

f_o=\frac{\partial F(x,y)}{\partial y} $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

F(x,y,p)=\frac{d \phi(x,y)}{dx} $$ $$ Substituting Eq. 8 into Eq. 7 results in
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f_o=\frac {\partial}{\partial y} \left[\frac{d \phi}{d x}\right]= \frac{\partial}{\partial y}\left[\frac{\partial \phi}{\partial x}+\frac{\partial \phi}{\partial y}\frac{d y}{d x}\right]=\phi_{xy} + \phi_{yy}\frac{d y}{d x} $$ $$ resulting in
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }
 * {| style="width:100%" border="0" align="left"


 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

f_o=\phi_{xy} + \phi_{yy}p $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

Solving Part 2:

Substituting n=1 into Eq. 5 yields
 * {| style="width:100%" border="0" align="left"

f_1 = \frac{\partial F}{\partial y^{(1)}} = \frac{\partial F}{\partial p} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

From Part 1, it can be seen that
 * {| style="width:100%" border="0" align="left"

F(x,y,p) = \frac{d\phi(x,y)}{dx} = \frac {\partial \phi}{\partial x} + \frac{\partial \phi}{\partial y}\frac{dy}{dx}=\phi_{x}+\phi_yp $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 12)
 * }
 * }

Subsituting Eq. 12 into Eq. 11 yields and differentiating
 * {| style="width:100%" border="0" align="left"


 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

f_1=\phi_y $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 13)
 * }
 * }

Solving Part 3: The following equation defines the 2nd Exactness condition
 * {| style="width:100%" border="0" align="left"

f_o- \frac {df_1}{dx}=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * }

Taking the derivative of the f_1 term yields
 * {| style="width:100%" border="0" align="left"

\frac{df_1}{dx} = \frac{\partial \phi_y}{\partial x} + \frac {\partial \phi_y}{\partial y}\frac {dy}{dx} = \phi_{yx} + \phi_{yy}p $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 15)
 * }
 * }

Substituting Eq. 10 and 15 into Eq. 14 yields
 * {| style="width:100%" border="0" align="left"

\phi_{xy} + \phi_{yy} p - \phi_{yx} - \phi_{yy} p=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 16)
 * }
 * }

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


 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

\phi_{xy} = \phi_{yx} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 17)
 * }
 * }

=Problem 5: Solving for N=2 form of Exactness Conditions=

Given
A non linear 2nd Order ODE is defined by
 * {| style="width:100%" border="0" align="left""

F(x,y,y',y'') = 0= \frac{d\phi}{dx}(x,y,y') $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

Find
Show:
 * {| style="width:100%" border="0" align="left""

1) f_1 = \frac{df_2}{dx} + \phi_y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left""
 * {| style="width:100%" border="0" align="left""

2) f_o = \frac{d\phi_y}{dx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * {| style="width:100%" border="0" align="left""
 * {| style="width:100%" border="0" align="left""

3) f_o - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2}= 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Secondly, relate Eq 4 to previously derived equations for exactness given by
 * {| style="width:100%" border="0" align="left"

f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

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

f_{xp} + pf_{yp} + 2f_{y} = g_{pp} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

Solution
From the basic definition of the function
 * {| style="width:100%" border="0" align="left""

F(x,y,p,q) = \frac {d}{dx}\phi(x,y,p) = \frac{\partial \phi}{\partial x}+\frac{\partial \phi}{\partial y}\frac{d y}{d x}+\frac{\partial \phi}{\partial p}\frac{d p}{d x} = \phi_x + \phi_y p + \phi_p q $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left""

p=y' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * {| style="width:100%" border="0" align="left""
 * {| style="width:100%" border="0" align="left""

q = y'' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

The second condition of exactness is given by
 * {| style="width:100%" border="0" align="left"

f_o - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2}-...+(-1)^n\frac{d^{n}}{dx^{n}}f_n=0 $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f_i :=\frac{\partial F}{\partial y^{i}} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

