User:Egm6321.f12.team5/Report 6

= Report 6, Team 5 =

Given
(From lecture notes [http://upload.wikimedia.org/wikiversity/en/d/db/Pea1.f12.sec31.djvu sec 31)

Find
Find $$ y_{xxxxx} $$ in terms of $$ y $$ with respect to $$ t $$

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. To find $$ y_{xxxxx} $$ we will take the derivative of $$ y_{xxxx} $$ with respect to $$ t $$

Applying chain rule and definition of derivatives of exponential functions, respectively, we get: $$ D{f(g(x))}=f'(g(x))g'(x)$$ $$ D{e^{(f(x))}}=f'(x)e^{(f(x))}$$

Since we know that $$ \frac{dt}{dx}=\left(\frac{dx}{dt}\right)^{-1} = e^{-t} $$, we can substitute it into the equation Then distribute and combine the like terms, and obtain the final solution:

Given
(From lecture notes sec 31) Euler(homogeneous)L2-ODE-VC

Its boundary conditions are

Find
Solve (6.2.1) using Method 2 with trial solution: and with boundary conditions. Plot the solution.

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

So we get

The (6.2.1) can be turned into

As $$x^r\neq0$$, we can simplify the equation

So $$r=1$$ or $$r=2$$

Hence two homogeneous solutions are $$y=x^2$$ and $$y=x$$. Combine together we get

According to the boundary conditions, we have

Therefore,

Use Wolfram Alpha plot the solution. (From Wolfram Alpha PLOT)

Given
An Euler Ln - ODE - VC, (From lecture notes sec 30)

and the two methods for solving the same, Method 1 Stage 1 : Transformation of variables

Stage 2 : Trial Solution

Method 2 Trial Solution

Find
Show that the trial solution in $$, Method 1 is equivalent to the combined solution $$, and $$, in Method 1.

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. To show that the two methods are equivalent, initially substituting the $$ x $$ and $$ y $$ from $$, and $$, into $$, From $$,

From $$, the 1st derivative of $$ y $$ is

Similarly, the 2nd derivative of $$ y $$ is

So, the ith derivative of $$ y $$ will be,

Now, substituting the general equations $$ x^i $$ and $$ y^{(i)} $$ in the Ln - ODE - VC

This equation can be further simplified as,

or,

We can prove that the two methods are equivalent if we obtain the same expression as above from the second Method. Considering Method 2

The first derivative of $$ y $$ is

The second derivative of $$ y $$ is

So, the ith derivative of $$ y $$ is

Substituting the above equation in the Ln - ODE - VC

The x term doesn't depend on i, So the above equation can be modified as,

We can see that $$, and $$, are the same. Thus we can conclude that the two methods for solving the Ln - ODE - VC are equivalent.

Given
Consider characteristic equation:

Where $$\lambda$$ is a given number: $$\lambda=5$$ or $$\lambda=7$$

And two Euler L2-ODE-VC:

Find
1.

1) Find a2, a1, and a0 such that (6.4.1) is the characteristic equation of (6.4.2).

2) 1st homogeneous solution:$$\displaystyle y_1 = x^\lambda$$

3) Complete solution : Find c(x) such that $$y(x)=c(x)y_1(x)$$

4) Find the 2nd homogeneous solution $$y_2(x)$$

2.Do the same thing with (6.4.3)

Solution
1.1

First, suppose:

Then we can find:

Substitute the equations into (6.4.2) yields:

So we can get:

Thus,

1.2

Substitute equation (6.4.9) back into (6.4.2) yields:

Thus, we can see that $$y_1=x^\lambda$$ is a solution of (6.4.2)

1.3

In order to find c(x) such that $$y(x)=c(x)y_1(x)$$

We can assume:

Then, substitute the equations above into (6.4.10), we can get:

Integrating (6.4.13), we can get:

1.4

Since the complete solution is

And $$y_1(x)=x^\lambda$$. Thus, the 2nd homogeneous solution should be:

2.1

First, suppose:

Then we can find:

Substitute the equations into (6.4.3) yields:

So we can get:

Thus,

2.2

Substitute equation (6.4.22) back into (6.4.3) yields:

Thus, we can see that $$y_1=e^{\lambda x}$$ is a solution of (6.4.3)

2.3

In order to find c(x) such that $$y(x)=c(x)y_1(x)$$

We can assume:

Then, substitute the equations above into (6.4.23), we can get:

Integrating (6.4.26), we can get:

2.4

Since the complete solution is

And $$y_1(x)=e^{\lambda x}$$. Thus, the 2nd homogeneous solution should be:

Given
(From lecture notes sec 32)

Recall p.12-2) R*3.3

Find
Use the same idea of variation of constants(parameters) to find the particular solution $$y_P(x)$$ after knowing the homogeneous solution $$y_H(x)$$, i.e., consider the following solution $$y(x) = A(x)y_H(x)$$ with $$A(x)$$ being the unknown to be found.

