User:Egm6321.f09.team5/HW2

 Insert 'EDIT' links for each problem so that I do not have to scroll through the entire assignment to provide feedback. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem Statement
Obtain the integrating factor, h, such that h = f(y).

Solution
If the condition for exactness is not satisfied: $$ \frac{\partial M}{\partial y}$$ does not equal to $$ \frac{\partial N}{\partial x} $$ The functions M(x,y) and N(x,y) may be multiplied by another function H(x,y) which is yet unknown. The result of the multiplication is as follows:

$$ \ H(x,y)[M(x,y)dx + N(x,y)dy] = 0 $$

$$ \ H(x,y)M(x,y) = M* $$

$$ \ H(x,y)N(x,y) = N* $$

$$ \ M*_{y}=h_{y}M+hM_{y} $$

$$ \ N*_{x}=h_{x}N+hN_{x} $$

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

If we set $$ \ h_{x}N=0$$, we are able to solve for H(x,y) as follows: $$ \ h_{y}M-h(N_{x}-M_{y})=0$$

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

If we integrate both side: $$ h(x)=exp\int_{}^{x}\frac{1}{M}(N_{x}-M_{y})$$  Good- concise. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem Statement
Verify that the integrating factor for the following equation is equal to x.

$$y' + \frac{1}{x}y = x^2$$

Solution
$$y' + \frac{1}{x}y = x^2$$

then cross x on the both sides of the equation we can get the below

$$ xy'+y=x*x^2$$-(1)

From the $$ hy'+h'y=hb $$ -(2)

compare the equation (1) and equation (2)

we can see that h(x)=x and b(x)=x^2

The integrating factor is calculated using the following equation. $$ h(x)=exp\int_{}^{x}a_0(s)ds$$

$$ h(x)=exp\int_{}^{x} \frac{1}{x}dx$$

$$ h(x)=exp\int_{}^{x} \frac{1}{x}$$

$$ h(x)=exp(ln(x))$$

Therefore, $$ h(x)=x$$  $$h(x)=x$$ was given in class. You were asked to show that the solution is $$y(x)=\frac{1}{4}x^3+\frac{1}{x}C$$. What is $$a_0$$ and $$b$$? You are using notation without stating where it comes from. It was used in class but without any explanation your solution has terms renamed and is not self contained. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem 3
This is problem 2 still

Problem Statement
Show that the solution of $$y' + \frac{1}{x}y = x^2$$ is $$y = \frac{x^3}{4} + \frac{C}{x}$$

Solution
$$y' + \frac{1}{x}y = x^2$$

$$y' +a_{0}y = b$$       (*from notes 8-2 eq. (1))

$$a_{0}(x) = \frac{1}{x}$$

$$b(x)=x^2$$

$$ y(x)=\frac{1}{h(x)}\int_{}^{x}h(x)b(x)dx$$      (*from notes 8-2 eq. (4))

$$ h(x)=exp\int_{}^{x}a_{0}(x)dx$$       (*from notes 8-1 referencing 6-3 eq. (3))

$$h(x)=exp\int_{}^{x}\frac{1}{x}dx=exp(log(x))=x$$

$$y(x)=\frac{1}{x}\int_{}^{x}(x)(x^2)dx=\frac{1}{x}(\frac{x^4}{4}+C)=\frac{x^3}{4}+\frac{C}{x}$$

So, $$y=\frac{x^3}{4}+\frac{C}{x}$$

Problem Statement
Show that $$\frac{1}{2}x^2y' + [x^4y+10]=0$$ is exact

Solution
To be exact, the problem must be in the form $$M(x,y) +N(x,y)y'=0$$ where $$M_{y}=N_{x}$$

For this case:

$$M_{y}=x^4$$

$$N_{x}=x$$

Since $$M_{y}$$ and $$N_{x}$$ do not equal, we do not have exactness. We can remedy this by using h(x,y):

$$h(x,y)[M(x,y) +N(x,y)y']=0$$

$$(hM) +(hN)y'=0$$ where $$hM=\overline{M}$$ and $$hN=\overline{N}$$

In order to be exact: $$\overline{M}_{y}=\overline{N}_{x}$$

Using $$h_{x}N-h_{y}M+h(N_{x}-M_{y})=0$$ (*from notes 6.3 eq. (1))

we can assume $$\overline{M}=(h_{y}M)=0$$ so that

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

We know the variables on the right hand side, and we can integrate to find our h(x) value:

$$\int_{}^{x}\frac{h_{x}}{h}dx=-\int_{}^{x}\frac{1}{\frac{1}{2}x^2}(x-x^4)dx$$

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

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

We multiply h(x) by the original equation to get exactness  You have not verified exactness. How do you know you calculated $$h$$ correctly? You need to verify $$\overline{M}_y=\overline{N}_x$$ --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem Statement
Show that $$(\frac{1}{3}x^3)(y^4)y' + (5x^3+2)(\frac{1}{5}y^5)=0$$ is an exact nonlinear 1st order ODE

$$a(x)=5x^3+2$$

$$b(x)=x^2$$

$$\overline{b}(x)=\frac{1}{3}x^3$$

$$c(y)=y^4$$

$$\overline{c}(y)=\frac{1}{5}y^5$$

Solution
To be exact, the problem must be in the form $$M(x,y) +N(x,y)y'=0$$ where $$M_{y}=N_{x}$$

