User:Egm6321.f09.Team1/HW3

 Nice organization in giving a title to each problem, making it easy to read the table of contents and to search for the desired problem (section). Please include your matlab code, as noted at the bottom of the page, then include a comment box with a green background color to note that you did so. You work could serve as a reference for other teams, so it would be nice to make it as complete as possible. After the above improvement, you can link the archived version of your improved HW3 report in the Report Table with the link name "Updated HW3" above your link to the previous archived HW3 (don't remove this link). Egm6321.f09 15:33, 15 October 2009 (UTC)

 It would be helpful for readers to provide a link to the lecture transparency where each problem was stated. See further comments below. Egm6321.f09 13:06, 13 October 2009 (UTC)

=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 $$\displaystyle (m,n) $$ such that Eq 1 is exact.

Solution
 One note about consistency: You define $$h(x,y)=x^my^n$$ and above and below you are using $$n$$ for the power of $$x$$ and $$m$$ for the power of $$y$$.--Egm6321.f09.TA 03:59, 15 October 2009 (UTC)

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 $$\displaystyle f $$ and $$\displaystyle g $$ 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 condition 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 $$\displaystyle f $$ function (Eq. 5) is not a function of $$\displaystyle p$$. Therefore all derivatives of the function $$\displaystyle f $$ wrt. $$\displaystyle p$$ are zero. In addition to this, $$\displaystyle g_p$$ is constant wrt. $$\displaystyle p$$ and hence $$\displaystyle g_{pp}$$ 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, $$\displaystyle m=0$$ 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 $$\displaystyle m=0$$, the function $$\displaystyle f$$ and $$\displaystyle g $$ 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 $$\displaystyle f $$ and $$\displaystyle g $$ 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 $$\displaystyle n$$ 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
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * }

The left hand side (LHS) of equation 14 must be zero, since there is nothing to balance a $$\displaystyle x^{-1.5+n}$$ on the right hand side (RHS). 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} $$ $$
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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 $$\displaystyle y(x)$$.

Solution
Rewriting Eq. 1 by dividing through by $$\displaystyle x$$


 * {| 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 $$\displaystyle exp(f(x))$$ 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 $$\displaystyle C$$ can be arbitrarily chosen, hence we choose $$C=0$$. Therefore plugging into $$\displaystyle exp(f(x))$$ 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 $$\displaystyle y$$ is solved for by dividing by $$\displaystyle exp(f(x))$$ and substituing for $$\displaystyle f(x) $$ and $$\displaystyle f(a)$$.
 * $$\displaystyle
 * $$\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 $$ $$ <div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Nice work - very clear solution. --Egm6321.f09.TA 23:43, 27 October 2009 (UTC)
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

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

Given
A class of exact Linear Second Order ODE with varying coefficients (L2_ODE_VC).

Find
Find the mathematical structure of $$\displaystyle \phi$$ that yields the above class (i.e. A way to obtain L2_ODEs_VC).

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)
 * }
 * }

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Actually, you would not need to use the second exactness condition to deduce that $$\displaystyle K (y) = {\rm constant}$$, as you did below, if you had noticed that $$\displaystyle [P_x (x) + C_y (x,y)] = Q (x)$$, thus $$\displaystyle C_y (x,y) = Q(x) - P_x (x) = T(x)$$, which is a function of $$\displaystyle x$$ only. Thus, $$\displaystyle K_y (y) = 0$$; it follows that $$\displaystyle K( \cdot )$$ must be a constant function. Egm6321.f09 13:06, 13 October 2009 (UTC)

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. 15)
 * }
 * {| 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. 16)
 * }
 * {| 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. 17)
 * }
 * }

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. 18)
 * }
 * }

We can write this


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

\frac{dK}{dy} - \frac{d^2K}{dy^2} \frac{dy}{dx} = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 19)
 * }
 * }


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

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


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

\left[1-\frac{d}{dx}\right]\frac{dK(y(x))}{dy} = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 21)
 * }
 * }

The above is true if $$K$$ is a constant. Hence


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

$$\displaystyle \phi(x,y,p) = P(x) p + T(x) y + K $$ $$
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 22)
 * }
 * }

