User:Egm6321.f09.team5.GV/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 $$ P=y^{'} $$

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 :

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

$$ 2m-1 = 0 $$

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

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:

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

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

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

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:

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

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

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

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:

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

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

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

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.

Problem Statement
Problem 4 From ([[media:Egm6321.f09.p13-3.png|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 $$ f_0-\frac{df_1}{dx}=0\Leftrightarrow \Phi_{xy}=\Phi_{yx}$$.

Problem Statement
Problem 5 From ([[media:Egm6321.f09.p13-3.png|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.

Problem Statement
Problem 6 From ([[media:Egm6321.f09.p14-2.png|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.) ([[media:Egm6321.f09.p10-3.png|p.10-3]]), Equations 4&5. 6.1b.) ([[media:Egm6321.f09.p14-1.png|p.14-1]]), Equation 5. 6.2 If it is not exact, see whether it can be made exact using the integrating factor with $$h(x,y)=x^my^n$$.

Problem Statement
Problem 7 From ([[media:Egm6321.f09.p14-3.png|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.

Problem Statement
Problem 8 From ([[media:Egm6321.f09.p15-2.png|p.15-2]]), plot the shape function $$N_{j+1}^{2}(x)$$.

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

Problem Statement
Problem 10 From (| p.16-4) Solve equation 1 on p.16-1, $$ x^2y''-2xy'+2y=0 $$ using the method of trial solution $$ y=e^{rx}$$ 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.

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.

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.

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

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

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