For this problem:

$$M(x,y) +N(x,y)y'=\overline{b}(x)c(y)y'+a(x)\overline{c}(y)=(\frac{1}{3}x^3)(y^4)y' + (5x^3+2)(\frac{1}{5}y^5)=0$$, which is the form for an exact, nonlinear, 1st order ODE

$$M_{y}=5x^3y^4+2y^4$$

$$N_{x}=x^2y^4$$

Since $$M_{y}$$ and $$N_{x}$$ do not equal, we do not have exactness. We can remedy this by using h(x,y):

$$h(x,y)[M(x,y) +N(x,y)y']=0$$

$$(hM) +(hN)y'=0$$ where $$hM=\overline{M}$$ and $$hN=\overline{N}$$

In order to be exact: $$\overline{M}_{y}=\overline{N}_{x}$$

Using $$h_{x}N-h_{y}M+h(N_{x}-M_{y})=0$$ (*from notes 6.3 eq. (1))

we can assume $$\overline{M}=(h_{y}M)=0$$ so that

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

Since our N and M terms are known:

$$\int_{}^{x}\frac{h_{x}}{h}dx=-\int_{}^{x}\frac{1}{(\frac{1}{3}x^3)(y^4)}(x^2y^4-(5x^3y^4+2y^4))dx$$

$$log(h(x))=\frac{3}{x^2}-15x+3log(x)$$

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

This multiplied by our original equation with yield an exact, nonlinear, 1st order ODE  Your integral for $$h$$ has some errors in it. Consequently your $$h(x)$$ is in error. You need to verify exactness, not just calculate $$h$$, otherwise you haven't shown that the equation can be written in an exact form. You have to show $$N_x=M_y$$ --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem Statement
Consider:

$$ \ xyy^{''} + x(y^{'})^{2} + yy^{'} = 0 $$ Given that the 1st exactness condition is satisfied show that the 2nd exactness condition is satisfied

Solution
By the 1st exactness condition the equation can be written as follows:

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

$$ F(x,y,p)= xy $$

$$ G(x,y,p)= x(y^{'})^{2} + yy^{'} $$

It is then needed to identify the following partial derivatives with respect to the different variables:

$$ F_{x}=y  \qquad \qquad  G_{y}= p $$

$$ F_{xx}=0 \qquad \qquad  G_{yp}= 1 $$

$$ F_{xy}=1 \qquad \qquad  G_{x}= (y^{'})^{2} $$

$$ F_{y}=x  \qquad \qquad  G_{xp}= 2y^{'} $$

$$ F_{yy}=0 \qquad \qquad  G_{p}= 2xp + y $$

$$ F_{yp}=0 \qquad \qquad  G_{pp}= 2x $$

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

The second exactness condition is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Eq. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Eq. 2 $$

Filling in for Equations 1 the results are as follows:

$$ \ 0 + 2p + 0 = 2p + p - p $$

$$ 2p=2p $$  '''Eq. 1''' is satisfied.

Filling in for Equations 2 the results are as follows:

$$ \ 0 + 0 + 2x = 2x $$

$$ \ 2x = 2x $$   '''Eq. 2''' is satisfied.

By satisfying Eq. 1 and Eq. 2, the 2nd order ODE is considered to be exact.  Follow the other groups' approach to equation numbering so that they show up visibly on the right hand side. Good solution. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

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

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

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

Derive the relation in Equation 1.

Solution
The form for an exact 2nd order ODE is as follows:

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

From this relation the following can be stated:

$$ \ F(x,y,p) = \phi_{p} $$

$$ \ G(x,y,p) = \phi_{x} + \phi_{y}p $$

Where p = y'

Since $$ ( \phi_{p} )_{y} = ( \phi_{y})_{p} $$

Rearranging the definitions of F and G, the previous equation can be written as:

$$ \ F_{y} = \left[ \frac{1}{p}(G - \phi_{x}) \right]_{p} $$

This can be evaluated to obtain the following:

$$ \ F_{y} = \frac{-1}{p^{2}}(G - \phi_{x}) + \frac{1}{p}(G_{p} - \phi_{xp}) $$  This does not connect to the previous step with your problem setup above. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Rearranging:

$$ \ F_{y}p^{2} = - (G - \phi_{x}) + p(G_{p} - \phi_{xp}) $$

$$ \phi_{x} = F_{y}p^{2} + G - p(G_{p} - F_{x}) \qquad \qquad Equation 3. $$

Taking a partial derivative of Equation 3. with respect to p

$$ \phi_{xp} = F_{y}2p + F_{yp}p^{2} + G_{p} - pG_{pp} - G_{p} + pF_{xp} + F_{x} $$

$$ \phi_{xp} = 2pF_{y} + p^{2}F_{yp} - pG_{pp} + pF_{xp} + F_{x} $$

$$ \phi_{xp} = P(2F_{y} + pF_{yp} - G_{pp} + F_{xp}) + F_{x} $$

$$ \phi_{xp} = P(2F_{y} + pF_{yp} - G_{pp} + F_{xp}) + \phi_{xp} $$

$$ \phi_{xp} = P(2F_{y} + pF_{yp} - G_{pp} + F_{xp}) + \phi_{xp} $$

$$ \ 0 = 2F_{y} + pF_{yp} - G_{pp} + F_{xp} $$

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

The result is indeed the second equation of the 2nd exactness condition, Equation 2.  Your solution could use more explanation; there are 7 lines of equations in a row without any text to explain them. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