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Good work. Could be shorter, but know how to use the single relation for exactness condition 2 involving $$\displaystyle f_i := \frac{\partial F}{\partial y^{(i)}}$$ to show $$\displaystyle K (y) = {\rm constant}$$. Egm6321.f09 13:06, 13 October 2009 (UTC)

=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 $$\displaystyle n=0$$ 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 $$\displaystyle n=1$$ 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)
 * }
 * }

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Good derivation -clean and precise. --Egm6321.f09.TA 12:05, 15 October 2009 (UTC)

=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 $$\displaystyle f_1$$

By substituting Eq. 7 into Eq. 11 using $$\displaystyle i=1 $$, 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 $$\displaystyle i=2 $$, 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 $$\displaystyle f_0$$


 * {| 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 $$\displaystyle F(x,y,p,q)$$ in terms of $$\displaystyle f(x,y,p)$$ and $$\displaystyle g(x,y,p)$$, 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 $$\displaystyle f_0$$, $$ \displaystyle f_1$$, and $$ \displaystyle f_2$$ in terms of $$\displaystyle f$$ and $$\displaystyle g$$
 * $$\displaystyle
 * $$\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 $$\displaystyle f_1$$ and $$\displaystyle f_2$$ with respect to $$\displaystyle x$$
 * $$\displaystyle
 * $$\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 $$\displaystyle q$$, 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} $$ $$ <div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Very nice solution. One thing: $$y''=q$$, so when you are taking partials,$$ \frac{\partial}{\partial x}q=0$$. The extra 2 terms in your expressions cancelled themselves out. You can think of $$q$$ as a variable which is independent of $$x $$. --Egm6321.f09.TA 01:13, 16 October 2009 (UTC)
 * <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)
 * }
 * }

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Avoid using the symbol $$\displaystyle n$$ in the integrating factor $$\displaystyle h(x,y) = x^m y^n$$ since $$\displaystyle n$$ may be confused with the order $$\displaystyle n$$ in the Legendre differential equation; use $$\displaystyle h(x,y) = x^a y^b$$ instead. Egm6321.f09 15:33, 15 October 2009 (UTC)

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 $$\displaystyle n $$ is a free variable, the function can be considered exact if Eq. 10 is solved for $$\displaystyle n $$. 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:

Eq 2 is an alternative exactness condition, so we can compute each term as follows:


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

f_0 = \frac{\partial F}{\partial y} = n (n + 1) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 13)
 * }
 * }


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

f_1 = \frac{\partial F}{\partial y'} = -2x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * }


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

f_2 = \frac{\partial F}{\partial y''} = 1-x^2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 15)
 * }
 * }

Now, we can take the total derivatives of $$\displaystyle f_1 $$, and $$\displaystyle f_2 $$


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

\frac{d f_1}{d x } = -2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 16)
 * }
 * }


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

\frac{d f_2}{d x } = -2x \quad \Rightarrow \quad \frac = - 2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 17)
 * }
 * }

Substituting Eqs 13, 16 and 17 into Eq 2 gives


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

n (n + 1)= 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 18)
 * }
 * }

The condition is not satisfied and therefore the equation is not exact. If $$\displaystyle n $$ is a free variable, the function can be considered exact if Eq 18 is solved for $$\displaystyle n $$. This results in


 * {| 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 as it was stated by using the 2 alternative relations of the second exactness condition in Part 1.
 * <p style="text-align:right;">$$\displaystyle (Eq. 19)
 * }
 * }

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Good work. For $$\displaystyle n = n_1 = 0$$ and $$\displaystyle n = n_2 = -1$$, you showed that the Legendre differential equation is exact; a first integral $$\displaystyle \phi$$ can be found by inspection, since the Legendre differential equation becomes $$\displaystyle F  = (1-x^2)y'' - 2xy' =  \frac{d}{dx} \left[ (1-x^2) y'  \right] =  0 $$, and thus $$\displaystyle \phi (x,y,y') = (1-x^2) y' = k = {\rm constant}$$. For $$\displaystyle n \ne n_1 = 0$$ and $$\displaystyle n \ne n_2 = -1$$, you showed that the Legendre differential equation is not exact; can you use the integrating factor method with the integrating factor $$\displaystyle h(x,y) = x^a y^b$$ to make the Legendre differential equation exact? Egm6321.f09 15:33, 15 October 2009 (UTC)

