# User:Egm6321.f09.team5/HW3

## Problem 1

### Problem Statement

Find $(m,n)$ such that $(X^{m}Y^{n})\left[\sqrt[]{x}y^{''}+2xy^{'}+3y \right]=0$ is exact.

### Solution

The first step is to identify what kind of ODE is the given equation. It is identified as a Non-Linear Second-Order ODE (N2.ODE). In order for a second order ODE to be exact the following two conditions must be met:

Condition 1: The ODE must be of the form:

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


Where $P=y^{'}$

You are mixing upper and lowercase letters for $f,g,F,G$ in your partial expressions and exactness criteria, which is generally treacherous ground. Some of your expressions below mix upper/lowercase in the same equation. Recall that on p.13-2 we defined $P(x)= \Phi_x$ and $p=y'$. You want to be consistent in your notation. --Egm6321.f09.TA 04:24, 15 October 2009 (UTC)

Condition 2: The following Conditions must be met:

    $\ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad$      Equation 2


    $\ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad$         Equation 3


By arranging the given equation so it satisfies the first exactness condition (Equation 1) The Following is obtained

    $F= x^{m}y^{n}\sqrt[]{x}y^{''} + 2x^{m+1}y^{n}P +3x^{m}y^{n+1}$


Where

    $f(x,y,p) = x^{m}y^{n}\sqrt[]{x}$              $g(x,y,p) = 2x^{m+1}y^{n}P +3x^{m}y^{n+1}$


In order to meet the Second Condition for exactness the following partial derivatives must be identified:

$f_{x}=(m+0.5)x^{(m-0.5)}y^{n} \qquad \qquad g_{y}= 2Pnx^{(m+1)}y^{(n-1)} + 3(n+1)x^{m}y^{n}$

$f_{xx}=(m^{2}-0.25)x^{(m-1.5)}y^{n} \qquad \qquad g_{yp}= 2nx^{(m+1)}y^{n-1}$

$f_{xy}=(m+0.5)nx^{(m-0.5)}y^{(n-1)} \qquad \qquad g_{x}= 2P(m+1)x^{m}y^{m} + 3mx^{(m-1}y^{(n+1)}$

$f_{y}=nx^{(m+0.5)}y^{(n-1)} \qquad \qquad \qquad g_{xp}= 2(m+1)x^{m}y^{n}$

$f_{yy}=n(n-1)x^{(m+0.5)}y^{(n-2)} \qquad \qquad g_{p}= 2x^{(m+1)}y^{n}$

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

By populating equation 3, the following is obtained:

$0 + P(0) + 2nx^{(m+0.5)}y^{(n-1)} = 0 \qquad$

Therefore

     $n=0$


Using the value for n and populating Equation 2, the following is found:

$(m^{2}-0.25)x^{(m-1.5)} + 2P(0) + P^{2}(0) = 2(m+1)x^{m} + P(0) - 3x^{m}$

Rearranging: $(m^{2}- 0.25)x^{(m-1.5)} = x^{m}(2m-1)$

$x^{-1.5} = \frac{(2m-1)}{(m^{2}-0.25)}$

In order for this equation to be satisfied, both sides of the equation must be equal to 0 :

$\frac{(2m-1)}{(m^{2}-0.25)} = 0$

$\ 2m-1 = 0$

    $m = \frac{1}{2}$


The exact ODE is:

    $(X^{0.5}Y^{0})\left[\sqrt[]{x}y^{''}+2xy^{'}+3y \right]=0$


You can leave out $y^0$ terms in the expression for the exact ODE. --Egm6321.f09.TA 04:24, 15 October 2009 (UTC)

## Problem 2

### Problem Statement

$\Phi(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y+k_1=k_2$ where $k_1,k_2$ are constants.

Solve for $y(x)$ Hint: L1.ODE.VC (int. Factor Method)

### Solution

The First step is to test the equation for exactness:

The two conditions for exactness of a L1.ODE.VC are:

The ODE must be of the form:

   $M(x,y) + N(x,y)y^{'} = 0$


The Second Condition is:

    $M_{y}=N_{x}$


Applying our equation to the condition test we find that the equation is not exact as follows:

$M(x,y)= 2(x^{\frac{3}{2}} - 1)y + K1 - K2$

$N(x,y)= x$

$M_{y} = (2x^{\frac{3}{2}}-2)$

$N_{x} = 1$

In order to find y(x) we must then use Euler's Integration Factor Method as Follows:

An integration factor h is identified and used as follows:

    $hMdx + hNdy=0 \qquad$

$h_{x}N - h_{y}M + h(N_{x}-M_{y})=0 \qquad$  Equation 1


At this point we will consider the particular case when the integration factor is only a function of x.

