User:Egm6321.f12.team2/Report6

=Introduction=

Summary
This report explores various topics including linear nth-order differential equations in various forms including varying and constant coefficients, different orders, hyper-geometric functions, and proofs of exactness.

Also explored are the use of trial solutions for solving differential equations.

Team Members
Zhe Wang Yitien Yan Ranjit Alexander Karl Van Wyk Anubhav Mohan

=R*6.1 - Find $$y_{xxxxx}$$=

Given
and,

Find
Find $$y_{xxxxx}$$

Solution
We must use chain rule to solve for $$y_{xxxxx}$$ as was done for ($$) and ($$). Since,

where,

so,

And further solving for $$y_{xxxxx}$$ we get,

using product rule,

and after simplification, the final solution is:

Given
and,

With the following boundary conditions:

Find
Find $$y(x)$$ and plot the solution for values of $$x$$.

Solution
Utilizing ($$) we can find $$ y' $$ and $$ y'' $$ :

Now replacing ($$), ($$) and ($$) into ($$) we get:

Which can be simplified to:

where, the $$x^r$$ can be divided out resulting in:

Now, to find $$ y(x) $$ we must use the following definition:

where, $$ r_1 $$ and $$ r_2 $$ are, the two roots of ($$), respectively.

Solving for the roots of ($$),

Which means that $$ r_1 = 1 $$ and $$ r_2 = 2 $$. Then, replacing these values into $$x$$ in ($$):

Now, placing boundary conditions on ($$), we get:

The values $$ c_1 $$ and $$ c_2 $$ can be solved for as two equations two unknowns, resulting in:

So the final equation is:

Which can be plotted as shown below:

The above image was plotted at FooPlot.com

Given
Two methods for solving Euler Ln-ODE-VC: Method 1: Stage 1: Transformation of variables Stage 2: Method 2: Trial Solution

Find
Show that Method 1 is equivalent to Method 2.

Solution
Substituting ($$) into ($$) yields ($$):

Given
The following characteristic equation,

the following Euler L2-ODE-VC,

the first homogeneous solution,

and the following Euler L2-ODE-CC, .

1.1
Find $$a_2,a_1,a_0$$ such that ($$) is indeed the characteristic equation of ($$)

1.2
Find the complete solution to ($$),

1.3
Find the second homogenous solution, $$y_2(x)$$

1.4
Find $$a_2,a_1,a_0$$ such that ($$) is indeed the characteristic equation of ($$)

1.5
Find the complete solution to ($$),

1.6
Find the second homogenous solution, $$y_2(x)$$

1.1
Rewriting the characteristic equation yields, Since ($$) is an Euler L2-ODE-VC, we must transform the variable with $$x=e^t$$ which implies that the characteristic equation corresponds to, Using the relations previously established in lecture, we know that from (2) p.30-7, and from (5) p.30-6, .

Rearranging ($$) and ($$) and using $$x=e^t$$ yields,

Substituting ($$) and ($$) into ($$) yields, .

Therefore, $$a_2=1,a_1=-9,a_0=25$$.

1.2
The complete solution is $$y(x)=c(x)y_1(x)$$ From the previously developed relations in lecture, we know the following:

From (2) p.34-5,

From (1) p.34-5,

From (4) p. 34-4,

Since ($$) is homogeneous, $$f(x)=0$$. Also, $$a_1(x)=\frac{-9}{x}$$. Therefore,

Furthermore,

and,

Finally,

1.3
Distributing ($$) yields,

where the second homogeneous solution is

1.4
Since ($$) has constant coefficients, the coefficients are obtained directly from the characteristic equation $$b_2=1,b_1=-10,b_0=25$$

1.5
The previously mentioned relations are used again to obtain the full solution,

Following,

Then,

Finally,

1.6
Distributing ($$) yields,

Therefore, the second homogeneous solution is

Find
The particular solution using Variation of Parameters method

Solution
From ($$), $$\displaystyle y_H(x)'= -P(x)\cdot exp[-\int P(x)dx]=-P(x)y_H(x)$$ Thus, we get Next, plug ($$) into ($$), we get: Thus, we get the particular solution:

Given
Nonhomogeneous L2-ODE-CC

1
Find the PDEs that govern the integrating factor $$\displaystyle h(t,y) $$ for equation ($$). Can you solve these PDEs for $$\displaystyle h(t,y) $$?

2
Trial solution for the integrating factor

where $$\displaystyle \alpha$$ is unknown to be determined.