=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 $$\displaystyle \alpha, \beta=1$$, 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 $$\displaystyle v = 0$$, 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)
 * }
 * }

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Good explanation. --Egm6321.f09.TA 04:00, 16 October 2009 (UTC)

= Problem 8: Shape Functions =

Given
The shape functions $$N_{J}^1$$, $$N_{J+1}^1$$, and $$N_{J}^2$$ are shown below for $$\displaystyle J=0$$. 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 $$\displaystyle A^{-1}$$. 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 $$\displaystyle J=0$$. 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 $$\displaystyle J=0$$


 * {| 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



Matlab code:

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Nice work. Did more than expected. Egm6321.f09 13:06, 13 October 2009 (UTC)

= Problem 9: Transform of Variable to solve 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 $$\displaystyle x$$ 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
Solving for 3rd Derivative

To solve for the 3rd derivative wrt $$\displaystyle x$$, the derivative wrt $$\displaystyle x$$ of Eq 2 is taken


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

y_{xxx} = \frac{d}{dx} (y_{xx}) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


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

y_{xxx} = \frac{dt}{dx} \frac{d}{dt} (y_{xx}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

Substituting in Eq. 2 yields


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

\begin{align} y_{xxx} & = \frac{dt}{dx} \frac{d}{dt} \left[e^{-2t}(y_{tt} - y_t) \right] \\ & = e^{-t} \left[-2e^{-2t}(y_{tt} - y_t) + e^{-2t}(y_{ttt} - y_{tt}) \right] \\ & = e^{-3t} \left[-2y_{tt} + 2y_t + y_{ttt} - y_{tt} \right] \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Collecting terms yields Eq. 4


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

$$\displaystyle y_{xxx} = e^{-3t}(y_{ttt} - 3y_{tt} + 2y_t) $$ $$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

Solving for the 4th Derivative


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

\begin{align} y_{xxxx} & = \frac{d}{dx} (y_{xxx}) \\ & = \frac{dt}{dx} \frac{d}{dt} (y_{xxx}) \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Substituting Eq 7 into the previous Eq yields,


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

\begin{align} y_{xxxx} & = \frac{dt}{dx} \frac{d}{dt} \left[e^{-3t}(y_{ttt} - 3y_{tt} + 2y_t) \right] \\ & = e^{-t} \left[-3e^{-3t}(y_{ttt} - 3y_{tt} + 2y_t) + e^{-3t}(y_{tttt} - 3y_{ttt} + 2y_{tt}) \right] \\ & = e^{-4t} \left[-3y_{ttt} + 9y_{tt} - 6y_t + y_{tttt} - 3y_{ttt} + 2y_{tt} \right]  \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Collecting terms yields Eq. 4 giving the solution
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y_{xxxx} = e^{-4t}(y_{tttt} - 6y_{ttt} + 11y_{tt} - 6y_t) $$ $$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;">

very good; clear write up. Egm6321.f09.TA 03:16, 28 October 2009 (UTC)

= Problem 10: Method of Trial Solutions =

Given
A 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)
 * }
 * }

with given boundary conditions
 * {| style="width:100%" border="0" align="left"

y(x=1) = 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"

y(x=2) = 4. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

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

y=e^{rx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Find
Solve Eq. 1 directly by using by using method of trial solution. Compare with transform of variable solution given Eq. 5 and given BC conditions
 * {| style="width:100%" border="0" align="left"

y(x) = C_1x^{r_1} + C_2x^{r_2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

Solution
Using the trial solution, it and its corresponding derivatives are calculated
 * {| style="width:100%" border="0" align="left"

y = e^{rx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

y' = re^{rx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

y'' = r^2e^{rx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

Substuting $$\displaystyle y$$ and its derivatives into Eq. 1 yields
 * {| style="width:100%" border="0" align="left"

x^2(r^2e^{rx})-2xre^{rx}+2e^{rx}=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