$\ h(x,y)=h(x)$

With this definition Equation 1, becomes:

$\ h_{x}N + h(N_{x} - M_{y}) = 0$

Which can then be re-arranged as follows:

$\frac{h_{x}}{h} = \frac{-1}{N}(N_{x} - M_{y})$

Substituting the known values into this relation we obtain:

$\frac{(2-2x^{\frac{3}{2}})}{x} = f(x)$

Using this expression an integration factor is found as follows:

$h(x) = \exp \int_{}^{x}f(s)ds$

$h(x) = \exp (\int_{}^{x}2ds - \int_{}^{x}2s^{\frac{1}{2}}ds$

$h(x) = x^{2} - \exp(\frac{4x^{\frac{3}{2}}}{3})$

This integrating factor is not correct. Egm6321.f09.TA 01:27, 28 October 2009 (UTC)

Using the newly identified integration factor the following can be defined:

The Original Equation can be re-written as:

    $P + \color{blue}{\frac{2(x^{\frac{3}{2}} - 1)}{x}} y = \color{red}{K_{2} - K_{1}} \qquad$   $\color{blue}{Blue=a_{0}(x)} \qquad \color{red}{Red=b(x)}$


A solution for an equation in this form can be identified as follows:

    $y(x)= \frac{1}{h(x)} \int{}^{x} h(s) b(s) ds$


By populating the previous equation with our known values the following is obtained:

$y(x) = \frac{1}{x^{2}-\exp(\frac{4x^{\frac{3}{2}}}{3})} \int{}^{x} \left[ s^{2} - exp(\frac{4s^{\frac{3}{2}}}{3}) \right] (K_{2}-K_{1})ds$

    $y(x) = \frac{K_{2}-K_{1}}{x^{2}-\exp(\frac{4x^{\frac{3}{2}}}{3})} \left[ \frac{x^{3}}{3} - \int{}^{x}exp(\frac{4s^{\frac{3}{2}}}{3}) \right]$


Your $h(x)$ produces an incorrect result. If you don't verify exactness (as a double check on the integrating factor) then you have to be extra careful when using the result. Egm6321.f09.TA 01:27, 28 October 2009 (UTC)

## Problem 3

### Problem Statement

Find the mathematical structure of $\Phi$ that will yield L2.ODE.VC

### Solution

First it is necessary to define the first integral as follows:

$F(x,y,y^{'},y^{''}) = \frac{d \phi(x,y,p)}{dx}$

    $F(x,y,y^{'},y^{''}) = \phi_{x} + \phi_{y}P + \phi_{p}P^{'} \qquad \qquad \qquad Where \quad P=y^{'}$      Equation 1


Why did you change $p$ to $P$ ? Egm6321.f09.TA 02:07, 28 October 2009 (UTC)

The next step is to define a 2nd Order ODE as follows:

    $F = W(x)y^{''} + Q(x)y^{'} + R(x)y$  Equation 2


Comparing Equation 2 with Equation 1 the following is identified:

$\phi_{p}=W(x)$

$\phi_{y}=Q(x)$

$\phi_{x}=R(x)y(x)$

Why are you introducing $W(x)$ into your notation? Egm6321.f09.TA 02:07, 28 October 2009 (UTC)

The next step is to integrate as follows:

$\phi= \int{}^{}W(x)dp = W(x)P + C_{1}$

$\phi= \int{}^{}Q(x)dy = Q(x)y + C_{2}$

$\phi= \int{}^{}R(x)y(x)dx = R(x)y(x) + C_{3}$

Adding all the definition for Phi the following is obtained:

    $\phi= W(x)P + T(x)y + K \qquad \qquad \qquad Where \quad T(x) = \left[R(x)+Q(x)\right]y+ (C_{2}+C_{3}) \qquad and \quad K = C_{1}+C_{2}+C_{3} \qquad \qquad$  Equation 3


In order to generate additional exact L2.ODE.VC, the first integral, $\phi$, must be of the form outlined in Equation 3.

This does not quite match the expected form. Egm6321.f09.TA 02:07, 28 October 2009 (UTC)

## Problem 4

### Problem Statement

Problem 4 From (p.13-3), for the case $n=1$ (N1_ODE)

$F(x,y,y')=0=\frac{d\Phi}{dx}(x,y)$.