Because of the integrating factor in exponential form, assume the LHS of ($$) take the form :

2.1
Find $$\displaystyle (\bar a_1,\bar a_0)$$ in terms of$$\displaystyle (a_0,a_1,a_2)$$.

2.2
Find the quadratic equation for

2.3
Reduced-order equation: ($$)and ($$) lead to

2.4
Use the IFM to solve ($$)

Find the solution y(t) for general excitation f(t).

2.5
Show that

Thus $$\displaystyle (\alpha,\beta)$$ are roots of the quadratic equation:

2.6
Deduce the particular solution $$y_p(t)\,$$ for general excitationf(t).

2.7
Verify result with table of particular solution for:

2.8
Solve the nonhomogeneous L2-ODE-CC ($$) with the Gaussian function:

For the coefficients $$(a_0,a_1,a_2)$$, consider two different characteristic equations:

2.9
For each case in ($$)and ($$), determine the fundamental period of undamped free vibration, and plot the solution for the excitation for about 5 periods, assuming zero initial conditions.

1
Multiply ($$) by the integrating factor h(t,y): Recall the two relations in the 2nd exactness condition for N2-ODEs:

Calculate the following components in the above exactness condition:

Plugging those into the 2nd exactness condition:

From the second equation, we can obtain that

which means that integrating factor $$\displaystyle h(t,y)=h(t)$$. Substituting the expression of h(t) into the first equation of 2nd exactness condition, we can obtain an L2-ODE-CC,

Therefore, that PDE reduces into an ODE, which can be solved.

2.1
Differentiate the RHS of($$),

Compare the coefficient with the integrand on LHS of ($$),

We can find $$\displaystyle (\bar a_1,\ \bar a_0)$$ in terms of $$\displaystyle (a_0,\ a_1,\ a_2)$$ that

2.2
Substituting $$\displaystyle \bar a_0$$ and $$\displaystyle \bar a_1$$ into $$\displaystyle \bar a_0+\alpha \bar a_1=a_1$$, we can get that <\br>

Rearrange it, we obtain the quadratic equation for $$\displaystyle \alpha$$:

2.3
Therefore, the reduced-order equation is obtained

Rewrite it,

2.4
Recall the IFM of general non-homogeneous L1-ODE-VC,

The integrating factor is

The solution is

Divided ($$) by $$\displaystyle \bar a_1$$,

Then we can obtain the integrating factor

Therefore, the solution of ($$)is

2.5
Substitute $$\displaystyle \bar a_1=a_2\,;\,\bar a_0+\alpha \bar a_1=a_1\,;\,\alpha \bar a_0=a_0$$ into $$\displaystyle \alpha$$ and$$\displaystyle \beta$$, we can obtain that

Thus, $$\displaystyle (\alpha,\beta)$$ are the roots of the following quadratic equtation:

Obviously, it is the same as the quadratic equation obtained by $$\mathbf{2.2}$$.

2.6
Rearrange the solution of ($$)

$$\displaystyle y_{(t)}=\frac{1}{e^{\beta t}}\int^t e^{\beta s}\cdot\frac{e^{-\alpha s}}{\bar a_1}\left[\int^s e^{\alpha \tau}f(\tau)d\tau\right]ds$$ $$\displaystyle =e^{-\beta t}[\int^t\frac{e^{(\beta-\alpha)s}}{\bar a_1}\int^s e^{\alpha \tau}f(\tau)d\tau]ds$$ $$\displaystyle =\frac{e^{-\beta t}}{\bar a_1}[\int^t e^{(\beta-\alpha)s}[\int e^{\alpha s}f(s)ds+k_1]ds]$$ $$\displaystyle =\frac{e^{-\beta t}}{\bar a_1}[\int^te^{(\beta-\alpha)s}[\int e^{\alpha s}f(s)ds]ds+\int^t e^{(\beta-\alpha)s}k_1ds]$$ $$\displaystyle =\frac{e^{-\beta t}}{\bar a_1}[\int e^{(\beta-\alpha)t}[\int e^{\alpha s}f(s)ds]dt+k_2+\frac{k_1}{\beta-\alpha}e^{(\beta-\alpha)t}]$$ $$\displaystyle =\underbrace{\frac{k_1}{\bar a_1(\beta-\alpha)}e^{-\alpha t}}_{\displaystyle C_1 y_H^1}+\underbrace{\frac{k_2}{\bar a_1}e^{-\beta t}}_{\displaystyle C_2 y_H^2}+\underbrace{\frac{e^{-\beta t}}{\bar a_1}\int e^{(\beta-\alpha)t}[\int e^{\alpha s}f(s)ds]dt}_{\displaystyle y_P}$$