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. First we can put the equation ref 5.8.1 into the particular form

Since we know that the function is a sum of it's particular and homogeneous solution, we can substitute that into ($$) to get

We consider the particular solution $$y(x) = A(x)y_H(x)$$, and substitute into the equation

Using chain rule, we can expand and distribute $$a_0(x)$$

Additionally, using the idea of variation of constants, we can group the terms in the equation so that the homogeneous counterpart is equal to zero (ie: $$y'+a_0(x)y=0$$

Which leaves us with the equation

Solving for A(x), we obtain

Now that $$ A(x) $$ has been found, then we can put it back into the original equation to find the particular solution as follows

Given
(From lecture notes sec 32)

Nonhomogeneous L2-ODE-CC

Find
1.Special IFM to solve for general $$ f(t) $$ 2.Find particular solutions for $$ f(t)=texp(bt)$$ 3.Find particular solutions for $$ f(t)=exp(-t^2)$$ For the coefficients $$ a_0,a_1,a_2 $$, consider two different characteristic equations: (1) $$ (r+1)(r-2)=0 $$ (2) $$ (r-4)^2=0 $$

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. Part 1 Setting the integrating factor has the form $$\begin{align} h(t,y) \end{align}$$ Multiply the integrating factor to both side of the ($$).

Then we can get:

where $$\begin{align} p:= y' \end{align} $$ Then, we can check the 2nd condition of exactness of ($$).

From ($$), we already have

Substitute them into ($$) and ($$), we can get:

Thus, assuming $$ \displaystyle h(t)={e}^{\alpha t} $$, where $$ \alpha $$ is a constant. Then ($$) can be expressed as:

Then we try to reduce ordre of ($$)

It is obvious that ($$) satisfies following equation:

where

Thus

Which can be further transformed as

Using $$ \overline{h}(t)=exp[\int \frac{\overline{a}_0}{\overline{a}_1} dt]=e^{\beta t} $$, where $$ \frac{\overline{a}_0}{\overline{a}_1}=\beta $$ We can get:

Which can be simplied as following:

(1) When $$ \alpha \neq \beta $$

(2) when $$ \alpha = \beta $$

Which can be expressed as: (1) When $$ \alpha \neq \beta $$

Thus $$ y_H(x)={{C}_{1}}{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\beta t}},y_P(x)=\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} $$ (2) when $$ \alpha = \beta $$

Thus $$ y_H(x)={{C}_{1}}{{e}^{-\beta t}}+{{C}_{2}}t{{e}^{-\beta t}},y_P(x)=\frac\int1{\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} $$

Part 2 Since we already get ($$) and ($$), for $$ f(t)=texp(bt) $$, we just substitute the $$ f(t) $$ in ($$) or ($$) by $$ texp(bt) $$ Then we can get:

We can integrate following parts as:

(1) Then when $$ \alpha \neq \beta $$

Further intergrated as:

Which can be got that when $$ \alpha \neq \beta $$ :

(2) When $$ \alpha = \beta $$

Thus when $$ \alpha = \beta $$

Part 3 From the ($$) and ($$), we know $$ \alpha $$ and $$ \beta  $$ satisfy that $$ \alpha \beta =\frac{a_0}{a_2} $$ and $$ \alpha +\beta =\frac{a_1}{a_2} $$ So $$ \alpha $$ and $$ \beta  $$ are two roots for the equation: $$ a_2\alpha ^2-a_1\alpha +a_0=0 $$ While the characteristic equations for ($$) should be $$ a_2\alpha ^2+a_1\alpha +a_0=0 $$ So $$ \alpha $$ and $$ \beta  $$ should equal to $$ -r_1  $$ and $$ -r_2  $$ (1) characteristic equations $$ (r+1)(r-2)=0 $$ and $$ f(t)=exp(-t^2) $$ Thus $$ \alpha $$ and $$ \beta  $$ equal to 1 and -2. Thus

(2) characteristic equations $$ (r-4)^2=0 $$ and $$ f(t)=exp(-t^2) $$ Thus $$ \alpha $$ and $$ \beta  $$ equal to -4 and -4. Thus

Comments
Intial solved by Enze

Given
(From lecture notes sec 35)

Equating Two Forms of $$y_P(x)$$
Two expressions are given below for $$y_P(x)$$,

where

is the Wronskian.

