User:Egm6321.f10.team4/HW 3

= Problem 1 - Equation of motion in 2 dimensions =

Given
Fig 1-1: This image (from Mtg. 13 p. 13-1) depicts a particle in motion, subjected to air resistance. The velocity of the particles is given by $$\displaystyle v$$, with $$\displaystyle x$$ and $$\displaystyle y$$ components $$\displaystyle v_x$$ and $$\displaystyle v_y$$, respectively. The air resistance is opposite the direction of $$\displaystyle v$$, and is given by $$\displaystyle kv^n$$, where $$\displaystyle k$$ and $$\displaystyle n$$ are real constants. The downward force impinging on the particle due to gravity is $$\displaystyle mg$$, where $$\displaystyle m$$ is the particle's mass and $$\displaystyle g$$ is the acceleration due to gravity. The initial velocities of the particle in the $$\displaystyle x$$- and $$\displaystyle y$$-directions are $$\displaystyle v_{x0}$$ and $$\displaystyle v_{y0}$$, respectively.



Fig 1-2: This image (from Mtg. 13 p. 13-2) shows the linear relationship for mass as a function of time, which will be used in part 3.2 of this problem.

Find

 * 1) Derive the equations of motion of the particle.
 * 2) For the particular case when $$\displaystyle k=0$$, verify that $$\displaystyle y(x)$$ is parabolic.
 * 3) Consider the case when $$\displaystyle k \neq 0, v_{x0}=0$$
 * 4) Find $$\displaystyle v_y(t), y(t)$$ for constant $$\displaystyle m$$.
 * 5) Find $$\displaystyle v_y(t), y(t)$$ for $$\displaystyle m=m(t)$$.

Part 1
The balance of forces in 2 dimensions for the dynamic system above is given by:

(Eq 1-2)
 * 
 * }

(Eq 1-3)
 * 
 * }

We can also derive a relationship between the velocity and its individual components (all magnitudes):

Part 2
For $$\displaystyle k=0$$, Eq. (1-2) becomes: {| style="width:100%" border="0"
 * style="width:95%" |
 * style="width:95%" |
 * $$\begin{align}

m \frac{dv_x}{dt}=-(0)v^ncos \alpha=0 \Rightarrow \frac{dv_x}{dt}=0 \end{align}$$

But:


 * $$\displaystyle v_x := \frac{dx}{dt} $$

Thus, from Eq. (1-5):


 * $$\displaystyle \int dx = v_{x0} \int dt $$
 * $$\displaystyle x(t) = v_{x0}t + x_0 $$

Rearrange to solve for $$ \displaystyle t $$:

Likewise, Eq. (1-3) becomes:


 * Note: There is one possible problem with the above equation; if $$ \displaystyle v_{x0}=0 $$, then $$ \displaystyle y(x) $$ explodes. Why is this?  The simple answer is that if there is zero initial velocity in $$ \displaystyle x $$, i.e., $$ \displaystyle v_{x0}=0 $$, then $$ \displaystyle x $$ is always the same, since no other forces act in this direction other than air resistance, which is zero when $$ \displaystyle v_x=0 $$.  Therefore, we can conclude that $$ \displaystyle y $$ is not a function of $$ \displaystyle x $$ when $$ \displaystyle v_{x0}=0 $$, so $$ \displaystyle y(x) $$ is undefined in this particular case.


 * We can also attempt to describe this mathematically, as follows.
 * Start by going back to Eq. (1-6):


 * $$\displaystyle \frac{dx(t)}{dt} = v_{x0} = 0 $$
 * Therefore $$\displaystyle x(t) $$ is a constant.
 * Looking back at Eq (1-9), if $$\displaystyle x $$ is a constant, then $$\displaystyle x-x_0 $$ in the numerator is zero, and $$\displaystyle 0/0=0 $$, thus we get:


 * $$\displaystyle y(t)=y_0 $$,


 * for the particular case that $$\displaystyle v_{x0}=0 $$, which is much "nicer" mathematically than dividing by zero.