The particular solution $$\displaystyle y_P(t)$$ for general excitation $$\displaystyle f(t)$$ is

2.7
The particular solution for an arbitrary $$\displaystyle f(x)$$ is given directly above in 2.6. The solutions for $$\displaystyle f(x)=t^2$$, as evaluated by Wolfram-Alpha is:

The solution as found in Table 3.6.1 on page 1 of the following link (note use $$\displaystyle s=0$$ accordingly):

Choosing the coefficients $$\displaystyle A$$ wisely yields a perfect match. Now check solutions for $$\displaystyle f(x)=te^{bt}$$, again use Wolfram-Alpha:

The solution for this case also is found in the same link above:

In a similar way, a careful choice of coefficients proves equivalency.

2.8
If $$\displaystyle f(t)=\exp (-{{t}^{2}})$$, then


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

$$ \displaystyle \begin{align} & y_{H}^{1}=\left\{ \begin{align} & {{e}^{-\alpha t}}....................................................................................(\alpha \ne \beta ) \\ & t{{e}^{-\alpha t}}...................................................................................(\alpha =\beta ) \\ \end{align} \right. \\ & y_{H}^{2}={{e}^{-\beta t}} \\ & {{y}_{P}}=\left\{ \begin{align} & \frac{\sqrt{\pi }\left[ {{e}^{\frac{4}-\alpha t}}erf(\frac{\alpha }{2}-t)-{{e}^{\frac{4}-\beta t}}erf(\frac{\beta }{2}-t) \right]}{2{{a}_{2}}(\alpha -\beta )}........................(\alpha \ne \beta ) \\ & -\frac{1}{2{{a}_{2}}}\sqrt{\pi }{{e}^{\frac{4}-\alpha t}}\left[ (t-\frac{\alpha }{2})erf(\frac{\alpha }{2}-t)-\frac{1}{\sqrt{\pi }}{{e}^{-\frac{4}}} \right].........(\alpha =\beta ) \\ \end{align} \right. \\ \end{align}$$
 * 
 * }

2.8.1
$$\displaystyle \alpha$$ and $$\displaystyle \beta$$ is the roots of quadratic equation:

Therefore let $$\displaystyle \alpha=-1\,;\,\beta=2$$ As for the excitation f(t) is Gaussian function:

For the general solution: where $$\displaystyle {{C}_{1}}$$ and $$\displaystyle {{C}_{2}}$$ are constants. This is calculated by www.wolframalpha.com.

2.8.2
The quadratic equation is:

Therefore let $$\displaystyle \alpha=\beta=4$$ As for the excitation f(t) is Gaussian function:

For the general solution

where $$\displaystyle {{C}_{1}}$$ and $$\displaystyle {{C}_{2}}$$ are constants. This is calculated by www.wolframalpha.com.

Find
Show that ($$) agrees with King 2003 p.8 (1.6), i.e.,

with

Hint: $$\displaystyle \left( \frac{u_2}{u_1} \right)' = \frac{1}{h} $$ Discuss the feasibility of the following choices for variation of parameters:

Solution
Define $$ G\left(x\right)=\frac{u_{2}\left(x\right)}{u_{1}\left(x\right)} $$ ,thus

$$ h\left(x\right)=\frac{u_{1}^{2}}{u'_{2}u_{1}-u_{2}u'_{1}} $$

From the formula of integration by parts: $$ \int F\left(x\right)G'\left(x\right)dx=F\left(x\right)G\left(x\right)-\int G\left(x\right)F'\left(x\right)dx $$ where

and

($$) can then be written as:

After substituting all the terms into ($$):

($$) and ($$) are identical, which shows that (1) p.34-6 agrees with King 2003 p.8 (1.6). For feasibility of case 1: $$ y\left(x\right)=U\left(x\right)\pm u_\left(x\right) $$, $$ y'={\frac{d}{dx}}U\left(x\right)\pm {\frac{d}{dx}}u_\left(x\right) $$, and $$ y''={\frac{d^{2}}{d{x}^{2}}}U\left(x\right) \pm {\frac{d^{2}}{d{x}^{2}}}u_\left(x\right) $$ $$ a_{0}y+a_{1}y'+y''=a_U\left(x\right) \pm a_{\frac{d}{dx}}U\left(x\right) \pm {\frac{d^{2}}{d{x}^{2}}}U\left(x\right) $$ For feasibility of case 2: $$ y={\frac {U \left( x \right) }{u_ \left( x \right) }} $$, $$ y'={\frac { \left( {\frac {d}{dx}}U \left( x \right) \right) u_ \left( x \right) -U \left( x \right) {\frac {d}{dx}}u_ \left( x \right) }{ \left( u_ \left( x \right)  \right) ^{2}}} $$, and