Show that $f_0-\frac{df_1}{dx}=0 \Leftrightarrow\Phi_{xy}=\Phi_{yx}$.

Hint: Use $f_1=\Phi_y$.

Specifically:
4.1) Find $f_0$ in terms of $\Phi$
4.2) Find $f_1$ in terms of $\Phi$($f_1=\Phi_y$)
4.3) Show that $f_0-\frac{df_1}{dx}=0\Leftrightarrow \Phi_{xy}=\Phi_{yx}$.

### Solution

Note: $F=\frac{d\Phi_{{(x,y^{(0)},...,y^{(n-1)})}}}{dx} =\Phi_{x}+\Phi_{y^{(0)}}y^{(1)}+...+\Phi_{y^{(n-1)}}y^{(n)}$

For n=1, $F= \Phi_{x}+\Phi_{y^{(0)}}y^{(1)}$

$f_{i}=\frac{dF}{dy^{(i)}}$

4.1)

$f_0=\frac{dF}{dy}=\frac{d(\Phi_{x}+\Phi_{y}y')}{dy}$

$f_0=\Phi_{xy}$


$f_0=\frac{\partial}{\partial y}\frac{d\phi}{dx}=\phi_{xy}+\phi_{yy}y'$. You are missing terms. The differential with respect to y should be a partial. --Egm6321.f09.TA 14:29, 15 October 2009 (UTC)

4.2}

$f_1=\frac{dF}{dy'}=\frac{d(\Phi_{x}+\Phi_{y}y')}{dy'}$

$f_1=\Phi_{y}$


4.3)

$f_0-\frac{df_1}{dx}=0$

$\Phi_{xy}-\frac{d\Phi_y}{dx}=0$

$\Phi_{xy}-\Phi_{yx} = 0$

$\frac{d\phi_y}{dx}=\phi_{yx}+\phi_{yy}y'$. --Egm6321.f09.TA 14:29, 15 October 2009 (UTC)

$\Phi_{xy} = \Phi_{yx}$


## Problem 5

### Problem Statement

Problem 5 From (p.13-3), for the case $n=2$ (N2_ODE) show:
5.1) Show $f_1=\frac{df_2}{dx}+\Phi_y$
5.2) Show $\frac{d}{dx}(\Phi_y)=f_0$
5.3) $f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$
5.4) Relate eqn. 5 to eqs. 4&5 from p.10-2.

### Solution

Note: $F=\frac{d\Phi_{{(x,y^{(0)},...,y^{(n-1)})}}}{dx} =\Phi_{x}+\Phi_{y^{(0)}}y^{(1)}+...+\Phi_{y^{(n-1)}}y^{(n)}$

For n=2, $F= \Phi_{x}+\Phi_{y^{(0)}}y^{(1)}+\Phi_{y^{(1)}}y^{(2)}$

$f_{i}=\frac{dF}{dy^{(i)}}$

$f_i=\frac{\partial F}{\partial y^(i)}$ -> partial wrt to y, not total derivative. --Egm6321.f09.TA 01:48, 16 October 2009 (UTC)

5.1)

$f_2=\frac{dF}{dy''}=\frac{d(\Phi_{x}+\Phi_{y}y'+\Phi_{y'}y''))}{dy''}=\Phi_{y'}$

$f_1=\frac{dF}{dy'}=\frac{d(\Phi_{x}+\Phi_{y}y'+\Phi_{y'}y''))}{dy'}=\frac{d(\Phi_{x}+\Phi_{y}y')}{dy'}+\frac{d(\Phi_{y'}y'')}{dy'}=\Phi_y+\frac{d(\frac{d}{dx}(\Phi_{y'}y'))}{dy'}=\Phi_y+\frac{d(\Phi_{y'})}{dx}$

$f_1=\frac{df_2}{dx}+\Phi_{y}$


This expression for $f_1$ is not correct. --Egm6321.f09.TA 01:48, 16 October 2009 (UTC)

5.2}

$f_0=\frac{dF}{dy}=\frac{d(\Phi_{x}+\Phi_{y}y'+\Phi_{y'}y'')}{dy}=\frac{d\Phi_y}{dx}+\frac{d(\Phi_yy')}{dy}+\frac{d(\Phi_yy'')}{dy'}=\frac{d\Phi_{y}}{dx}$

$f_0=\frac{d\Phi_{y}}{dx}$


Your partial differentiations are incorrect. The correct expression was obtained by complimentary errors. --Egm6321.f09.TA 01:48, 16 October 2009 (UTC)