3.1: m is constant
For $$\displaystyle k \neq 0, v_{x0}=0 $$, and starting from Eq. (1-1), reproduced below:

{| style="width:100%" border="0"
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle

m \frac{dv_y}{dt}=-k(v_y)^n-mg $$

which is another form of Eq. (1-3). For simplicity, let us define:

Thus:

{| style="width:100%" border="0"
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle

z'=z'(t) = \frac{dv_y}{dt} $$

Eq. (1-1) can then be written as:

which is now in the familiar form satisfying the First Condition of Exactness, where we have defined:

The Second Condition of exactness is satisfied if

We will use the Integrating Factor Method to solve for $$\displaystyle z $$, in which we multiply Eq. (1-11) by a newly created function, $$\displaystyle h=h(t,z) $$:

{| style="width:100%" border="0"
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle

\underbrace{hM}_{\overline{M}} + \underbrace{hN}_{\overline{N}}z'=0 $$

We can now write the Second Condition of Exactness (Eq. (1-14)) as:
 * $$\displaystyle \overline{M}_z = \overline{N}_t $$, or, substituting $$\displaystyle h $$ back in:

It is readily seen that $$\displaystyle N_t = \frac{dN}{dt}=0 $$, since $$\displaystyle N=m= constant $$. We will now make the simplifying assumption that $$\displaystyle h $$ is not a function of $$\displaystyle z $$, meaning that $$\displaystyle h_z = 0 $$

Therefore we can rewrite the above as


 * $$\displaystyle h M_z = h_t N $$, or:

Substitute for $$\displaystyle M $$ and $$\displaystyle N $$ from Eq.s (1-12) and (1-13), respectively, and rewrite $$\displaystyle h_t $$ in its formal form:


 * $$\displaystyle \frac{1}{h} \frac{dh}{dt} = \frac{nk }{m} z^{n-1} $$
 * $$\displaystyle \int \frac{1}{h} dh = \frac{nk }{m} \int z^{n-1} dt $$
 * $$\displaystyle ln(h) = \frac{nk }{m} \int z^{n-1} dt $$

But if $$\displaystyle z=v_y $$ is a function of $$\displaystyle t $$ only, then the integral is easy to solve:

where C is a constant of integration.

Now we can substitute $$\displaystyle h $$ from Eq. (1-18) back into Eq. (1-15) to get:
 * $$\displaystyle exp \left[ \frac{k}{m} z^n +C \right] M + exp \left[ \frac{k}{m} z^n +C \right] N z' = 0 $$

But it is clear that the exponential terms simply cancel. We will resort to another method, as taught in class (Mtg. 10, p. 10-2 and 10-3), where it was shown that if the original equation (Eq. 1-11) were arranged as follows:


 * $$\displaystyle mz'+kz^n=-mg $$

then the equation would be of the form:


 * $$\displaystyle hz' + h'z= bh = (hz)' $$

where in this case, $$\displaystyle b=-g $$. From this equation, $$\displaystyle z $$ could be readily solved.

In our case, $$\displaystyle h' $$ is found from Eq. (1-19), (as a function of $$\displaystyle z $$):

Thus multiplying both sides of Eq. (1-20) by $$\displaystyle h $$ gives: $$\displaystyle hz'+ \underbrace{ \frac{k}{m}h}_{h'}z^n=-gh $$ So:

However, in contrast to the example in class, we are left with an exponent (not equal to 1) over $$\displaystyle z$$, making it more difficult to solve for $$\displaystyle z $$, and thus also $$\displaystyle v_y $$ and $$\displaystyle y $$:

For the sake of finding a possible solution for $$\displaystyle v_y $$ and $$\displaystyle y $$, we can make the assumption that $$\displaystyle n=1 $$. Eq. (1-22) now becomes:

(Recall: we assumed earlier that $$\displaystyle h $$ is a function of $$\displaystyle t $$ only.) Note the similarity to the form: $$\displaystyle y(x) = \frac{1}{h} \int h(x)b(x) dt $$. In our case, the only difference is that $$\displaystyle b=-g $$ has been moved outside the integral. Continuing, we can solve for $$\displaystyle y $$ from:

3.2: m = m(t)
For $$\displaystyle m=m(t) $$:
 * $$\displaystyle N_t = \frac{dm}{dt} \neq 0 $$  If we make the same assumption that $$\displaystyle h $$ is not a function of z, then Eq. (1-16) becomes:


