User:Egm6321.f12.team06.gu/report6

==Problem *6.3: Equivalence between two methods for solving Euler Ln-ODE-VC ==

Problem Statement
Show that the trial solution (5) p.30-4 in Method 2:
 * $$ y=x^r$$, r=constant

is equivalent to the combined trial solutions (3)-(4) p.30-4 in Method 1:

Stage 1: Transformation of variables
 * $$ x=e^t$$

Stage 2: Trial solution
 * $$ y=e^{rt} $$, r=constant

Given
Euler Ln-ODE-VC

And its compact form

Nomenclature

 * $$ y'=y^{(1)}=\frac{dy}{dx}, y=y{(0)} $$
 * $$ y_t=\frac{dy}{dt} $$

Ln-ODE-VC: Nth order linear ODE with varying coefficients

Ln-ODE-CC: Nth order linear ODE with constant coefficients

Step 1 Method 1
First substitue $$ x $$ with $$ e^t$$, and it has been proved in the lecture notes that


 * $$ \cdots $$

Then substitute $$ y(t) $$ with the trial solution $$ e^{rt}$$ into ($$), which yields

Bringing ($$) to ($$) into ($$) further gives


 * $$ \cdots $$

Following the pattern from ($$) to ($$), we could derive the general expression for a single term in($$):

Thus after using trial solution in Method 1,Euler Ln-ODE-VC($$) is transformed into

Dividing ($$) by $$ e^{rt} $$ yields

Which is the characteristic equation for Euler Ln-ODE-VC

Step 2 Method 2
Using the trial solution $$ y=x^r$$ gives

Following the pattern from ($$) to ($$), we could also derive the general expression for a single term in($$):

Thus after using trial solution in Method 2,Euler Ln-ODE-VC($$) is transformed into

Dividing ($$) by $$ x^r $$ yields

Which is the characteristic equation for Euler Ln-ODE-VC


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Obviously, ($$) is the same as ($$), we can conclude that the trial solution in Method 2 is equivalent to the combined solutions in Method 1
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Part 2 n=3
2.1 Find z(t)

When n=3, it is difficult to solve the integral in ($$) by hand, thus we need to use hypergeometric function to approximate the result

In ($$), we are given that


 * $$ \int\frac{dz}{az^n+b}= \frac{1}{b}\,z\,_2F_1 \left(1,\frac{1}{n};1+\frac{1}{n};-a\,\frac{z^n}{b}\right)+k $$

Substituting a,n,b with 2,3,10 yields

Now ($$)can be re-written as:

Next, we want to calculate the value for each term in ($$)

Use Pochhammer's symbol to expand hypergeometric function into power series, and truncate to approximate:
 * $$ \frac{1}{10}\,z\,_2F_1 \left(1,\frac{1}{3};\frac{4}{3};-\frac{z^3}{5} \right)= \frac{1}{10}\,z\,[1-\frac{1\cdot \frac{1}{3}}{\frac{4}{3}}\frac{\frac{z^3}{5}}{1!}+ \frac{1\cdot2\cdot\frac{1}{3} \cdot\frac{4}{3}}{\frac{4}{3}\cdot \frac{7}{3}}\frac{(\frac{z^3}{5})^2}{2!}-\frac{1\cdot2\cdot3\cdot\frac{1}{3}\cdot\frac{4}{3}\cdot\frac{7}{3}}{\frac{4}{3}\cdot\frac{7}{3}\cdot\frac{10}{3}}\frac{(\frac{z^3}{5})^2}{3!}+\cdots] $$

Initial value for the L.H.S. of ($$) can be calculated in Wolfram Alpha by typing "50/10 2F1(1,1/3;4/3;-1/5 50^3)":

($$) is re-written as

Since there is no explicit form of z in terms of t, we cannot solve for z(t) analytically.

Alternatively, we could solve for a corresponding value of z for each value of t in Wolfram Alpha:
 * $$ t=0, z=50 $$
 * $$ t=0.08, z=1.439 $$
 * $$ t=0.09, z=1.2779 $$
 * $$ t=0.1, z=1.14 $$
 * $$ t=0.15, z=0.5722 $$
 * $$ t=0.2, z=0.067 $$
 * $$ t=0.3, z=-0.8976 $$
 * $$ t=0.4, z=-1.5147 $$
 * $$ t=0.5, z=-1.7948 $$
 * $$ t=0.6, z=-1.9516 $$
 * $$ t=0.7, z=-2.0577 $$
 * $$ \cdots $$

Invoke Matlab to plot z against t:

Figure: Velocity vs. time



Let's use MotionGenesis and Matlab to verify the accuracy of previous steps

Numerical values for velocity z(t):
 * {| class="prettytable"


 * Time t (s)
 * Velocity z(t) (m/s)


 * 0
 * 50


 * 0.0010
 * 15.0712


 * 0.0020
 * 10.9025


 * 0.0030
 * 8.9679


 * 0.0040
 * 7.7923


 * }

Figure: Velocity vs. time



From the above figure, one can tell that hypergeoetric approximation is accurate after 0.08 sec, however there is an obvious error during the time interval 0-0.08 sec between hypergeoetric approximation and computer results. This is caused by failure in deriving any value of z when t is less than 0.08 sec from ($$) using Wolfram Alpha.

2.2 Find y(t)

Add the following codes to matlab

Numerical values for altitude y(t):
 * {| class="prettytable"


 * Time t (s)
 * Altitude y(t) (m)


 * 0.5080
 * 0.0030


 * 0.5090
 * 0.0013


 * 0.5100
 * -0.0003


 * 0.5110
 * -0.0020


 * 0.5120
 * -0.0037


 * }

Figure: Altitude vs. time




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From the above figure and table of numerical values, we can tell that when $$ t=0.51 s, y=-0.00003496\approx 0 $$, which means that the particle returns to ground at the time 0.51s
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