User:Egm6321.f12.team4.hong/Report3

Problem R*3.3 – Homogeneous Solution of L1-ODE-VC
sec12-2

Statement
Instead of identifying $$y_h(x)$$ from $$ h(x) = \exp[\int^x a_0(s)ds + k_1 ] $$  and  $$ y(x) = \frac{1}{h(x)}\left[\int^x h(s)b(s)ds + k_2 \right] $$, solve the homogeneous counterpart of

Solution
To solve this problem, the separation of variable method is used. We can rephrase the above equation as below.

Then we can solve for y by using the separation of variable method.

s is a dummy variable. $$K_2$$ and $$K_1$$ are some constants. Therefore let, $$ K_3 = K_2 - K_1 $$ then,

Since $$ \displaystyle \exp[K_3] $$ is a constant, let $$ \displaystyle K = \exp[K_3] $$ then,

Author and References

 * Solved and Typed by -- Seong Hyeon Hong
 * Reviewed by --

Problem R*3.9 - Deriving Solving Equations of Motion
sec14-2

Statement
1. Derive the equations of motion

2. Particular case k = 0: Verify that y(x) is parabola.

3. Consider the case $$ k \neq 0 $$ and $$ v_{x0} $$

''3.1. Is (3.9.0), for n=1,2, either exact or can be made exact using IFM? Find $$v_y(t)$$ and $$y(t)$$ for m constant.''

3.2. Find $$v_y(t)$$ and $$y(t)$$ for $$m = m(t)$$

Solution
1. Looking at the figure 3.9.1, we can see that there are two force factors acting on a point which are:$$ kv^n, mg $$. However we want to express them in v which is not a direct force factor. v must be differentiated with respect to time and multiplied by mass to become a force factor. (F = ma = m*(dv/dt))

We can easily derive equations (1) and (2) by taking the sum of all the force factors in x and y direction respectively.

Sum of forces acting in x direction.

Sum of forces acting in y direction.

Negative signs indicate that the forces are acting in opposite directions of vx and vy.

The third equation can be proved by the Pythagoras theorem which is,

When C is the hypotenuse of sides A and B which are right angle to each other forming right angle triangle. If you look at the figure it is clear that the three vectors form a right angle triangle such that magnitude of v is a hypotenuse of the addition of magnitude of vx and magnitude of vy.

Although v, vx, vy are vectors when they are squared they become scalar values and therefore the Pythagoras theorem can be used to derive the equation 3. Therefore,

The last equation can be derived by a simple trigonometry relationship. If there is a right angle triangle in such a way that C is the hypotenuse, A is the adjacent, B is the opposite side, and let theta be the angle between A and C then,

With the above relationship and the relationship of angle alpha and v, vx, vy we can make replacements as below.

Now we have the equation,

2. Recall equation,

Since k = 0, the equation now becomes,

Before we actually solve for y(x) it is really important to know that g is a constant which have a values of gravitational force of Earth. We will integrate equation 3.9.8 with respect to time twice to find y(x).

However, the right hand side of the equation, we must have a function of x, since we know that y(x) is a function of x. Therefore on the right hand side, instead of integrating it with respect to time we will use the relationship,

Now we can integrate the right hand side with respect to x and integrate left hand side with respect to time. There is an assumption that the v_x is a function of time, not x.

The equation (3.9.12) shows that y(x) is a parabola.

3.

There are two exactness conditions to check whether the equation is exact or not. Here are the two exactness conditions.

Let's rewrite the equation (3.9.0) in more general form such that we can apply these conditions.

Since we can have the equation (3.9.0) in the form of (3.9.13) as it is stated in equation (3.9.14) it can be said that the exactness condition 1 is satisfied regardless of n values.

Let's now apply the 2nd exactness condition. We can define $$M_{v_y}$$ and $$N_{t}$$ as follows,

Before we continue applying the second exactness condition we need to consider the fact that the mass function m(t) is not a constant but a time dependent function. If we look at the figure 3.9.2 there is a graph of m with respect to time. We can clearly see that the function m(t) can be defined as below,

Since m(t) is a straight line function and a constant depending on the time given, we need to check the 2nd exactness condition for both cases.

We cannot directly say that the equation (3.9.17) does not satisfy the 2nd exactness condition because we do not know the function $$v_y(t)$$. But since $$m'(t)$$ is just a negative number according to the graph, $$nk(v_y)^{n+1}(t)$$ must be a same negative number in order to be exact.
 * When $$ 0 \leq t < t_1, m(t) = -\frac{m_1 - m_0}{t_1 - t_0}t + m_0 $$

The right hand side of the equation becomes 0 since $$N_t = 0 $$ For any n if the constant k is zero, the equation becomes exact. If k is not zero, then n must equal zero to make the equation exact.
 * When $$ t\geq t_1, m(t) = m_1 $$


 * We cannot find integrating factor for both cases mentioned above when n=2 or n=0 because the equation cannot be written in the form,

However depending on the equation we might be able to solve for $$v_y(t)$$ and $$y(t)$$

3.1: m constant
 * n = 0,


 * n=1


 * n=2

Cannot solve this differential equation with Integrating Factor Method.

3.2


 * n=0

Let $$ \frac{1}{m(t)} = n(t) $$,


 * n=1

Let $$ \frac{1}{m(t)} = n(t) $$,


 * n=2

Cannot solve this differential equation with Integrating Factor Method.

Author and References

 * Solved and Typed by -- Seong Hyeon Hong
 * Reviewed by --
 * Two Figures are from the below links: