User:Egm6321.f09.Team6/HW2

Problem 1
Given the Equation hxN-hyM+h(Nx-My)=0

Consider the case where hxN=0

Since N can not be 0, then h(y)

So you get -hyM+h(Nx-My)=0

Rearranging you are left with hy/h=(1/M)(Nx-My)=-g(y)  So what is the final form of $$ h(y)$$ ?--Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 2
Given the Equation y'+(1/x)y=x2

Solve for h(x), where h(x)=

$$\int$$a0dx

and a0=(1/x)

So h(x)=

$$\int$$1/x=elnx=x

Therefore h(x)=x

 This does not verify exactness. You were asked to find the solution $$y(x)$$. --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 3
Given the equation hy'+h'y=hb where

h=x and b=x2

The resulting equation is xy'+y=x3

which can also be written as d/dx(xy)=x3

One can now find y(x)=

(1/x)$$\int(x)$$3

=1/x(x4/4+C)

So y(x)=x3/4+C/x

 This is still problem 2 ..... --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

 What happend to the actual problem #3? --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 4:
Statement: 

Show the following is an exact N1-ODE:

$$(\frac{1}{3})x^3y^4y'+(5x^3+2)(\frac{1}{5})y^5=0 $$

Solution: 

First, we need to see if the ODE has the form

$$M+Ny'=0$$

It does, so the first condition for exactness is satisfied.

Next, we need to see if

$$M_y=N_x$$

We know that

$$M_y=a(x)c(y)$$ $$N_x=b(x)c(y)$$

From the problem statement, we can determine the following: $$a(x)=5x^3+2$$ $$bi(x)=(\frac{1}{3})x^3$$ $$b(x)=x^2$$ $$c(y)=y^4$$ $$ci(y)=(\frac{1}{5})y^5$$

Therefore:

$$M_y=5x^3y^4+2y^4$$ $$N_x=x^2y^4$$

These are not equal! Thus, we must use the Euler integrating factor method. This class of exact N1-ODE is related to case 1. That means $$h_y$$ is equal to 0. Thus: $$(\frac{1}{N})(N_x-M_y)=(\frac{1}{bi(x)c(y)})(b(x)c(y)-a(x)c(y))$$ $$(\frac{1}{N})(N_x-M_y)=(\frac{1}{bi(x)})(b(x)-a(x))$$ $$(\frac{1}{N})(N_x-M_y)=-f(x)$$

We now know that h(x,y) is only a function of x. Therefore: $$h(x)=exp\int{f(x)dx}$$

where: $$f(x)=(\frac{1}{bi(x)})(b(x)-a(x))$$

So: $$h(x)=exp\int{\frac{x^2-5x^3-2}{\frac{1}{3}x^3}}dx$$

$$h(x)=exp\frac{3(x^2ln(x)-5x^3+1)}{x^2}$$

$$h(x)=abs(x^3)exp(\frac{3}{x^2}-15x)$$

The initial equation is now exactly integratable by the Euler integrating factor method: $$hM+hNy'=0$$

 This is not the correct $$h(x)$$. You have an absolute value in the expression? Additionally you have not verified exactness. You must show $$\overline{N}_x=\overline{M}_y$$--Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 5:
Statement: 

Show that the second condition is satisfied for the following N2-ODE:

$$(xy)y''+x(y')^2+yy'=0$$

Solution: 

We know that in order for the second condition of exactness to be satisfied, the following two equations must hold true:

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

Solving for the partial differential equations gives: $$f_x=y$$ → $$f_{xx}=0$$ $$f_x=y$$ → $$f_{xy}=1$$ $$f_y=x$$ → $$f_{yy}=0$$ $$g_x=p^2$$ → $$g_{xp}=2p$$ $$g_y=p$$ → $$g_{yp}=1$$ $$g_y=p$$

When entered into the first equation, we get: $$0+2p(1)+p^2(0)=2p+p(1)-p$$ $$2p=2p$$

The first equation holds true! Now let's check the second equation. First, solving for the partial differential equations gives: $$f_x=y$$ → $$f_{xp}=0$$ $$f_y=x$$ → $$f_{yp}=0$$ $$f_y=x$$ $$g_p=2xp+y$$ → $$g_{pp}=2x$$

When entered into the second equation, we get: $$0+p(0)+2x=2x$$ $$2x=2x$$

The second equation also holds true!

Thus, $$(xy)y''+x(y')^2+yy'=0$$ satisfies the second condition of exactness for N2-ODEs.

 good work. --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 6
Problem Statement:

Show that the Second Condition of Exactness is satisfied for the following ordinary differential equation:

$$(xy)y''+x(y')^2+yy'=0$$ (6.1)

Second Condition of Exactness:

$$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$ (6.2)

$$f_{xp}+pf_{yp}+2f_y=g_{pp}$$ (6.3)

Solution:

First, we must solve for all of the partial differentials of Equation 6.1, which are given below.

$$f_x=y \rightarrow f_{xx}=0, f_{xy}=1, f_{xp}=0$$

$$f_y=x \rightarrow f_{yp}=0, f_{yy}=0$$

$$g_x=p^2 \rightarrow g_{xp}=2p$$

$$g_y=p \rightarrow g_{yp}=1$$

$$g_p=2xp+y \rightarrow g_{pp}=2x$$

By substituting all of the partial differentials of Equation 6.1 into Equations 6.2 and 6.3, we find that since the left and right halves of the equations are equal, the Second Condition of Exactness is satisfied for Equation 6.1. This substitution is shown below.