For feasibility of case 3: $$ y={\frac {u_ \left( x \right) }{U \left( x \right) }} $$, $$ y'=-{\frac { \left( {\frac {d}{dx}}U \left( x \right) \right) u_ \left( x \right) -U \left( x \right) {\frac {d}{dx}}u_ \left( x \right) }{ \left( U \left( x \right)  \right) ^{2}}} $$, and

None of the given trial solutions are feasible for substituting the trial solutions into a Non-homogenous L2-ODE-VC can not produce a general non-homogenous L1-ODE-VC. Several terms contain $$U(x)$$ still exist after the substitutions.

Given
where $$ \displaystyle r_2(x) = \frac{1}{x-1} $$, and the trial solution: $$ y=e^{r_2(x)} $$

Find
Explain why $$ r_2(x) $$ is not an valid root.

Solution
Substitute $$y,y',y''$$ into ($$): $$ r_2(x) $$ is not an valid root because any root of the characteristic equation has to be a constant. And as we can see, ($$) is not equal to zero for every $$ x $$.

Given
For the given ODE, Select a valid valid homogeneous solution and call it $$u_1$$

Find
Find the second homogeneous solution $$u_2(x)$$ by variation of parameters and compare it to $$e^{xr_2(x)}$$.

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. Let $$u_1=e^{rx}$$, So inputing this into equation 6.9.2, gives Eliminating $$e^{rx}$$ gives This implies $$r=1$$ or $$\frac{1}{x-1}$$. Taking the constant value here gives $$r=1$$.Hence To find $$u_2(x)$$,rearranging equation 6.9.1 gives, Here $$a_1(x)=\frac{-x}{x-1}$$ and $$a_0(x)=\frac{1}{x-1}$$ and the equation for u_2(x) is, To evaluate this integral, Substituting in the eqn232, gives Doing integration by parts gives, Comparing $$u_2(x)$$ and $$e^{xr_2{x}}$$ gives, Applying Log on both sides gives This implies $$r_2(x)$$ is

Given
The governing equation for a particle moving with air resistance with initial velocity,$$z(0)=50$$ is

This relation comes from equating the unbalanced forces on the particle.ie$$mz'+kz^n=mg$$, where $$m$$ is mass, $$z$$ is the verticle velocity,$$v_y$$,$$k$$, the proportionality constant for air resistance and $$g$$ is the acceleration due to gravity. Here $$a=\frac{k}{m}=2,b=g=10$$

Also the solution to the integral is given as

Where

For each value of $$n$$,find the vertical velocity $$z(t)$$ v/s $$t$$. Plot the function.Also find and plot the altitude $$y(t)$$ v/s time,$$t$$, then find the time that the projectile returns to the ground.

If an explicit expression for $$z(t):- v_y(t)$$ cannot be obtained, use a numerical method to find $$z(t)$$ for each given value of time t for $$n=2,3.$$

Find
For each value of $$t$$, solve for ,$$z(t)$$ and then plot $$z(t)$$ versus $$t$$ for values of $$n=1,2,3$$. Find the time when the projectile hits the ground in each case.

For n=1
Here the equation of motion becomes becomes, Substituting values for $$\frac{k}{m}$$ and $$g$$ gives, Solving for the vertical velocity,$$z(t)$$ gives, To derive the equation of motion of the particle, i.e the altitude with respect to time we write equation ($$) as,

Integrating both sides gives From the Altitude v/s Time graph, it can be noticed that there is a maximum altitude after which it comes down. Also from the Velocity v/s Time graph it can be noticed after a sufficient period of time the velocity is virtually constant. This velocity is the "terminal velocity", which occurs when the two opposing forces of gravity and air resistance effectively cancels each other. In a body moving which terminal velocity, there is no net force acting on the body and so it moves in a constant velocity.

Equating the velocity to zero, we can get the time at which the altitude is maximum.Thus,from ($$), we get Taking log on both sides, The altitude at this time is got by inputing t=1.2 in eqn 343, Thus, the projectile reaches a maximum height of 19.0053 after a time of 1.2 units. The terminal velocity can be found by doing a limit of time tending to infinity in the equation for velocity. Thus the terminal velocity is -5 units.(Negative since the velocity is pointing downwards). The time when the particle returns to the ground is found by letting y=0 and solving for t in ($$). Doing so gives t=5.5 units. The velocity when the body returns to the ground by substituting t=5.5 units in ($$) Hence the body returns with the terminal velocity.