Factoring out $$\displaystyle e^{rx}$$ yields Substituting $$\displaystyle y$$ and its derivatives into Eq. 1 yields
 * {| style="width:100%" border="0" align="left"

e^{rx}\left[ x^2r^2-2xr+2 \right]=0 $$ $$ Dividing yields
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

x^2r^2-2xr+2=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

This equation still cannot be solved so a new trial solution is guessed
 * {| style="width:100%" border="0" align="left"

y=x^r $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 12)
 * }
 * }

Computing the 1st and 2nd derivatives results in
 * {| style="width:100%" border="0" align="left"

y'=rx^{r-1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 13)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

y''=(r^2-r)x^{r-2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * }

Substiuting Eq. 12, 13, and 14 into Eq 1 yields
 * {| style="width:100%" border="0" align="left"

x^2(r^2-r)x^{r-2} - 2xrx^{r-1} + 2x^r = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 15)
 * }
 * }

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

e^r \left[ r^2-r - 2r + 2 \right] = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 16)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

r^2- 3r + 2 = 0 $$ $$ Solving for the roots(r) yields
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 17)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

r_1 = 1; r_2 = 2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 18)
 * }
 * }

It has been shown that the form of a homogeneous solution takes the form of
 * {| style="width:100%" border="0" align="left"

y(x) = \sum^{n_{roots}}_{i=1} C_ix^{r_i} $$ $$ Therefore for two roots, the solution takes the form
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 19)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y(x) = C_1x^{r_1}+C_2x^{r_2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 20)
 * }
 * }

Substituting Eq. 18
 * {| style="width:100%" border="0" align="left"

y(x) = C_1x^{1}+C_2x^{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 21)
 * }
 * }

Applying the boundary condition given by Eqs. 2 and 3 respectively yields
 * {| style="width:100%" border="0" align="left"

3= C_11^{1}+C_21^{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 22)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

4 = C_12^{1}+C_22^{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 23)
 * }
 * }

Using Linear algebra to solve Eq. 22 and Eq. 23 for $$\displaystyle C_1$$ and $$\displaystyle C_2$$ yields
 * {| style="width:100%" border="0" align="left"

C_1 = 4; C_2=-1 $$ $$ Therefore substiting the results from Eq. 24 into Eq. 21
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 24)
 * }
 * }
 * {| style="width:100%" border="0" align="left"


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

y(x) = 4x-x^2 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 25)
 * }
 * }

'''Solving Eq. 5 for $$\displaystyle C_1, C_2$$'''

Note that Eq. 5 is the same as Eq. 21, therefore applying the same BCs will result in the same answer as Eq. 25. But for diligence, we will continue through using Eq. 5 Substituting Eq. 2 and Eq. 3 into Eq. 5 results in
 * {| style="width:100%" border="0" align="left"

3 = C_1 + C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 26)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

4 = C_12^1 + C_22^2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 27)
 * }
 * }

Solving Eq. 26 and Eq. 27 results in
 * {| style="width:100%" border="0" align="left"

C_1 = 4; C_2=-1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 28)
 * }
 * }

Therefore substituting the results from Eq. 28 into Eq. 5
 * {| style="width:100%" border="0" align="left"


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

y(x) = 4x-x^2 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 29)
 * }
 * }

The solution is ploted below

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;">

very good. Egm6321.f09.TA 03:46, 28 October 2009 (UTC)

= Problem 11: Integrating Factor Method =

Given
A L1_ODE_VC is given by


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

u_1(x)Z'(x) + \left[ a_1(x)u_1(x) + 2u_1'(x)\right]Z(x) = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

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

Z:=U', \quad y(x) = U(x) u_1(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }

Find
Solve Eq. 1 using the integrating factor method to show that
 * {| style="width:100%" border="0" align="left"

Z(x)=\frac{c}{u_{1}^2}\exp\left(-\int^x a_1(\xi)d\xi\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