Part 1: Solving for $$f_1$$

By substituting Eq. 7 into Eq. 11 using i = 1, then differentiating yields
 * {| style="width:100%" border="0" align="left""

f_1 = \frac{\partial F}{\partial p} = \phi_{xp} + \phi_{yp} p + \phi_y + \phi_{pp} q $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 12)
 * }
 * }

The same procedure is follow substituting Eq. 7 into Eq. 11 using i = 2, then differentiating yields
 * {| style="width:100%" border="0" align="left""

f_2 = \frac{\partial F}{\partial q} = \phi_p $$ $$ Differentiating Eq. 13 yields
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 13)
 * }
 * }
 * {| style="width:100%" border="0" align="left""

\frac{d f_2}{dx} = \phi_{px} + \phi_{py} \frac{dy}{dx} + \phi_{pp} \frac{dp}{dx} = \phi_{px} + \phi_{py} p + \phi_{pp} q = f_1 - \phi_y $$ $$ which can be rewritten
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * }


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


 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

f_1 = \frac{d f_2}{dx} + \phi_y $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 15)
 * }
 * }

Part 2: Solving for $$f_0$$


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

f_0 = \frac{\partial }{\partial y} \left[\frac{d \phi}{d x} \right] = \frac{\partial }{\partial y} \left[\frac{\partial \phi}{\partial x} + \frac{\partial \phi}{\partial y} p + \frac{\partial \phi}{\partial p} q\right] = \frac{\partial^2 \phi}{\partial x \partial y} + \frac{\partial^2 \phi}{\partial y^2} p + \frac{\partial^2 \phi}{\partial p \partial y} q = \frac{\partial \phi_y}{\partial x} + \frac{\partial \phi_y}{\partial y} p + \frac{\partial \phi_y}{\partial p} q = \frac{d \phi_y}{dx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 16)
 * }
 * }

Or just
 * {| style="width:100%" border="0" align="left"


 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

f_0 = \frac{d \phi_y}{dx} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 17)
 * }
 * }

Part 3: Proving Exactness Condition 

Eq. 4 can be written
 * {| style="width:100%" border="0" align="left""

f_0 - \frac{d}{dx}\left[ f_1 - \frac{d f_2}{dx}\right] $$ $$ Using Eq. 2, we can rewrite this as
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 18)
 * }
 * }
 * {| style="width:100%" border="0" align="left""

f_0 - \frac{d}{dx}\left[\phi_y\right] $$ $$ Using Eq. 3, this becomes
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 19)
 * }
 * }
 * {| style="width:100%" border="0" align="left""

\frac{d\phi_y}{dx} - \frac{d}{dx}\left[ \phi_y\right] = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 20)
 * }
 * }

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


 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

0=0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 21)
 * }
 * }

'''Part 4: Relating Eq. 4 to Eq. 5 and 6'''

First, we can write $$F(x,y,p,q)$$ in terms of $$f(x,y,p)$$ and $$g(x,y,p)$$, by rewriting equation 7
 * {| style="width:100%" border="0" align="left""

F(x,y,p,q) = \underbrace{\phi_p}_{f(x,y,p)} q + \underbrace{\phi_x + p \phi_y}_{g(x,y,p)} = f y'' + g $$ $$ Next, we want to put $$f_0$$, $$f_1$$, and $$f_2$$ in terms of $$f$$ and $$g$$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 22)
 * }
 * }
 * {| style="width:100%" border="0" align="left""

f_0 = \frac{\partial F(x,y,p,q)}{\partial y} = f_y y'' + g_y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 23)
 * }
 * {| style="width:100%" border="0" align="left""
 * {| style="width:100%" border="0" align="left""

f_1 = \frac{\partial F(x,y,p,q)}{\partial p} = f_p y'' + g_p $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 24)
 * }
 * {| style="width:100%" border="0" align="left""
 * {| style="width:100%" border="0" align="left""

f_2 = \frac{\partial F(x,y,p,q)}{\partial q} = f $$ $$ Next, we need to differentiate $$f_1$$ and $$f_2$$ with respect to $$x$$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 25)
 * }
 * }
 * {| style="width:100%" border="0" align="left""