 * $$\displaystyle h_t N = h(M_z-N_t) $$
 * $$\displaystyle \frac{h_t}{h} = \frac{1}{N}(M_z-N_t) $$
 * $$\displaystyle \frac{1}{h} \frac{dh}{dt} = \frac{nkz^{n-1}-m_t}{m} $$
 * $$\displaystyle ln(h) = \int \frac{nkz^{n-1}-m_t}{m} dt $$
 * $$\displaystyle h = exp \left[ \int \frac{nkz^{n-1}-m_t}{m} dt \right] $$
 * $$\displaystyle h = exp \left[ \int \frac{nkz^{n-1}}{m} dt - \int \frac{m_t}{m} dt \right] $$


 * The loss of mass over time is assumed to have a linear relationship (see Given). So we can substitute:
 * $$\displaystyle m(t)= m_0 + \frac{m_1-m_0}{t_1}t $$
 * $$\displaystyle m_t(t)= \frac{m_1-m_0}{t_1} $$

Then:
 * $$\displaystyle h = exp \left[ \int \frac{nkz^{n-1}}{m_0 + \frac{m_1-m_0}{t_1}t} dt - \int \frac{\frac{m_1-m_0}{t_1}}{m_0 + \frac{m_1-m_0}{t_1}t} dt \right] $$
 * $$\displaystyle h = exp \left[ \int \frac{nkz^{n-1}}{m_0 + \frac{m_1-m_0}{t_1}t} dt - \int \frac{m_1-m_0}{m_0 t_1 + (m_1-m_0)t} dt \right] $$

Note: When $$\displaystyle m $$ is a function of $$\displaystyle t $$, we are unable to attain a equation of the form of Eq. (1-23). The loss of mass over time makes this problem much more complicated, so we need some new method to solve this problem.

Egm6321.f10.team4.ejm 20:41, 6 October 2010 (UTC) Author

= Problem 2 : Linear Time variant system(SC_L1_ODE_VC) = From Meeting 13, p 13-3.

Given
Linear Time Variant Equation(SC_L1_ODE_VC)

cf. Time Invariant Equation is consisted with constant A and B, instead of A(t) and B(t) respectively.

Equation of Connected double pendulum - Application of Enigneering : Control

where,


 * $$\displaystyle

{m}_{1}, {m}_{2} $$ : mass of the pendulum 1 and 2 respectively


 * $$\displaystyle

g $$ : acceleration of gravity


 * $$\displaystyle

{\theta}_{1}, {\theta}_{2} $$ : the angle of pendulum 1 nd 2 respectively


 * $$\displaystyle

l $$ : length of the pendulums


 * $$\displaystyle

a $$ : length of the pendulum up to connection point


 * $$\displaystyle

k $$ : spring constant


 * $$\displaystyle

{u}_{1}, {u}_{2} $$ : applied forces

Find
1. Derive (Eq 2-2)

2. Transform (Eq 2-2) to the form of (Eq 2-1) - Find A(t), B(t), U(t)

where,

Solution for Derivation of Equation of Motion
 Required Background information 1.Linear Approximation 2.Hooke's law(Spring Force)

where,
 * $$\displaystyle

F $$ : restoring force


 * $$\displaystyle

k $$ : spring constant


 * $$\displaystyle

x $$ : displacement from the equilibrium position

The moment of inertia of the pendulums can be rewritten by summing all the applied torques on the pendulum 1 and 2

where,
 * $$\displaystyle \tau $$ : torque
 * $$\displaystyle I={mv}^{2}$$ : moment of inertia
 * $$\displaystyle \alpha= \frac{{d}^{2}}{{dt}^{2}} \theta=\ddot{\theta} $$ : angular acceleration

torque by the spring force by Linear Approximation

torque by the gravity by Linear Approximation

torque by the applied force(AF) by Linear Approximation

Combining the equations from (Eq 2-3) to eqn(Eq 2-7) and applying the process to pendulum2 yields

Solution for System Matrix Equation
Transform the 2nd order ODEs to a system matrix equation,

First, reduce this system of second-order ODEs to a first-order differential equation by introducing the vectors.

Then,

Developing the equation(Eq 2-2) by making the coefficient of $$ \ddot{\theta} $$ to 1 and gathering variables with respect to $$\displaystyle{\theta}_{n} $$ yield,

Thus,

Transforming the rearranged equation to System Matrix equation form becomes,

Egm6321.f10.team4.Yoon 00:16, 3 October 2010 (UTC-Estern Time) - Primary Author & Editor

= Problem 3 - Linear Time Invariant system(SC_L1_ODE_CC)= From Meeting 14, p 14-1.