Solution
Divide equation 1 by $$\displaystyle u_1(x)$$


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

Z' + \underbrace{\left[ a_1 + 2\frac{u_1'}{u_1}\right]}_{f'(x)} Z = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Multiply by $$\displaystyle e^{f(x)}$$


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

e^{f(x)} Z' + e^{f(x)} f'(x) Z = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

Which can be rewritten as


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

\left[ e^{f(x)} Z \right]' = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

Integrating the above yields


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

e^{f(x)} Z = c $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

where $$\displaystyle c$$ is a constant. We can rewrite Eq. 7 as


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

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

Next, we integrate $$\displaystyle f'(x)$$


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

f(x) = \int^x f'(\xi) \, d\xi = \int^x a_1(\xi) + 2\frac{u_1(\xi)'}{u_1(\xi)} \, dx = 2 \ln u_1(x) + \int^x a_1(\xi) \, d\xi $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

and then


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

e^{f(x)} = \exp\left(2 \ln u_1(x) + \int^x a_1(\xi) \, d\xi\right) = u_1(x)^2 \exp\left(\int^x a_1(\xi) \, d\xi\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

we substitute this into equation 8


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

$$\displaystyle Z = \frac{c}{u_1^2}  \exp\left(-\int^x a_1(\xi) \, d\xi\right)
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |

$$ $$ <div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;">
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

very good. Egm6321.f09.TA 06:31, 28 October 2009 (UTC)

= Problem 12: Reduction of Order Method from given algebric operations =

Given
To solve the Homogeneous L2-ODE-VC,


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

y'' + a_1(x)y' + a_0(x)y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

which has one solution as


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

y = u_1(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }

where the complete solution is given by,
 * {| style="width:100%" border="0" align="left"

y = k_1u_1(x) + k_2u_2(x) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Find
Attempt to develop a reduction of order method using the following algebraic expressions,

Part 1: $$\displaystyle y(x)=U(x)\pm u_1 (x)$$ Part 2: $$y(x)=\frac{U(x)}{u_1 (x)}$$ Part 3: $$y(x)=\frac{u_1 (x)}{U(x)}$$

Solution
Part 1:

Using the first algebraic expression,


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

y(x)=U(x)\pm u_1 (x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

On differentiating the Equation (3), we get the following first derivative,


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

y'(x)=U'(x)\pm u_1'(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Second derivative,


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

y(x)=U(x)\pm u_1''(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

Using Equations (3), (4) and (5) in Equation (1),


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

y + a_1(x)y' + a_0(x)y = \left[U(x)\pm u_1''(x) \right] + \left[U'(x)\pm u_1'(x) \right] a_1(x) +  \left[U(x)\pm u_1(x) \right] a_0(x)  = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

\begin{align} \Rightarrow y + a_1(x)y' + a_0(x)y = & \left[U(x) + U'(x)a_1(x) + U(x)a_0(x)\right] \pm \underbrace{\left[u_1''(x) + u_1'(x)a_1(x) + u_1(x)a_0(x) \right]}_{=0\ as\ per\ Eq(1)} \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

\begin{align} \Rightarrow y + a_1(x)y' + a_0(x)y = & \left[U(x) + U'(x)a_1(x) + U(x)a_0(x)\right] = 0 \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

In the above equation, the dependent variable U(x) is present and not missing. The Reduction of Order Method is possible only when dependent variable is missing.


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




 * style="width:70%; padding:10px; border:2px solid #8888aa" |

So, the assumption that $$\displaystyle y(x)=U(x)\pm u_1 (x)$$ does not reduce the order of Eq. 1.
 * style= |
 * }
 * }

Part 2: 

Using the second algebraic expression,


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

y(x)=\frac{U(x)}{u_1 (x)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

On differentiating the Equation (6), we get the following first derivative,


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

y'=\frac{U'u_1 - Uu_1'}{u_1^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

Second derivative,


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

y =\frac{1}{u_1^2}(U'u_1' + Uu_1 - Uu_1'' - U'u_1') - \frac{u_1'}{u_1^3}(U'u_1 - Uu_1') $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