\frac{df_1}{dx} = \left( f_{px} + f_{py} p + f_{pp} q \right) q + f_{p} y''' + g_{px} + p g_{py} + q g_{pp} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 26)
 * }
 * {| style="width:100%" border="0" align="left""
 * {| style="width:100%" border="0" align="left""

\frac{df_2}{dx} = f_x + f_y p + f_p q $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 27)
 * }
 * {| style="width:100%" border="0" align="left""
 * {| style="width:100%" border="0" align="left""

\frac{d^2f_2}{dx^2} = f_{xx} + p f_{xy} + q f_{xp}+ \left(f_{yx} + f_{yy} p + f_{yp} q \right) p + \left(f_y + f_{px} + p f_{py} + q f_{pp} \right) q + f_p y''' $$ $$ Now, we can substitute Eqs. 23, 26, and 28 into Eq. 4, which yields
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 28)
 * }
 * }
 * {| style="width:100%" border="0" align="left""

f_y y'' + g_y - g_{px} - p g_{py} - q g_{pp} +f_{xx} + p f_{xy} + q f_{xp} + \left(f_{yx} + f_{yy} p + f_{yp} q \right) p + f_y q = 0 $$ $$ Rearranging the terms yields
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 29)
 * }
 * }
 * {| style="width:100%" border="0" align="left""

f_{xx} + 2 p f_{xy} + p^2 f_{yy} - g_{px} - p g_{py} + g_y + \left(f_{xp} + p f_{yp} + 2 f_y - g_{pp} \right)q = 0 $$ $$ For the above equation to be valid for arbitrary $$q$$, both the following equations must be satisfied for Eq 30 to be satisfied
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 30)
 * }
 * }


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


 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

f_{xx} + 2 p f_{xy} + p^2 f_{yy} = g_{px} + p g_{py} - g_y $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 31)
 * }
 * }


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


 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

f_{xp} + p f_{yp} + 2 f_y = g_{pp} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 32)
 * }
 * }

=Problem 6=

Given
The Legendre differential equation is given by
 * {| style="width:100%" border="0" align="left"

F = (1-x^2)y'' - 2xy' +n(n+1)y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

Find
First: Verify the exactness of this equation using the 2 exactness conditions given by:


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



1: f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)


 * }
 * }

and


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

2: f_{xp} - pf_{yp} + 2f_{y} = g_{pp} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

Second: Also veryify exactness by
 * {| style="width:100%" border="0" align="left"

f_0 - \frac {df_1}{dx} + \frac {d^2f_2}{dx^2}=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

If not exact, then see wether it can be made exact by
 * {| style="width:100%" border="0" align="left"

h(x,y) = x^{m}y^{n}. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

Solution
Part 1: By definition, the function must have the form


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

F = f(x,y,p)y'' + g(x,y,p) $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

p = y'. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

Therefore, it is seen that
 * {| style="width:100%" border="0" align="left"

f(x,y,p) = 1-x^2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

g(x,y,p) = -2xp+n(n+1)y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

Taking the partial derivatives yields
 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} {f_x = -2x} & {f_{xx}  = -2} & {g_x  = -2p} & {g_{xp}  = -2}  \\ {} & {f_{xy} = 0} & {g_y  = n(n+1)} & {g_{yp}  = 0}  \\ {} & {f_{xp} = 0 } & {g_p  = -2x} & {g_{pp}  = 0 }  \\ {f_y = 0} & {f_{yy}  = 0} & {} & {}  \\ {} & {f_{yp} = 0} & {} & {}  \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle

Subsituting into Eq 2 to verify exactness condition 1 yields
 * }
 * }


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


 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

-2 + 0 + 0 = -2 + 0 - n(n+1) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

Subsituting into Eq 3 to verify exactness condition 2 yields
 * {| style="width:100%" border="0" align="left"


 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

0 + 0 + 0 = 0 $$ $$ Currently, Eq 10 and 11 prove condition 2 is satisfied but condition 1 is not and therefore the equation is not exact. If n is a free variable, the function can be considered exact if Eq. 10 is solved for n. This results in
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }


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


 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

n_1 = 0, \quad n_2 = -1 $$ $$ and an exact function.
 * <p style="text-align:right;">$$\displaystyle (Eq. 12)
 * }
 * }

