User:Egm6321.f12.team5.report 4

= Report 4, team 5 =

Given
Considering the following ODE Sec21.

Find
Verify the exactness of ($$).

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

For an equation to be exact, it must satisfy two conditions Sec16. 1st Exactness Condition for N2-ODEs: The N2-ODE must have the particular form:

where

To reduce the order of the equation, we can use the following substitutions

to rewrite($$) in reduced order form.

From inspection, we can identify the following:

Therefore ($$) satisfies the 1st Exactness Condition. 2nd Exactness Condition for N2-ODEs: The N2-ODE must satisfy the following conditions:

Evaluating the derivatives of $$ f(x,y,p) $$ and $$ g(x,y,p) $$ from ($$) and ($$) we can evaluate the 2nd Exactness conditions ($$) and ($$) as follows

Conclusion Since the second exactness condition was not satisfied in ($$), we can conclude that this N2-ODE is not exact.

Given
N2-ODE below is from lecture notes Sec21.

Find
Find $$m, n \in \mathbb R$$ such that the ($$) becomes exact and show that its first integral is the L1-ODE-VC shown below:

where,

Then solve the first integral ($$) for y(x).

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Find $$m, n \in \mathbb R$$ to make ($$) exact. Equation ($$) can be rearranged as:

Using the notation of lecture notes Sec16-4, ($$) can be rewritten as

where,

and $$ p = y'$$ (as defined in the Find section above) and subscripts indicate partial derivatives w.r.t. the subscripted variable. Equations ($$) and ($$) show that 1st exactness condition is satisfied regardless of the values of m and n. The 2nd exactness condition for N2-ODEs is defined in lecture notes Sec16-5 as,

The partial derivatives of $$f$$ and $$g$$ can be found by taking the derivatives of ($$) and ($$).

Substituting in the results from ($$)~($$) into ($$) results in,

which tells us that $$n = 0$$, and that integrating factor does not a $$y$$ term ($$y^0 = 1$$). Given the above information about $$n$$, updating $$f_{xx}, f_{xy}, f_{yy}, g_{xp}, g_{yp},$$ and $$g_{y}$$ gives,

Substituting in the updated $$f_{xx}, f_{xy}, f_{yy}, g_{xp}, g_{yp},$$ and $$g_{y}$$ into ($$) results in,

Since the powers of the variable $$x$$ are different, the only way for ($$) to be true is if the coefficients are equal to zero. Although both $$m = \frac{1}{2}$$ and $$m = -\frac{1}{2}$$ work for the left side of the equation, only $$m = \frac{1}{2}$$ satisfies the zero coefficient requirement for the right side of the equation. Therefore $$m = \frac{1}{2}$$, and the integrating factor is,

Show that the first integral of the N2-ODE ($$) is: $$\phi(x,y,p) = xp + (2x^{\frac{3}{2}}-1)y = k$$ Since we know that the integrating factor from ($$), equation ($$) becomes,

Integrating $$\phi_p$$ gives,

Taking the total derivative of ($$),

Comparing ($$) and ($$) shows that

Integrating ($$) and ($$) gives

Equating ($$) and ($$),

shows that

Therefore,

and substituting ($$) into ($$) shows that

Solve the first integral ($$) for y(x). Inspection of ($$) shows that it is a L1-ODE-VC, which can be solved using the Euler integrating factor method (IFM). Since the variable constants are all functions of $$x$$ only, it can be assumed that the integrating factor,$$h(x,y)$$, is a function of $$x$$ only as well ($$N, N_x, M_y$$ are all function of $$x$$ only, so that $$h(x,y) = h(x)$$) or that $$h_y(x,y) = 0$$. [See lecture notes Sec11.] First rearrange ($$) as shown below.

Lecture notes, Sec11-4, show that the integrating factor $$h$$ can be found using the following formula.

For this case,

so that ($$) becomes

$$y(x)$$ is obtain by using the equation below. (See lecture notes, Sec11-5.)

Substituting in for $$h(x)$$ and $$b(s)$$ gives,

Comments
Initial solution by --Egm6321.f12.team5.kim (talk) 04:31, 14 October 2012 (UTC)

Given
A general class of L2 - ODE - VC represented by the following equation,

where,

and the following function which creates a class of exact L2 - ODE - VC

(From lecture notes sec 21)

Find
Show that ($$) and ($$) leads to ($$)

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

From ($$) and ($$) we can write,

Since it is given that $$ P(x) = \phi_p $$, ($$) becomes

The above equation represents an L2 - ODE - VC which satisfies the first exactness condition. Now, We know that,

Therefore,

Differentiating the above expression for $$ \phi $$ with respect to $$ x $$ to get the L2 - ODE - VC (G)

Since $$ p = y' $$ and $$ p' = y'' $$, we can write the above equation as,

Comparing the above equation with ($$) and ($$), We get,

Now substituting the above equation into Equation ($$)

Setting $$ \int^x R(s)ds = T(x) $$, The above equation can be modified as,

Differentiating the above equation with respect to $$ y $$.

But from Equations ($$) and ($$), We know that,

Equating the two equations above,

which gives,

or

Substituting in ($$),

Hence it is proved that ($$) and ($$) leads to ($$)

Comments
Initial Solution by Vishnu

Given
(From lecture notes sec 21) L2-ODE-VC

Find
1. Show (4.4.1) is exact 2. Find $$\phi$$ 3. Solve for $$y(x)$$

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Part one
If we substitute $$p=y'$$

Then we get

which is in a particular form, so the operator safisfies the first exactness condition.

The operator satisfies the second exactness condition, if

In this case,

$$f_{xx}=\frac{\partial ^2f(x,y,p)}{\partial x^2}=-\cos x$$

$$pf_{xy}=p\frac{\partial ^2f(x,y,p)}{\partial x \partial y}=0$$

$$p^2f_{yy}=p^2\frac{\partial ^2f(x,y,p)}{\partial y^2}=0$$

$$g_{xy}=\frac{\partial ^2g(x,y,p)}{\partial x \partial y}=2x-\cos x$$

$$pg_{py}=p\frac{\partial ^2g(x,y,p)}{\partial p \partial y}=0$$

$$-g_{y}=-\frac{\partial g(x,y,p)}{\partial y}=-2x$$

$$-\cos x=-\cos x$$, so (4.4.5) is satisfied

$$f_{xp}=\frac{\partial ^2f(x,y,p)}{\partial x \partial p}=0$$

$$pf_{py}=p\frac{\partial ^2f(x,y,p)}{\partial p \partial y}=0$$

$$2f_{y}=2\frac{\partial f(x,y,p)}{\partial y}=0$$

$$g_{pp}=\frac{\partial ^2g(x,y,p)}{\partial p^2}=0$$

$$0=0$$, so (4.4.6) is satisfied

So the function G satisties both first exactness condition and the second exactness condition. (4.4.1) is exact.

Part two
So $$h_x+h_yp=px^2+2xy$$

By comparing left side and right side, we got $$h_x=2xy$$ and $$h_y=x^2$$

So $$k'_1(y)=0$$. Thus $$k_1(y)=k_1$$

Part three
Substitute $$\frac{x^2}{\cos x}=a_0(x)$$ and $$\frac{k}{\cos x}=b(x)$$

We get

(From lecture notes Sec11, Pg 4 and 5.)

Comments
Initial Solution by Yu

Given
(From lecture notes sec 22)

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

From the equation 4.5.2,we know:

Comments
= Contributors = Report Leader: Yu Hou Report Co-Leaders : Yi Zhongwen