Using Equations (6), (7) and (8) in Equation (1),


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

y + a_1(x)y' + a_0(x)y = \frac{U}{u_1}a_0 + \left[\frac{U'}{u_1} - \frac{U'u_1'}{u_1^2}\right] a_1+ \left[\frac{U}{u_1} - \frac{Uu_1''}{u_1^2} - \frac{U'u_1'}{u_1^2} + \frac{Uu_1'^2}{u_1^3}\right]  = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

In the above equation, the dependent variable $$\displaystyle U(x)$$ is present and not missing. The Reduction of Order Method is possible only when dependent variable is missing.


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



So, assumption that $$\displaystyle y(x)=\frac{U(x)}{u_1 (x)}$$ does not reduce the order of Eq. 1.
 * style="width:70%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

Part 3: 

Using the third algebraic expression,


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

y(x)=\frac{u_1(x)}{U(x)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

It is obvious from the above Part 2 section, if we let $$\displaystyle U(x) = u_1(x)$$ and $$\displaystyle u1(x) = U(x)$$, then the same argument holds that Equation (9) cannot be used as a full homogeneous solution for Equation (1). Following is the derivation, anyways.

On differentiating the Equation (9), we get the following first derivative,


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

y'=\frac{u_1'U - u_1U'}{U^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

Second derivative,


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

y =\frac{1}{U^2}(u_1'U' + u_1U - u_1U'' - u_1'U') - \frac{U'}{U^3}(u_1'U - u_1U') $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

using Equations (9), (10) and (11) in Equation (1),


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

y + a_1(x)y' + a_0(x)y = \frac{u_1}{U}a_0 + \left[\frac{u_1'}{U} - \frac{u_1'U'}{U^2}\right] a_1+ \left[\frac{u_1}{U} - \frac{u_1U''}{U^2} - \frac{u_1'U'}{U^2} + \frac{u_1U'^2}{U^3}\right]  = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

In the above equation, the dependent variable $$\displaystyle U(x)$$ is present and not missing. The Reduction of Order Method is possible only when dependent variable is missing.


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



So, assumption that $$\displaystyle y(x)=\frac{u_1(x)}{U (x)}$$ does not reduce the order of Eq. 1.
 * style="width:70%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Nice work. Egm6321.f09 13:06, 13 October 2009 (UTC)

= Problem 13: Using given trial solutions to solve function =

Given
Given the following homegenious L2_ODE_VC
 * {| style="width:100%" border="0" align="left"

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

The two test trial solutions are
 * {| style="width:100%" border="0" align="left"

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

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

Find
Find $$\displaystyle u_1(x)$$ and $$\displaystyle u_2(x)$$ of Eq. 1 using two given trial solutions.

Compare the two solutions using boundary conditions y(0) = 1 and y(1) = 2 and compare to the solution by reduction of order method 2. Plot the solutions in Matlab.

Solution
Differentiate Eq 2


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

u_1 = a x^{b} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

u_1'= b a x^{b-1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

u_1''=(b^2-b)ax^{b-2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

Plug into Eqs. 4, 5, and 6 into Eq. 1


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

(1-x^2)(b^2-b)ax^{b-2} - 2x b a x^{b-1} + 2 a x^b = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

Divide by $$\displaystyle a x^b$$ and rearrange


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

(b^2-b)x^{-2}-(b^2-b) - 2 b + 2 = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

We note that the first term is 0 for $$\displaystyle b=0$$ and $$\displaystyle b=1$$. Since the first term is the only $$\displaystyle x$$-dependent term, it must be 0. We then note that only $$\displaystyle b=1$$ is valid for the entire LHS to be 0. Hence,


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


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

u_1 = a x $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

Next, we differentiate Eq. 3


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

u_2 = e^{rx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

u_2' = r e^{rx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

u_2'' = r^2 e^{rx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 12)
 * }
 * }

Substituting Eqs. 10, 11, and 12 into Eq. 1


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

(1 - x^2) r^2 e^{rx} - 2 x r e^{rx} + 2 e^{rx} = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 13)
 * }
 * }

Divide by $$\displaystyle e^{rx}$$


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

(1 - x^2) r^2 - 2 x r + 2 = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 14)
 * }
 * }

Since $$\displaystyle r$$ is not a function of $$\displaystyle x$$, this is a dead end.