5.3)

$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$

$\frac{d\Phi_{y}}{dx}-\frac{d(\frac{df_2}{dx}+\Phi_{y})}{dx}+\frac{d^2f_2}{dx^2}=0$

$\frac{d\Phi_{y}}{dx}-\frac{d^2\Phi_{y'}}{dx^2}-\frac{d\Phi_{y}}{dx}+\frac{d^2\Phi_{y'}}{dx^2}=0$

$0=0$


I am not sure how you arrived at this result. It is not clear what you have done. --Egm6321.f09.TA 01:48, 16 October 2009 (UTC)

5.4)

Equation (4) from 10-2:

$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$

Equation (5) from 10-2:

$f_{xp}+pf_{yp}+2f_y=g_{pp}$

$F(x,y,y',y'')=\Phi_{x}+\Phi_{y}y'+\Phi_{y'}y''$

$F(x,y,p,q)=\Phi_{x}+\Phi_{y}p+\Phi_{p}q=(\Phi_{p})q+(\Phi_x+\Phi_{y}p)=fq+g$

Find $f_0, f_1, f_2$ using $F(x,y,p,q)$

$f_0=\frac{d(fq+g)}{dy}=f_yq+g_y$

$f_1=\frac{d(fq+g)}{dp}=f_pq+g_p$

$f_2=\frac{d(fq+g)}{dq}=f$

Plug into $f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$

$(f_yq+g_y)-((f_{px}+f_{py}p+f_{pp}q)q+f_pq'+(g_{px}+g_{py}p+g_{pp}q))+\frac{d(f_x+f_yp+f_pq)}{dx}=$

$(f_yq+g_y)-((f_{px}+f_{py}p+f_{pp}q)q+f_pq'+(g_{px}+g_{py}p+g_{pp}q))+(f_{xx}+f_{xy}p+f_{xp}q)+((f_{yx}+f_{yy}p+f_{yp}q)p+f_yq)+((f_{px}+f_{py}p+f_{pp}q)q+f_pq')$

Which can be simplified down to:

$f_{xx}+2pf_{xy}+p^2f_{yy}+(f_{xp}+pf_{yp}+2f_y-g_{pp})q-(g_{xp}+pg_{yp}-g_y)=0$


The previous equation is valid if the following equations are satisfied:

$f_{xp}+pf_{yp}+2f_y=g_{pp}$

$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$


good work. --Egm6321.f09.TA 01:51, 16 October 2009 (UTC)

## Problem 6

### Problem Statement

Problem 6 From (p.14-2), for the Legendre differential equation $F=(1-x^2)y''-2xy'+n(n+1)y=0$,
6.1 Verify exactness of this equation using two methods:
6.1a.) (p.10-2), Equations 4&5.
6.1b.) (p.14-1), Equation 5.
6.2 If it is not exact, see whether it can be made exact using the integrating factor with $h(x,y)=x^my^n$.

### Solution

6.1a)

Equation (4) from 10-2:

$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$

Equation (5) from 10-2:

$f_{xp}+pf_{yp}+2f_y=g_{pp}$

The Legendre differential equation has the form:

$F=f(x,y,p)y''+g(x,y,p)$

so:

$f(x,y,p)=(1-x^2)$

and

$g(x,y,p)=(-2xy'+n(n+1)y)$

To use equation (4) and (5), we must find all the partial derivatives:

$f_x=-2x$

$f_{xx}=-2$

$f_{xy}=0$

$f_{xp}=0$

$f_y=0$

$f_{yy}=0$

$f_{yp}=0$

$g_x=-2p$

$g_{xp}=-2$

$g_y=n(n+1)$

$g_{yp}=0$

$g_p=-2x$

$g_{pp}=0$

Plugging these derivatives into equations (4) and (5) we get:

$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y \Rightarrow -2+0+0=-2+0-n(n+1)$


Which is not exact, (unless n equals 0 or -1).

$f_{xp}+pf_{yp}+2f_y=g_{pp} \Rightarrow 0-0+0=0$


Which is exact.

We can't show exactness because equation (4) fail to show exactness.

6.1b)

For an alternate method of finding exactness we can use equation (5) from page 14-1:

$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$

$f_0=\frac{dF}{dy}=n(n+1)$

$f_1=\frac{dF}{dy'}=-2x$

$f_2=\frac{dF}{dy''}=(1-x^2)$

Plugging these into equation (5) we get:

$n(n+1)-\frac{d(-2x)}{dx}+\frac{d^2(1-x^2)}{dx^2}=0$

$n(n+1)+2-2=0$

$n(n+1)=0$


The exactness condition still isn't satisfied, but it could be if n were to equal 0 or -1.

