User:Egm6321.f09.team5/HW3

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 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 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 Statement
Problem 4 From, 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: $$F=\frac{d\Phi_}{dx} =\Phi_{x}+\Phi_{y^{(0)}}y^{(1)}+...+\Phi_{y^{(n-1)}}y^{(n)} $$

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 Statement
Problem 5 From, 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: $$F=\frac{d\Phi_}{dx} =\Phi_{x}+\Phi_{y^{(0)}}y^{(1)}+...+\Phi_{y^{(n-1)}}y^{(n)} $$

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 Statement
Problem 6 From, 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.), Equations 4&5. 6.1b.) , 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 Statement
Problem 7 From, 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 Statement
Problem 8 From, plot the shape function $$N_{j+1}^{2}(x)$$.

Solution


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

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

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

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

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

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

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

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

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

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

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

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

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 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.

<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 03:51, 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:

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:

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

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

We've already stated that:

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

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:

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

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

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

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.

<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 04:17, 28 October 2009 (UTC)

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

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)