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Actually, you may want to find the expressions for $$\displaystyle r$$ before declaring that "this is a dead end", since in some cases, one of the solution $$\displaystyle r$$ may be a constant, which is not a "dead end", even though the coefficients in the characteristic equation may be a function of $$\displaystyle x$$. See a future lecture and the related HW problems. Egm6321.f09 13:06, 13 October 2009 (UTC)

Let's instead assume that


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

y(x) = U(x) u_1(x), \quad Z(x)=U'(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 15)
 * }
 * }

Then, we can use Eq. 4 from Problem 11:


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

Z'+\left[a_1(x) + 2\frac{u_1'}{u_1}\right] Z = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 16)
 * }
 * }

where


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

a_1(x) = \frac{2 x}{x^2-1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 17)
 * }
 * }

Then, we can skip forward to Eq. 11 in Problem 11


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

Z = \frac{\bar c}{u_1^2} \exp \left( - \int^x a_1(\xi) d\xi \right) = \frac{\tilde c}{x^2} \exp \left( - \int^x \frac{2 \xi}{\xi^2 - 1} d\xi \right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 18)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

Z = \tilde c \frac{1}{x^2} \exp \left( - \left[ \ln(x^2 - 1) \right]^x \right) = c \frac{1}{x^2(x^2 - 1)} = U' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 19)
 * }
 * }

To simplify integration, we rewrite


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

U' = c \frac{1}{x^2(x^2 - 1)} = c\left[\frac{A}{x^2} + \frac{B}{x-1} + \frac{C}{x+1} \right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 20)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

1 = A(x^2 - 1) + B x^2(x + 1) + C x^2(x - 1) = x^3 (B + C) + x^2 (A + B + C) + (-A + B - C) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 21)
 * }
 * }


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

B + C = 0, \quad -A = 1, \quad A + B - C = 0 \quad \Rightarrow \quad A = -1, \,\, B=\frac{1}{2}, \,\, C=-\frac{1}{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 22)
 * }
 * }

So we can write Eq. 20


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

U' = c \frac{1}{x^2(x^2 - 1)} = c\left[- \frac{1}{x^2} + \frac{1}{2(x-1)} - \frac{1}{2(x+1)} \right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 23)
 * }
 * }

Now, $$U$$ can be integrated


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

U = \bar c_1\left[\frac{1}{x} + \frac{1}{2} \ln (x-1) - \frac{1}{2} \ln (x+1) + c_2 \right] = \bar c_1\left[\frac{1}{x} + \frac{1}{2} \ln \frac{x-1}{x+1} + c_2 \right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 24)
 * }
 * }

Then, since $$y(x) = u_1(x) U(x)$$


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

y = c_1\left[1 + \frac{x}{2} \ln \left( \frac{x-1}{x+1}\right) \right] + u_1(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 25)
 * }
 * }

Hence


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

$$\displaystyle u_2 = c_1\left[1 + \frac{x}{2} \ln \left( \frac{x-1}{x+1}\right) \right] $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 26)
 * }
 * }

Next, we should plot $$y$$ with the boundary conditions $$y(0) = 1$$ and $$y(1) = 2$$. The problem is $$u_2(1)$$ a singular point, hence the boundary conditions can not be met. In addition to this, $$u_2$$ is complex for $$x<1$$. Instead we use the boundary conditions $$y(4)=5$$, $$y(10) = 10$$. The analytical solution is compared to that from Matlab's ode45 solver:



Matlab code: problem13.m

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Nice work. You want to include your matlab code for the benefits of the readers. Egm6321.f09 13:06, 13 October 2009 (UTC)

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

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

Egm6321.f09.Team1.vasquez 22:43, 6 October 2009 (UTC)

Egm6321.f09.Team1.andy 23:10, 6 October 2009 (UTC)