6.2)

Make $F = (1-x^2)y''-2xy'+n(n+1)y=0$ an exact equation by using an integrating factor of the form $x^my^n$

$x^my^n((1-x^2)y''-2xy'+n(n+1)y) = 0$

    $f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$           Equation 1

    $f_{xp}+pf_{yp}+2f_y=g_{pp}$                             Equation 2


Using Equations 1 and 2 we check the for exactness of the function F multiplied by the integrating factor. Equation 2 yields:

$0+0+2(nx^my^{n-1}-n^{m+2}y{n-1}=0$

Therefore n = 0.

Populating Equation 1 with n=0 yields.

$m(m-1)x^{m-2}-(m+2)(m+1)x^m=-2(m+1)x^m$

Which can be rearranged to produce:

$x^{-2}= \frac{m(m+1)}{m(m-1)}$

Both sides of the equation must equal zero, therefore

$m=-1$

The exact ODE is:

   $x^{-1}((1-x^2)y''-2xy'+n(n+1)y)=0$


## Problem 7

### Problem Statement

Problem 7 From (p.14-3), Show that equations 1 and 2, namely
7.1 $\forall u,v$ functions of $x$, $L(u+v)=L(u)+L(v)$. and
7.2 $\forall \lambda\in\mathbb{R},L(\lambda u)=\lambda L(u)\forall$ functions of $x$.
are equivalent to equation 3 on p.3-3.

### Solution

Use:

$L(u+v)=L(u)+L(v)$ and $L(\lambda u)=\lambda L(u)$

to show that:

$L(\alpha u+\beta v)=\alpha L(u)+\beta L(v)$

Using the additive property we can show that:

$L(\alpha u+\beta v)=L(\alpha u)+L(\beta v)$

If α and β are equal to 1, this would give us the additive equation:

$L(u+v)=L(u)+L(v)$

Using the multiplicative property we can state that:

$L(\alpha u)= \alpha L(u)$

and

$L(\beta v)= \beta L(v)$

If both α and β are equal to λ, the multiplicative property is apparent as both equations take the form of the multiplicative equation:

$L(\lambda u)=\lambda L(u)$

Putting all of this together, the additive and multiplicative properties show that:

$L(\alpha u+\beta v)=\alpha L(u)+\beta L(v)$


good explanation, although you do not need to choose values for $\alpha,\beta$. --Egm6321.f09.TA 04:05, 16 October 2009 (UTC)

## Problem 8

### Problem Statement

Problem 8 From (p.15-2), plot the shape function $N_{j+1}^{2}(x)$.

### Solution

correct. Egm6321.f09.TA 03:07, 28 October 2009 (UTC)

## Problem 9

### Problem Statement

Problem 9 From (p.16-2), show that
$y_{xxx}=e^{-3t}\left(y_{ttt}-3y_{tt}+2y_t\right)$
$y_{xxxx}=e^{-4t}\left(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t\right)$

### Solution

$\frac{d^3y}{dx^3}=(\frac{dt}{dx}(\frac{d}{dt}))(\frac{dt}{dx}(\frac{d}{dt}))(\frac{dt}{dx}(\frac{d}{dt})y)$

$\ y_{xxx} = (\frac{dt}{dx}(\frac{d}{dt}))(\frac{dt}{dx}(\frac{d}{dt}))e^{-t}y_t$

$\ = e^{-t}(\frac{d}{dt}(-e^{-2t}y_t+e^{-2t}y_{tt})$

$\ = e^{-t}(2e^{-2t}y_t-e^{-2t}-2e^{-2t}y_{tt}+e^{-2t}y_{ttt})$

$\ y_{xxx} = e^{-3t}(y_{ttt}-3y_{tt}+2y_t)$

$\frac{d^4y}{dx^4}=(\frac{dt}{dx}(\frac{d}{dt}))(\frac{dt}{dx}(\frac{d}{dt}))(\frac{dt}{dx}(\frac{d}{dt}))(\frac{dt}{dx}(\frac{d}{dt})y)$

$y_{xxxx} = (\frac{dt}{dx}(\frac{d}{dt})y_{xxx})$

$= (\frac{dt}{dx}(\frac{d}{dt})e^{-3t}(y_{ttt}-3y_{tt}+2y_t))$

$= (\frac{dt}{dx})(-3e^{-3t}(y_{ttt}-3y_{tt}+2y_t)+e^{-3t}(y_{tttt}-3y_{ttt}+2y_{tt})$