$$0+2p(1)+p^2(0)=2p+p(1)-p \rightarrow 2p=2p$$

$$0+p(0)+2(x)=2x \rightarrow 2x=2x$$

 This appears to be a second solution for problem #5. --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

 Where is the real problem #7 ? --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 7
Problem Statement:

Derive Equation 7.1 by differentiating Equation 7.2. Where 7.1 and 7.2 are given as:

$$f_{xp}+pf_{yp}+2f_y=g_{pp}$$ (7.1)

$$g(x,y,p):=\phi_x(x,y,p)+\phi_y(x,y,p)p$$ (7.2)

Solution:

First, we take the first derivative with respect to $$p$$ of Equation 7.2, leaving out the arguments for simplicity's sake.

$$\frac{d}{dp}[g=\phi_x+\phi_yp] \rightarrow g_p = \phi_{xp}+\phi_y(1)+\phi_{yp}$$

This results in the following relationship.

$$\rightarrow g_p = f_x +\phi_y +f_yp$$

Now, we take the second derivative of Equation 7.2 with respect to $$p$$.

$$\frac{d^2}{dp^2}[g_p = f_x +\phi_y +f_yp] \rightarrow g_{pp}=f_{xp}+\phi_{yp}+f_y(1)+f_{yp}p$$

$$\rightarrow g_{pp}=f_{xp}+f_y+f_y+f_{yp}p$$

This resulting relationship, Equation 7.3, is equivalent to Equation 7.1 and the derivation is complete.

$$\rightarrow g_{pp}=f_{xp}+f_y+f_{yp}p$$ (7.3)  Good work. --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 8
$$ \ F = 8x^{5}p \qquad \qquad G = 2x^{2}p + 20x^{4}(y^{'})^{2} + 4xy $$

Show that equations (4) & (5) on 10-2 for exactness are satisfied for the above equations?

Solution:

The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. (4) $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. (5) $$

The partial derivatives are identified as,

$$ F_{x}=40x^{4}p  \qquad $$

$$ F_{xx}=160x^{3}p \qquad $$

$$ F_{xy}=0 \qquad $$

$$ F_{xp}=40x^{4}  \qquad $$

$$ F_{y}=0 \qquad $$

$$ F_{yy}=0 \qquad $$

$$ F_{yp}=0 \qquad $$

$$ G_{x}= 4xp + 80x^{3}p^{2} + 4y \qquad $$

$$ G_{xp}= 4x + 160x^{3}p \qquad $$

$$ G_{y}= 4x \qquad $$

$$ G_{yp}= 0 \qquad $$

$$ G_{p}= 2x^{2} + 40x^{4}p \qquad $$

$$ G_{pp}= 40x^{4} \qquad $$

Substituting in Equ.(4), we get

$$ \ 160x^{3}p + 2p (0) + p^{2} (0) = 4x + 160x^{3}p + p (0) - 4x $$

$$ \ 160x^{3}p = 160x^{3}p $$

Eq. (4) is satisfied.

Applying the results in Eq. (5), we get

$$ \ 40x^{4} + p(0) + 2(0) = 40x^{4} $$

$$ \ 40x^{4} = 40x^{4} $$

Eq. (5) is also satisfied.

Since Equation (4) and Equation (5) are both satisfied, then the 2nd exactness condition is satisfied. It is then concluded that the NL.ODE is exact.

 good work. --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Problem 9
$$ \sqrt[]{x}y^{''} + 2xy^{'} + 3y = 0 $$

Verify exactness of above ODE.

Solution

Exactness Condition 1:

$$ \ F(x,y,p)y^{''} + G(x,y,p) = 0, \qquad where \quad p = y^{'} $$

$$ \ F(x,y,p) = \sqrt[]{x} \qquad \qquad G(x,y,p) = 2xp + 3y $$

The 2nd exactness condition for a second order ODE is,

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. (4) $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. (5) $$

The following partial derivatives are found:

$$ F_{x}= -\frac{1}{2}x^{\frac{-1}{2}}  \qquad $$

$$ F_{xx}= -\frac{1}{4}x^{\frac{-3}{2}}  \qquad $$

$$ F_{xy}= 0 \qquad $$

$$ F_{xp}= 0  \qquad $$

$$ F_{y}= 0 \qquad $$

$$ F_{yy}=0 \qquad $$

$$ F_{yp}=0 \qquad $$

$$  G_{x}= 2p  \qquad $$

$$ G_{xp}= 2 \qquad $$

$$ G_{y}= 3 \qquad $$

$$ G_{yp}= 0 \qquad $$

$$ G_{p}= 2x \qquad $$

$$ G_{pp}= 0 \qquad $$

Substituting these values in Eq. (4):

$$ \frac{-1}{4}x^{\frac{-3}{2}} + 2p(0) + p^{2}(0) = 2 + p(0) -3 $$

$$ \frac{-1}{4}x^{\frac{-3}{2}} = -1 $$

Equation (4) is not satisfied therefore the ODE is not exact.

But it does satisfy Eq.(5), which is shown below

$$ \ 0 = 0 + p(0) + 2(0) $$

$$ \ 0 = 0 $$

But in order for the ODE to be exact it must satisfy both Eq.(4) and Eq.(5), since it doesn't satisfy equ.(4) it is concluded that the ODE is not exact.  Correct. --Egm6321.f09.TA 04:46, 28 September 2009 (UTC)

Contributing Team Members
Egm6321.f09.Team6.PaulMoore 22:58, 22 September

Egm6321.f09.Team06.sada 04:59, 23 September 2009 (UTC)

Egm6321.f09.Team6.dianafoster 15:16, 23 September 2009 (UTC)