For n=2
Here the equation of motion becomes, Substituting values for $$\frac{k}{m}$$ and $$g$$ gives, Solving for the vertical velocity,$$z(t)$$ gives, By partial fractions the integral becomes, Integrating,we get Re-arranging, we get the expression for $$z(t)$$ as,

To derive the equation of motion of the particle, i.e the altitude with respect to time we write ($$) as,

Integrating both sides gives

Evaluating this integral is beyond me.An alternate expression for $$z(t)$$ is obtained from wolfram alpha for ($$) as,

($$) is the expression for velocity. From the plot of the graph, it can be noticed it is not continuous everywhere.It has discontinuities at intervals of $$\frac{\pi}{2\sqrt 5}$$.Also a closed form solution for the altitude can be obtained by integrating ($$) as follows, From $$\int tan(t)=-log(cos(t))$$, we get, This equation will also have discontinuities at intervals of $$\frac{\pi}{2\sqrt 5}$$.The plots of both Velocity v/s Time and Altitude v/s Time is given below. From the velocity plot, it is obvious that the particle reaches zero velocity when the term within the tan function in velocity ($$) becomes zero.That is, the velocity is zero, and hence the altitude is maximum at $$1.526-2t\sqrt5=0$$. This is at t=0.3412 units and this is evident in the velocity graph.Now considering the fact that the velocity graph is symmetrical about the y-axis at t=0.3412, the velocity graph extends to the right for a t=$$\frac{\pi}{4\sqrt 5}=$$0.3512 more.Hence the time when the particle reaches the ground is $$t=0.3412+0.3512-(0.3512-0.3412)=0.6824$$ units. The value $$(0.3512-0.3412)$$ is reduced from the total to account for the fact that the velocity starts from 50 units and not from infinity.The velocity of the particle when it reaches the ground is given by entering the value of time when the particle hits the ground to the velocity equation.Hence,

For n=3
From the values of $$a=2, b=10, n=3$$ the solution for $$z(t)$$ is

Then,

This approximates to,

The equation ($$), is an implicit equation in $$z(t)$$.An explicit equation cannot be obtained fro this relation for $$z(t)$$. The graph of ($$),which is the Velocity v/s Time graph, for all real values of y are given below.OF our interest, is the part of the plot drawn in red. For points ear zero, the plot cannot be obtained because the drop in velocity is so steep that the slope tends to infinity. So,the point closest to the y axis is obtained at t=0.0712 units. So the graph is only relevant from t=0.0712 to the time when the particle reaches the ground. From the graph it is also obtained the time,t=0.2067 units, when the velocity is zero(This is also obtained by entering z(t)=0 in ($$) . Hence, the maximum altitude is given by the area under the curve from t=.0712 to 0.2067 which is obtained by approximating the graph between these two points as a line.Various values of z(t) for differing values of t are obtained in the table below.

As can be seen from the values, velocity decreases as time increases.The velocity profile can be approximated to a line graph.From the above table of values the equation is line regressed and is approximated to z=-10.25t+2.15. This is illustrated in the figure below by plotting $$z(t)$$ versus time $$t$$. From the approximation, the equation for altitude is obtained by integrating the velocity function. Thus To obtain the time at which it reaches the ground equate ($$) to 0. Thus the body returns back after 0.42 time units.The maximum altitude is given by inputing the value of t=0.02067 in the altitude equation ,($$),which gives y=0.225 units.



Solution
Since $$ P_{2}\left(x\right) $$ is a homogeneous solution to the Legendre's equation with $$ n=2$$, the method of reduction of order can be applied to find the second homogeneous solution to the Legendre's equation. The Legendre's equation takes the form

From the reduction of order formula, King 2003 p.6 (1.3),i.e.,

where $$ u_{1}\left(x\right) $$ in this case is equal to $$ P_{2}\left(x\right) $$ The result shows that $$u_{2}\left(x\right)$$ derived from the method of reduction of order is the same as $$Q_{2}(x)$$ given in the problem.

Given
The following non-homogeneous Legendre equation L2-ODE-VC,

and its first homogeneous solution, .

Find
The final solution, $$y(x)$$ by variation of parameters

Solution
First, ($$) is rewritten as the following,

The complete solution is $$y(x)=U(x)P_1(x)$$. From the previously developed relations in lecture, we know the following:

From (2) p.34-5,

From (1) p.34-5,

From (4) p. 34-4,

Therefore,

Following,

Next,

Finally,