Given
Linear Time Invariant Equation(Appl of Eng'ring:Control)

cf. time Variant Equation is consisted with constant A(t) and B(t), instead.

Find
Solve the given equation of Linear time invariant system(SC_L1_ODE_CC)

Solution
Rearranging the equation (Eq 3-1) by the reduction order with respect to x,

Since the given coefficient of $$\displaystyle \dot{x} $$ is 1, we can directly get the first order Euler intergrating factor h(x),

By Multiplying the integrating factor $$\displaystyle h(x) $$ to (Eq 3-2) to simplify

Simplifying the left hand side of the equation of (Eq 3-4)

integrating the (Eq 3.5) by interval $$ \left[ {t}_{0},t \right] $$

Rearranging the equation above(Eq 3-7) yields,

By deviding the (Eq 3-8) by $$\displaystyle {e}^{-at} $$, we can get the desired solution of 

Egm6321.f10.team4.Yoon 00:16, 3 October 2010 (UTC-Estern Time) - Primary Author & Editor

= Problem 4 - Taylor Series Expansion= From Meeting 14, p 14-2.

Given
Part A


 * {| style="width:100%" border="0" align="left"

$$f(x) = exp (x) $$ $$
 * $$\displaystyle (Eq. 4-1)
 * }
 * }

Part B


 * {| style="width:100%" border="0" align="left"

$$[f(x)] = \exp ([A]) $$ $$
 * $$\displaystyle (Eq. 4-2)
 * }
 * }

Find
Determine the special case of the Taylor Series expansion known as the Maclaurin Series for Eq. (4-1) and Eq. (4-2) based on the following definition


 * {| style="width:100%" border="0" align="left"

$$ f(x) = f(a) + \frac (x - a) + \frac (x - a)^2 + ... $$ $$
 * $$\displaystyle (Eq. 4-3)
 * }
 * }

Note: The Maclaurin Series is an special case, where the Taylor Series is expanded around $$a=0$$.

Solution
Part A

We will first begin by computing the general form of the first three terms in the series. We will do so by starting with Eq. (4-1), first deriving it, then replacing $$x$$ with a and finally computing its value for the case of $$a=0$$.


 * {| style="width:100%" border="0" align="left"

$$ f(x) = e^x \to f(a) = e^a  \to f(0) = e^0  = 1$$
 * 
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$ f'(x) = e^x \to f'(a) = e^a  \to f'(0) = e^0  = 1$$
 * 
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$ f(x) = e^x \to f(a) = e^a  \to f''(0) = e^0  = 1$$
 * 
 * }
 * }

If the above terms are inserted into Eq. (4-3) and we set $$a=0$$, the expansion will take the following form


 * {| style="width:100%" border="0" align="left"

$$ f(x) = f(0) + \frac (x - 0) + \frac (x - 0)^2 + ...$$
 * 
 * }
 * }

evaluating we will get he final form


 * {| style="width:100%" border="0" align="left"

$$ f(x) = 1 + \frac{1} x + \frac{1} x^2 + ... = \sum\limits_{n = 0}^\infty {\frac{1} x^n } $$ $$
 * $$\displaystyle (Eq. 4-4)
 * }
 * }

The Taylor Series expansion is a rather simple but powerful tool that allows a function to be expressed as an infinite sum of its derivates. In engineering it is widely used, due to its practicality.

Part B

For the special case of the Matrix Exponential function we will begin by expanding a squared matrix $$ A_{nxn} $$

While it is not intuitive, as $$n$$ gets larger, the matrix in Eq. (4-5) will get closer and closer to $$ e^A$$. Thus,

Just like in the previous part, it is important to note the importance of the Taylor Series expansion for the Matrix Exponential. It is specially important when solving linear differential equations.

Egm6321.f10.team4.petralanda.n 02:23, 5 October 2010 (UTC)Egm6321.f10.team4.petralanda.n

= Problem 5: Generalization= From Meeting 15, p 15-1.