Part 2:

=Problem 7 - Linearity Proof=

Given
The two properies of linearity conisist of an additive and multiplicative property shown in the following equations respectively:


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

L(u+v) = L(u) + L(v) $$ $$ and
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }


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

L(\lambda u) = \lambda L(u) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }

Find
Show that these equations can be combined to form
 * {| style="width:100%" border="0" align="left"

L(\alpha u+ \beta v) = \alpha L(u) + \beta L(v) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

Solution
Using Eq. 1, we can show that The first step is to define two new relationships using Eq. 2 with the constants alpha and beta shown below.


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

L(\alpha u + \beta v) = L(\alpha u) + L(\beta v) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Then, using Eq. 2 and Eq. 4, we can shot that
 * {| style="width:100%" border="0" align="left"

L(\alpha u) + L(\beta v) = \alpha L(u) + \beta L(v) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

Combining Eq. 4 and 5 yields:
 * {| style="width:100%" border="0" align="left"

L(\alpha u + \beta v) = \alpha L(u) + \beta L(v) $$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

Hence


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


 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

\left.\begin{matrix} L(u + v) = L(u) + L(v) \\ L(\alpha u) = \alpha L(u) \end{matrix}\right\} \Rightarrow L(\alpha u + \beta v) = \alpha L(u) + \beta L(v) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

Next, with $$\alpha, \beta=1$$, Eq. 3 becomes


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

L(u + v) = L(u) + L(v) $$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

Which means that


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


 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

L(\alpha u + \beta v) = \alpha L(u) + \beta L(v) \Rightarrow L(u + v) = L(u) + L(v) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

Next, with $$v = 0$$, Eq. 3 becomes


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

L(\alpha u) = \alpha L(u) $$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

Hence


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


 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

L(\alpha u + \beta v) = \alpha L(u) + \beta L(v) \Rightarrow L(\alpha u) = \alpha L(u) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

Combining Eqs. 7, 9, and 11 yields:


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


 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

\left.\begin{matrix} L(u + v) = L(u) + L(v) \\ L(\alpha u) = \alpha L(u) \end{matrix}\right\} \Leftrightarrow L(\alpha u + \beta v) = \alpha L(u) + \beta L(v) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 12)
 * }
 * }

= Problem 8: Shape Functions =

Given
The shape functions $$N_{J}^1$$, $$N_{J+1}^1$$, and $$N_{J}^2$$ are shown below for $$J=0$$. They are third order polynomials, and


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

\begin{matrix} N_{J}^1(J)                = 1 & N_{J}^2(J)                 = 0 & N_{J+1}^1(J)                 = 0 \\ \left[N_{J}^1(J)\right]'  = 0 & \left[N_{J}^2(J)\right]'   = 1 & \left[N_{J}^1(J)\right]'     = 0 \\ N_{J}^1(J+1)              = 0 & N_{J}^2(J+1)               = 0 & N_{J+1}^1(J+1)               = 1 \\ \left[N_{J}^1(J+1)\right]' = 0 & \left[N_{J}^2(J+1)\right]' = 0 & \left[N_{J+1}^1(J+1)\right]' = 0 \end{matrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

Find
Plot $$N^2_{J+1}(x)$$ for $$ x \in [J, J+1]$$

Solution
We have


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

N_{J}^1(x)     = C_{11} + C_{21} x + C_{31} x^2 + C_{41} x^3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

N_{J}^2(x)     = C_{12} + C_{22} x + C_{32} x^2 + C_{42} x^3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

N_{J+1}^1(x)   = C_{13} + C_{23} x + C_{33} x^2 + C_{43} x^3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

N_{J+1}^2(x)   = C_{14} + C_{24} x + C_{34} x^2 + C_{44} x^3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