$y_{xxxx} = e^{-4t}(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)$

It is not clear from your expressions which terms your differential operators are acting upon. You need some parenthesis or brackets (or alternate expressions) to clarify this. Egm6321.f09.TA 03:26, 28 October 2009 (UTC)

## Problem 10

### Problem Statement

Solve:

    $x^2y''-2xy'+2y=0$  Equation 1


using the method of trial solution $y=e^{rx} \quad and \quad y=x^{r}$ directly for the boundary conditions $\left\{ \begin{array}{rl} y(1)=&3\\ y(2)=&4\\ \end{array}\right.$
Compare the solution with equation 10 on p.16-3. Use matlab to plot the solutions.

### Solution

Given that the solution to this ODE is of the form:

    $y(x) = C_{1}x^{r_{1}} + C_{2}x^{r_{2}}$  Equation 2


PART 1

Since the trial solution is:

$y=x^{r}$

The following can be determined:

$y^{'}=rx^{r-1}$

$y^{''}=r(r+1)x^{r-2}$

Using these results and using them in Equation 1, the following is obtained:

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

Simplify to find:

$r^{2} - 3r + 2= 0$

This previous result is the characteristic equation of the ODE. The next step is to find the roots "r".

$r_{1}= 2 \qquad r_{2}=1$

With these roots and comparing to Equation 2, the following is the particular solution:

$y(x) = C_{1}x^{2} + C_{2}x^{1}$

The constants (C1 and C2) can be then found using the known values of the function y(x):

since: $y(1)= 3 \quad and \quad y(2)=4$

$y(1)= 3 = C_{1} + C_{2}$ Equation 3.

$y(2)= 4 = 4C_{1}+ 2C_{2}$ Equation 4.

Solving Equations 3 and 4 simultaneously it yields:

$C_{1} = -1 \qquad and \qquad C_{2} = 4$

Then the solution is of the form:

    $y(x)= -x^{2} + 4x$    Equation 5


The Plot of the Solution is shown in Figure 1.

Description Graph of the Solution using Matlab 10/7/09 G.V G.V

very good. Egm6321.f09.TA 03:51, 28 October 2009 (UTC)

PART 2

Since the trial solution is:

$y= \exp^{rx}$

The following can be determined:

$y^{'}=re^{rx}$

$y^{''}=r^{2}e^{rx}$

Using these results and using them in Equation 1, the following is obtained:

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

Simplify to find:

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

The previous equation cannot be solved without additional information therefore it is assumed to be a bad guess for a solution.

By Using the values for the roots as found in the previous part, the Equation becomes of the form:

$y(x)= C_{1}\exp^{2x} + C_{2}\exp^{x}$

Using the known values for y(x) the constants are found as follows:

$C_{1}= -.1204 \qquad C_{2}= 1.4309$

It then follows that the solution is of the form:

    $Y(x)= -.1204\exp^{2x} + 1.4309\exp^{x}$


The plot of the solution is shown in Figure 2: File:Noow.jpg

Description Graph of the Solution using Matlab 10/7/09 G.V G.V

## Problem 11

### Problem Statement

Prolbem 11 From (p.17-4 ) obtain equation 2 from p.17-3

$Z(x)=\frac{c}{u_{1}^2}\exp\left(-\int^x a_1(s)ds\right)$

using the integrator factor method and equation (1) from page 17-3.

$u_1(x)Z'+(a_1(x)u_1(x)+2u_1'(x))Z=0$

### Solution

$u_1(x)Z'+(a_1(x)u_1(x)+2u_1'(x))Z=0$

Divide by u_1(x) to get:

$Z'+(a_1(x)+\frac{2u_1'(x)}{u_1{x}})Z=0$

Multiply by an integrating factor h(x):

$h(x)Z'+h(x)(a_1(x)+\frac{2u_1'(x)}{u_1{x}})Z=0$

$(a_1(x)+\frac{2u_1'(x)}{u_1{x}})=-\frac{Z'}{Z}$

We can say that:

$\frac{d(h(x)Z(x))}{dx}=h(x)Z'(x)+h'(x)Z(x)=0$

thus

$\frac{h'(x)}{h(x)}=-\frac{Z'(x)}{Z(x)}$

After some algebraic manipulation of equation (1) multiplied by h(x):