Given

 * {| style="width:100%" border="0" align="left"

$$ X'(t) = Ax(t) + Bu(t)$$ $$
 * $$\displaystyle (Eq. 5-1)
 * }
 * }

The associated Matrix dimensions are as follows


 * {| style="width:100%" border="0" align="left"

$$

X(t) = \left[ {X(t)} \right]_{nx1} $$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$ A = \left[ A \right]_{nxn} $$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$ B = \left[ B \right]_{nxn} $$
 * }
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$

u(t) = \left[ {u(t)} \right]_{nx1} $$
 * }
 * }
 * }

Find

 * {| style="width:100%" border="0" align="left"

$$X(t) = e^{A(t - t_0 )} X(t_0 ) + \int\limits_{t_0 }^t {e^{A(t - \tau )} Bu(\tau )d\tau } $$ $$
 * $$\displaystyle (Eq. 5 -2)
 * }
 * }

Solution
Let's rewrite Eq. (5 -1) to met the first exactness condition


 * {| style="width:100%" border="0" align="left"

$$ F(t,x,x') = X'(t) - AX(t) = Bu(t)$$ $$
 * $$\displaystyle (Eq. 5 -4)
 * }
 * }

We will now multiply it by and integrating factor to force the second exactness condition upon Eq. (5 -4)


 * {| style="width:100%" border="0" align="left"

$$ \underbrace h_NX'(t)\underbrace { - hAX(t)}_M = hBu(t)$$ $$
 * $$\displaystyle (Eq. 5 -5)
 * }
 * }

Implementing the second exactness condition


 * {| style="width:100%" border="0" align="left"

$$ \frac = \frac $$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$ \frac = \frac $$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$h_t =  - Ah - AXh_x $$
 * }
 * }

If we now assume that $$ h=f(t)$$, we will obtain the following


 * {| style="width:100%" border="0" align="left"

$$ \frac {h} = - A $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5 -6)
 * }
 * }

Integrating


 * {| style="width:100%" border="0" align="left"

$$

\int {\frac {h}} = \int { - Adt} $$
 * }
 * }
 * }

yields


 * {| style="width:100%" border="0" align="left"

$$ h = e^{ - At} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5 -7)
 * }
 * }

With the integration factor we can now go back to Eq. (5 -5) and replace $$ h$$ by $$ e^(-At)$$


 * {| style="width:100%" border="0" align="left"

$$ e^{ - At} x'(t) - e^{ - At} Ax(t) = e^{ - At} Bu(t)$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5 -8)
 * }
 * }

By inspection of Eq. (5 -8) we can make the following connection


 * {| style="width:100%" border="0" align="left"

$$ \underbrace {\underbrace {e^{ - At} }_hX'(t)\underbrace { - e^{ - At} A}_{h'}X(t)}_{(hX)'} = e^{ - At} Bu(t)

$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5 -9)
 * }
 * }

Integration of Eq. (5- 9) will take the following form


 * {| style="width:100%" border="0" align="left"

$$ \int\limits_{t_0 }^t {\frac } d\tau = \int\limits_{t_0 }^t {e^{ - A\tau } } Bu(\tau )d\tau $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5 -10)
 * }
 * }

Evaluating the left hand side of Eq. (5 -10)


 * {| style="width:100%" border="0" align="left"

$$ e^{ - At} X(t) - e^{ - At_0 } X(t_0 ) = \int\limits_{t_0 }^t {e^{ - A\tau } } Bu(\tau )d\tau $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5 -11)
 * }
 * }

Taking the following intermediate algebraic step


 * {| style="width:100%" border="0" align="left"

$$

e^{ - At} X(t) = e^{ - At_0 } X(t_0 ) + \int\limits_{t_0 }^t {e^{ - A\tau } } Bu(\tau )d\tau $$
 * }
 * }
 * }

will allow us to see that by dividing both sides of the equation with $$ e^(-At)$$ will yield the solution in Eq. (5 -2)


 * {| style="width:100%" border="0" align="left"

$$ X(t) = e^{A(t - t_0 )} X(t_0 ) + \int\limits_{t_0 }^t {e^{A(t - \tau )} Bu(\tau )d\tau } $$
 * }
 * }
 * }

Note: Because the integration factor is nothing more than a scalar multiplication to the matrixes, the dimensions of the different components remain the same throughout the computations.

Egm6321.f10.team4.petralanda.n 02:24, 5 October 2010 (UTC)egm6321.f10.team4.petralanda.n

= Problem 6: Solving SC_L1_ODE simultaneously  = From Meeting 15, p 15-2.