Which must satisfy the conditions


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

\begin{matrix} N_{J}^1(J)                = 1 & N_{J}^2(J)                 = 0 & N_{J+1}^1(J)                 = 0 & N_{J+1}^2(J)                 = 0 \\ \left[N_{J}^1(J)\right]'  = 0 & \left[N_{J}^2(J)\right]'   = 1 & \left[N_{J}^1(J)\right]'     = 0 & \left[N_{J+1}^2(J)\right]'   = 0 \\ N_{J}^1(J+1)              = 0 & N_{J}^2(J+1)               = 0 & N_{J+1}^1(J+1)               = 1 & N_{J+1}^2(J+1)               = 0 \\ \left[N_{J}^1(J+1)\right]' = 0 & \left[N_{J}^2(J+1)\right]' = 0 & \left[N_{J+1}^1(J+1)\right]' = 0 & \left[N_{J+1}^2(J+1)\right]' = 1 \end{matrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

This can be put in the equation system


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

\underbrace{ \left[ \begin{matrix} 1 & J  &   J^2   &    J^3    \\ 0 & 1  &   2 J   &   3 J^2   \\ 1 & J+1 & (J+1)^2 & (J+1)^2  \\ 0 & 1  & 2 (J+1) & 3 (J+1)^2 \end{matrix} \right] }_A \underbrace{ \left[ \begin{matrix} C_{11} & C_{12} & C_{13} & C_{14} \\ C_{21} & C_{22} & C_{23} & C_{24} \\ C_{31} & C_{32} & C_{33} & C_{34} \\ C_{41} & C_{42} & C_{43} & C_{44} \end{matrix} \right] }_C = \underbrace{ \left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right] }_I $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

The coefficients can be solved for by multiplying both sides by $$A^{-1}$$. This yields


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

C = A^{-1} I = A^{-1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

Hence, the coefficients are


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

\underbrace{ \left[ \begin{matrix} C_{11} & C_{12} & C_{13} & C_{14} \\ C_{21} & C_{22} & C_{23} & C_{24} \\ C_{31} & C_{32} & C_{33} & C_{34} \\ C_{41} & C_{42} & C_{43} & C_{44} \end{matrix} \right] }_C = \underbrace{ \left[ \begin{matrix} -(2 J - 1) (J + 1)^2 &   -J (J + 1)^2 &  J^2 (2 J + 3) & -J^2 (J + 1) \\ 6 J (J + 1) & 3 J^2 + 4 J + 1 & -6  J (J + 1) &  J (3 J + 2) \\ - 6 J - 3 &      - 3 J - 2 &        6*J + 3 &    - 3 J - 1 \\ 2 &              1 &             -2 &            1 \end{matrix} \right] }_{A^{-1}} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

So, we can simplify equation 9 by assuming $$J=0$$. This only shifts the location of the functions along the x-axis, it does not change it's shape. The simplification yields the coefficients


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

C = \left[ \begin{matrix} 1 & 0 &  0 &  0 \\ 0 &  1 &  0 &  0 \\ -3 & -2 &  3 & -1 \\ 2 &  1 & -2 &  1 \end{matrix} \right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

Hence, with the simplification $$J=0$$


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

N_{J+1}^2(x) = -x^2 + x^3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

The plot of the above equation is shown below



= Problem 9: Finding solution to Homogeneous L2-ODE-VC =

Given
A Homogeneous L2-ODE-VC is given by
 * {| style="width:100%" border="0" align="left"

x^2y_{xx} - 2xy_{x} + 2y = 0. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

The second derivative wrt x has been found to be
 * {| style="width:100%" border="0" align="left"

y_{xx} = e^{-2t}(y_{tt} - y_t) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }

It should be noted that in order to solve for the second derivative, a transform of variable was used as shown in the following equation
 * {| style="width:100%" border="0" align="left"

x=e^{t} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

Find
Determine the 3rd and 4th derivatives wrt x shown in the following equations:


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

y_{xxx} = e^{-3t}(y_{ttt} - 3y_{tt} + 2y_t) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }


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

y_{xxxx} = e^{-4t}(y_{tttt} - 6y_{ttt} + 11y_{tt} - 6y_t) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

Solution
=Contributing Team Members= Egm6321.f09.Team1.AH 19:43, 4 October 2009 (UTC)

Egm6321.f09.Team1.sallstrom 21:34, 4 October 2009 (UTC)