$(a_1(x)+\frac{2u_1'(x)}{u_1{x}})=-\frac{Z'}{Z}$

$(a_1(x)+\frac{2u_1'(x)}{u_1{x}})=\frac{h(x)'}{h(x)}$

Integrating both sides we get:

$\int(a_1(x)+\frac{2u_1'(x)}{u_1{x}})=\int\frac{h(x)'}{h(x)}$

$\int^x(a_1(s)ds)+ln(u_1)^2+d=1+ln(h(x))+e$

We can combine the constants d, e, and 1 into another constant b:

$\int^x(a_1(s)ds)+ln(u_1)^2+b=ln(h(x))$

$h(x)=exp(\int^x(a_1(s)ds)+ln(u_1)^2+b)$

$h(x)Z'(x)+h(x)'Z(x)=0$

which is the same as:

$(h(x)Z(x))'=0$

Integrating both side we get:

$h(x)Z(x)=k$

Plugging in h(x) into this we get:

$exp(\int^x(a_1(s)ds)+ln(u_1)^2+b)Z(x)=k$

solving for Z(x) we get:

$Z(x)=\frac{k}{exp(\int^x(a_1(s)ds)+ln(u_1)^2+b)}=\frac{c}{exp(\int^x(a_1(s)ds)+ln(u_1)^2)}=\frac{c}{u_1^2exp(\int^x(a_1(s)ds)}$

Which can be expressed as:

$Z(x)=\frac{c}{u_{1}^2}\exp\left(-\int^x a_1(s)ds\right)$


very good. long but straight forward. Egm6321.f09.TA 04:08, 28 October 2009 (UTC)

## Problem 12

### Problem Statement

Develop reductions of order method 2 using different algebraic Operations:

1) $y(x) = U(x)\pm u_{1}(x)$ 2) $y(x) = U(x) / u_{1}(x)$ 3) $y(x) = u_{1}(x) / U(x)$

Objective: Proof you will not obtain a solution with a missing dependent variable.

### Solution

The form of a homogeneous L2.ODE.VC is as follows:

    $y^{''} + a_{1}y^{'} + a_{0}y = 0$   Equation 1.


1a)

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

$y^{'}= U^{'} + u_{1}^{'}$

$y^{''} = U^{''} + u_{1}^{''}$

Substituting into Equation 1

$0 = U^{''} + u_{1}^{''} + a_{1}(U^{'} + u_{1}^{'}) + a_{0}(U + u_{1})$

Rearranging yields:

$\color{blue}{U^{''} + a_{1}U^{'} + a_{0}(U)} + \color{red}{u_{1}^{''} + a_{1}u_{1}^{'} + a_{0}(u_{1})} =0$

Realizing that the Blue and the Red Part of the previous equations are homogeneous solutions we realize that we cannot reach a form of Equation 1, with a missing dependent Variable.

1b)

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

$y^{'}= U^{'} - u_{1}^{'}$

$y^{''} = U^{''} - u_{1}^{''}$

Substituting into Equation 1

$0 = U^{''} - u_{1}^{''} + a_{1}(U^{'} - u_{1}^{'}) + a_{0}(U - u_{1})$

Rearranging yields:

$\color{blue}{(U^{''} + a_{1}U^{'} + a_{0}(U))} - \color{red}{(u_{1}^{''} + a_{1}u_{1}^{'} + a_{0}(u_{1}))} =0$

Realizing that the Blue and the Red Part of the previous equations are homogeneous solutions we realize that we cannot reach a form of Equation 1, with a missing dependent Variable.

2)

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

$y^{'}= \frac{u_{1}U^{'} - Uu_{1}^{'}}{u_{1}^{2}}$

$y^{''} = \frac{u_{1}U^{''} - U^{'}u_{1}^{'}}{u_{1}^{2}} - \left[ U(\frac{u_{1}^2u_{1}^{''} - u_{1}^{'}(u_{1}^{2})^{'}}{u_{1}^{4}} + \frac {u_{1}U^{'}}{u_{1}^{2}} \right]$

Substituting into Equation 1:

$\frac{u_{1}U^{''} - U^{'}u_{1}^{'}}{u_{1}^{2}} - \left[ U(\frac{u_{1}^2u_{1}^{''} - u_{1}^{'}(u_{1}^{2})^{'}}{u_{1}^{4}} + \frac {u_{1}U^{'}}{u_{1}^{2}} \right] + \frac{a_{1}U^{'}}{u_{1}} - a_{1}\frac{Uu_{1}^{'}}{u_{1}^{2}} + a_{0} \left[ \frac{Uu_{1}^{'}}{u_{1}^{2}} - \frac{Uu_{1}^{'}}{u_{1}^{2}} \right]$

From the previous expression it can be seen that an expression with a missing dependent variable is not obtained.