Given
159.178.245.223 15:22, 6 October 2010 (UTC)egm6321.f10.team4.petralanda.n

Find
Linear Time Variant Equation(SC_L1_ODE_VC)

159.178.245.223 15:22, 6 October 2010 (UTC)egm6321.f10.team4.petralanda.n

Solution
Equations 6-1, 6-2, and 6-3 can be easily be put in the form of equation 6-4 by defining $$\dot{\textbf{x}}$$ as

159.178.245.223 15:22, 6 October 2010 (UTC)egm6321.f10.team4.petralanda.n

Egm6321.f10.team4.Auerbach 17:38, 6 October 2010 (UTC) - Proof Reader

= Problem 7 - Alternative expressions for $$\displaystyle h(x,y) $$ = From Meeting 17, p 17-1.

Given
$$\displaystyle h_x + h_y p = 0 $$<p style="text-align:right;">$$\displaystyle (Eq. 7-1) $$

where

$$\displaystyle p = dy/dx $$.

Find
an alternative expression for $$\displaystyle h(x,y) $$.

Solution
Originally, we assumed that $$\displaystyle h_x = h_y = 0 $$, so that we could derive an expression for the first integral, $$\displaystyle \Phi $$, with relative ease.

If we instead rewrite the equation (7-1) as,

$$\displaystyle h_x = - h_y p $$<p style="text-align:right;">$$\displaystyle (Eq. 7-2) $$

we can see that $$\displaystyle h(x,y) $$ can be a function, and still give us a valid expression.

If we let $$\displaystyle h(x,y) = e^{x-y}$$,

then we still get a valid expression for equation (7-1).

$$\displaystyle h_x = e^{x-y}$$

$$\displaystyle h_y = -e^{x-y}$$

Therefore $$\displaystyle h_x + h_y p = 0 $$

128.227.50.129 18:48, 6 October 2010 (UTC)Egm6321.f10.team4.osentowski - Author

= Problem 8 = From Meeting 17, p 17-2.

Given
the following N2-ODE:

Find
Show that Eq (8-1) is exact.

Solution
In order to satisfy the first condition of exactness, a N2-ODE must be able to put in the form:

Eq. (8-1) can easily be put into this form if:

and

where $$\displaystyle p:=y' $$


 * {| style="width:40%" border="0"

Thus the first condition of exactness (Eq. 8-2) is met.
 * style="width:80%; padding:10px; border:2px solid #8888aa" |
 * style="width:80%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

Given the form of Eq. (8-2), the second condition of exactness requires the following two criteria to be met:

Solving for the derivatives individually:

{| style="width:100%" border="0" $$\begin{array}{*{20}l} {f_x = -30p^4xsinx^2} & {f_{xx}=-30p^4sinx^2-60p^4x^2cosx^2} & {g_x=6y^2p-6p^5sinx^2-12x^2p^5cosx^2} & {g_{xp}=6y^2-30p^4sinx^2-60x^2p^4cosx^2}  \\ {} & {f_{xy} = 0} & {g_y  = 12xyp+6y^2} & {g_{yp}=12xy}  \\ {} & {f_{xp} = -120p^3xsinx^2 } & {g_p  =6xy^2-30xp^4sinx^2} & {g_{pp}  = -120xp^3sinx^2 }  \\ {f_y = 0} & {f_{yy}  = 0} & {} & {}  \\ {} & {f_{yp} = 0} & {} & {}  \\ \end{array}$$
 * style="width:95%" |
 * style="width:95%" |

To check part (1) of the second condition of exactness, substitute the derivatives above into Eq. (8-5) to get:

Now to verify that part (2) of the second condition of exactness is satisfied, Eq. (8-6) becomes:

Note: The variable $$p$$ is treated as independent of $$x$$ and $$y$$, thus simplifying what would otherwise be a rather long and much more tedious derivation. This treatment is consistent with that of previous classes, e.g. F09 HW 3 Problem 6.

Egm6321.f10.team4.ejm 16:49, 4 October 2010 (UTC) Author

Egm6321.f10.team4.Auerbach 05:05, 6 October 2010 (UTC) - Proof Reader

= Problem 9  - Generating an expression for $$\displaystyle \Phi $$ from $$\displaystyle h_y$$ and $$\displaystyle h_x$$= From Meeting 18, p 18-3.