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

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

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

Derive the relation of Equation 2

Solution
The form for an exact 2nd order ODE is as follows:

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

From this relation the following can be stated:

$$ \ F(x,y,p) = \phi_{p} $$

$$ \ G(x,y,p) = \phi_{x} + \phi_{y}p $$

Where p = y'

$$ \ (\phi_{x})_{p} = G_{p} - (p \phi_{y})_{p} $$

Since: $$ (\phi_{x})_{p} = (\phi_{p})_{x} $$

$$ \ F_{x}= G_{p} - (p \phi_{y})_{p} $$

$$ \ G_{p} - F_{x} = (p \phi_{y})_{p} $$

$$ \ G_{p} - F_{x} = p \phi_{yp} + \phi_{y} $$

$$ \ \phi_{y} = G_{p} - F_{x} - p \phi_{yp} $$

$$ \ \phi_{y} = G_{p} - F_{x} - p (F_{y}) \qquad Equation 3 $$

Since $$ (\phi_{p})_{y} = F_{y} \qquad $$ By the definition of F.

From the definition of G, the following can be identified:

$$ \phi_{x} = G - p \phi_{y} \qquad Equation 4 $$

Using Eqs. 1 and 2 the following is found:

$$ \phi_{xy} = \phi_{yx} $$

$$ \left[G - p \phi_{y} \right]_{y} = G_{px} - F_{xx} - p(F_{yx}) $$

$$ \ G_{y} - p (\phi_{yy}) = G_{px} - F_{xx} - p(F_{yx}) $$

$$ \ G_{y} - \left[ p (G_{py} - F_{xy} - pF_{yy}) \right] = G_{px} - F_{xx} - p(F_{yx}) $$

$$ \ G_{y} - p(G_{py}) + pF_{xy} + p^{2}F_{yy}) = G_{px} - F_{xx} - p(F_{yx}) $$

Rearranging Yields:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation 5. $$

Equation 5 is equivalent to the Equation 2 from the exactness condition.  Your solution could be condensed into half a dozen lines. This is an awkward (albeit correct) solution. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem Statement
Given:

$$ (8x^{5}y^{'})y^{''} + 2x^{2}y^{'} + 20x^{4}(y^{'})^{2} + 4xy = 0 $$

Show that the Nonlinear 2nd-Order ODE is exact

Solution
Exactness Condition 1:

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

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

It is concluded that it satisfies the first exactness condition.

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

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

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

The Following partial derivatives are then identified:

$$ F_{x}=40x^{4}p  \qquad \qquad  G_{x}= 4xp + 80x^{3}p^{2} + 4y $$

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

$$ F_{xy}=0 \qquad \qquad  G_{y}= 4x $$

$$ F_{xp}=40x^{4}  \qquad \qquad  G_{yp}= 0 $$

$$ F_{y}=0 \qquad \qquad  G_{p}= 2x^{2} + 40x^{4}p $$

$$ F_{yy}=0 \qquad \qquad  G_{pp}= 40x^{4} $$

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

Applying the results in Eq. 1:

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

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

Eq. 1 is then satisfied.

Applying the results in Eq. 2:

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

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

Eq. 2 is then satisfied.

Since Equation 1 and Equation 2 are both satisfied, then the 2nd exactness condition is satisfied. It is then concluded that the N2.ODE is exact.

 Good- clear and concise.--Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Problem Statement
Given

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

Prove that the equation is not exact.

Solution
Exactness Condition 1:

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

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

It is concluded that it satisfies the first exactness condition.

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

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

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

The following partial derivatives are found:

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

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

$$ F_{xy}= 0 \qquad \qquad \qquad  G_{y}= 3 $$

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

$$ F_{y}= 0 \qquad \qquad \qquad  G_{p}= 2x $$

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

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

Using these values in Eq. 1:

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

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

Equation 1. is not satisfied therefore the ODE is not exact.

Equation 2. can be populated to find:

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

$$ \ 0 = 0 $$

Equation 2. is satisfied.

In order for the ODE to be exact it must satisfy both Eq. 1 and Eq. 2, since it fails to do this it is concluded that the ODE is not exact.  correct. --Egm6321.f09.TA 04:22, 28 September 2009 (UTC)

Contributing Authors
--Egm6321.f09.team5.GV 01:39, 23 September 2009 (UTC)

--Egm6321.f09.team5.Jason(Zhichao Gong) 00:17, 23 September 2009 (UTC)

--Egm6321.f09.team5.bear 12:27, 23 September 2009 (UTC)

--Egm6321.f09.team5.risher 14:55, 23 September 2009 (UTC)