3)

This expression is similar to part 2 and the conclusion is that an expression with a missing dependent variable cannot be obtained.

very good. Egm6321.f09.TA 04:17, 28 October 2009 (UTC)

## Problem 13

### Problem Statement

Find $u_{1}(x)$ and $u_{2}(x)$ of

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


using trial solutions:

1) $y=ax^{b}$ where a,b are coefficients to be determined.

2) $y=e^{rx}$

Compare the two solutions using the boundary conditions:

$y(0)=1$ $y(1)=2$

and also compare to the solutions by reduction of order method 2. Plot Solutions using Matlab

### Solution

A solution to the 2nd Order ODE would be of the following form:

    $y(x) = K_{1}U_{1}(x) + K_{2}U_{2}(x)$   Equation 2


PART 1

A trial solution will be used as follows:

$y = ax^{b}$

"a" and "b" are coefficients to be determined using initial conditions.

using the trial solution the following is identified:

$y^{'}(x) = abx^{b-1}$

$y^{''}(x) = ab(b-1)x^{b-2} = (ab^{2} - ab)x^{b-2}$

Replacing the newly defined equations into equation 1, the following is obtained:

$(1-x^{2}) \left[ (ab^{2} -ab)x^{b-2}) \right] -2x \left[abx^{b-1} \right] + 2 \left[ax^{b}\right] = 0$

Dividing by $x^{b}$:

$(1-x^{2})(ab^{2}-ab)x^{-2} - 2x \left[ abx^{-1} \right] + 2a = 0$

Rearranging:

$(1-x^{2})(a)(b^{2}-b)x^{-2} - 2xabx^{-1} + 2a = 0$

Dividing by "a":

$(x^{-2}-1)(b^{2}-b) - 2b + 2 = 0$

From here we can then identify the following:

$U_{1}(x) = (x^{-2}-1) \qquad \qquad K_{1}=(b^{2}-b)$

$U_{2}(x) = (1) \qquad \qquad \qquad K_{2}=(-2b+2)$

    $y(x)= (b^{2}-b)(x^{-2}-1) + (-2b+2)(1)$


The next step is to identify the coefficient "b". If for both solutions known the value is computed to be the same then the trial solution is determined to be a solution to the ODE.

Since

$y(0)=1$

Applying this to Equation 2 with our identified values:

$y(0) = 1 = (0^{-2} -1)(b^{2}-b) + (-2b +2)$

The coefficient b cannot be identified as the first part of the equation yields to an undefined solution.

When

$y(1) = 2$

$y(1) = 2 = (1^{-2} -1)(b^{2}-b) + (-2b +2)$

$2 = -2b + 2$

$b=0$

The Coefficient "b" is then identified. It is concluded that this trial solution is not a solution to the ODE.

PART 2

A trial solution will be used as follows:

$y = \exp^{rx}$

"r" is a coefficient which represents the roots of the ODE's Characteristic Equation.

using the trial solution the following is identified:

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

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

Replacing the newly defined equations into equation 1, the following is obtained:

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

Dividing by $\exp^{rx}$

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

Rearranging:

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

From here we can then identify the following:

$U_{1}(x) = (1-x^{2}) \qquad \qquad K_{1}=(r^{2})$

$U_{2}(x) = (xr+1) \qquad \qquad K_{2}=(-2)$

    $y(x) = (r^{2})(1-x^{2}) + 2 (xr+1)$


The next step is to identify the coefficient "r". If for both solutions known the value is computed to be the same then the trial solution is determined to be a solution to the ODE.

Since

$y(0)=1$

Applying this to Equation 2 with our identified values:

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

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

$r = \pm 3$

When

$y(1) = 2$

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

$2 = -2(r+1)$

$r= -2$

Since the solutions for "r" do not match then the Trial Solution is not a solution to the ODE.

You should be able to make one of these trials work. Egm6321.f09.TA 05:43, 28 October 2009 (UTC)

## Contributing Authors

--Egm6321.f09.team5.GV 04:40, 7 October 2009 (UTC)

--Egm6321.f09.team5.risher 16:39, 7 October 2009 (UTC)

--Egm6321.f09.team5.bear 18:15, 7 October 2009 (UTC)