Given
$$\displaystyle (6xy^2)y' + 2y^3 = (h_y + 0)y' + h_x$$<p style="text-align:right;">$$\displaystyle (Eq. 9-1) $$

and

$$\displaystyle \Phi_p = 15p^4 cos x^2 $$<p style="text-align:right;">$$\displaystyle (Eq. 9-2) $$

Find
The expression for the first integral $$\displaystyle \Phi(x,y,p) $$

Solution
Two methods can be used to find the expression for the first integral. The first was shown in Meeting 18 p 18-2,18-3. The second will be shown here:

First, we will assume that

$$\displaystyle h_y y' = 2y^3 $$<p style="text-align:right;">$$\displaystyle (Eq. 9-3) $$

where

$$\displaystyle y' = \frac{dy}{dx}$$.

Then by applying the chain rule,

$$\displaystyle h_y dy = 2y^3 dx $$<p style="text-align:right;">$$\displaystyle (Eq. 9-4) $$

Integrating both sides yields

$$\displaystyle \int^y h_y dy = \int^x 2y^3 dx $$

$$\displaystyle h(x,y) = 2xy^3 + k_1(y) $$<p style="text-align:right;">$$\displaystyle (Eq. 9-5) $$

where $$\displaystyle k_1(y) $$ is a function with respect to y only as a result of the integration.

If we then differentiate with respect to x, we get the expression

$$\displaystyle h_x = 2y^3 + 0 $$.<p style="text-align:right;">$$\displaystyle (Eq. 9-6) $$

From our original equations (9-1)and (9-3), it follows that

$$\displaystyle h_x = 6xy^2 y' $$

Setting this result equal to our result in (9-6), we get

$$\displaystyle 2y^3 = 6xy^2 y' $$.<p style="text-align:right;">$$\displaystyle (Eq. 9-7) $$

Again, we use the chain rule, and integrate both sides, as before.

$$\displaystyle \int^x 2y^3 dx = \int^y 6xy^2 dy $$

$$\displaystyle 2xy^3 + k_1(y) = 2xy^3 + k_2(x) $$<p style="text-align:right;">$$\displaystyle (Eq. 9-8) $$

If $$\displaystyle k_1 $$ is a function with only y, and $$\displaystyle k_2 $$ is a function with only x, then logically these two functions can only be equal to one another if both $$\displaystyle k_1 $$ and $$\displaystyle k_2 $$ are constants. Furthermore, they must be the same constant. Therefore, only the constant $$\displaystyle k_1 $$ will be used.

It can now be established that

$$\displaystyle h(x,y) = 2xy^3 + k_1 $$<p style="text-align:right;">$$\displaystyle (Eq. 9-9) $$

The expression for the first integral $$\displaystyle \Phi $$, is defined as

$$\displaystyle \Phi(x,y,p) = h(x,y) + \int^p f(x,y,p) dp$$,

where $$\displaystyle f(x,y,p) = \Phi_p $$

Therefore,

$$\displaystyle \Phi(x,y,p) = 2xy^3 + k_1 + \int^p 15p^4 cos x^2 dp = k_3 $$<p style="text-align:right;">$$\displaystyle (Eq. 9-10) $$

$$\displaystyle \Phi(x,y,p) = 2xy^3 + k_1 + 3p^5 cos x^2 = k_3$$<p style="text-align:right;">$$\displaystyle (Eq. 9-11) $$

$$\displaystyle \Phi(x,y,p) = 2xy^3 + 3p^5 cos x^2 = k_3 - k_1$$<p style="text-align:right;">$$\displaystyle (Eq. 9-12) $$

$$\displaystyle \Phi(x,y,p) = 2xy^3 + 3p^5 cos x^2 = k_4$$<p style="text-align:right;">$$\displaystyle (Eq. 9-13) $$

where $$\displaystyle k_4 $$ is a constant.

This is the same expression for $$\displaystyle \Phi $$ as was originally given for Example #2, in Meeting 17 p 17-1.

68.17.105.125 04:09, 5 October 2010 (UTC)Egm6321.f10.team4.osentowski - Author

Egm6321.f10.team4.Auerbach 04:55, 6 October 2010 (UTC) - Proof Reader