Considering Alternate Solution Forms for Variation of Parameters
Given a linear, 2nd order ordinary differential equation with variable coefficients

assuming a solution of the form

where $$u_1(x)$$ is a known homogeneous solution, the method of variation of parameters can be used to reach the form

Since $$U$$ is missing in ($$), it has been effectively reduced in order and the equation can be solved by the integrating factor method from sec 11.

Equating Two Forms of $$y_P(x)$$
Using the hint that

Show that ($$) and ($$) are equivalent.

Considering Alternate Solution Forms for Variation of Parameters
Discuss the feasibility of the following choices as potential solution forms for use in variation of parameters:

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

Equating Two Forms of $$y_P(x)$$
We start by expanding the left side of the hint ($$) and applying it to ($$). Applying the quotient rule, the left side of ($$) becomes

and substitution for $$h$$ in ($$) gives

A little rearrangement shows that ($$) is in a form that makes it possible to use integration by parts to obtain a form that has single integrals rather than double integrals.

Since integration by parts tells us that

For this problem

So that

and ($$) can be rewritten as shown below

Cancelling the $$u_1(s)$$ terms in the numerators and denominators of the functions inside the integrals of the equation results in

Since $$u_1(x)$$ and $$u_2(x)$$ are not functions of $$s$$ they can be placed inside the integral; functions of $$x$$ are unaffected during the integration w.r.t. $$s$$.

Combining the two integrals into one gives

Realizing that the denominator is the Wronskian as defined by ($$)

we can see that ($$) and ($$) are indeed equivalent.

Considering Alternate Solution Forms for Variation of Parameters
If $$y(x) = U(x) \pm u_1(x)$$, then

Substitution back to ($$) gives

Rearranging to group like terms

($$) shows that we are back at the starting point without gaining anything. Therefore, $$y(x) = U(x) \pm u_1(x)$$ is not a feasible choice of a solution form for use with the method of variation of performance. If $$y(x) = U(x)/u_1(x)$$, then

Substitution back to ($$) gives

which is even more complicated than the original equation. Therefore, $$y(x) = U(x)/u_1(x)$$ is not a wise choice for the solution form either. If $$y(x) = u_1(x)/U(x)$$, we can see that we are back to the original solution form used in variation of parameters if we consider that there is a function $$W(x)$$ such that

To find $$U(x)$$, we would solve the problem using variation of parameters to determine $$W(x)$$ and find its reciprocal. Ultimately, $$y(x) = u_1(x)/U(x)$$ is equivalent to $$y(x) = u_1(x)U(x)$$, and is a feasible choice for the variation of parameters method.

Comments
initial soln by --Egm6321.f12.team5.kim (talk) 04:38, 10 November 2012 (UTC)

Given
An L2 - ODE - VC of the form, (From lecture notes sec 35)

Its first homogenous solution is,

The trial solution is

and, the characteristic equation is,

The roots of the characteristic equation are

Find
Explain why $$ r_2(x) $$ is not a valid root, i.e. $$ u_2(x) $$ is not a valid solution.

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. $$ r_2(x) $$ is not a valid solution of the characteristic equation because, we initially derive the roots by making the assumption that $$ r $$ is a constant. So if $$ r $$ becomes a function of $$ x $$ then the roots become invalid. We know that,

Therefore,

Now, substituting the values of $$ y' $$ and $$ y'' $$ obtained from the above equation into the L2 - ODE - VC Differentiating with respect to $$ x $$

And,

Now, substituting the above equations in the L2 - ODE - VC $$,

This is the characteristic equation for $$ r_2(x) $$ We can see that the equation $$, is not of the form of the characteristic equation $$, So we can conclude $$ r_2 $$ is not a valid root.

Given
(From lecture notes sec 35) For the L2-ODE-VC

Find
Select a valid homogeneous solution, and call it $$u_1$$. Find the 2nd homogeneous solution $$u_2(x)$$ by variation of parameters, and compare 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.

Part one
Assume

So

So the (6.9.1) becomes

As $$e^{rx}\neq0$$, we get

So So $$r=1$$ or $$r=\frac{1}{x-1}$$

So we get a valid homogeneous solution

Part two
Variation of parameters method

(From lecture notes sec 34) (From lecture notes sec 34-5) Use Wolfram Alpha to solve, we get

(From Wolfram Alpha Wolfram Alpha)

Use Wolfram Alpha to solve, we get

(From Wolfram Alpha Wolfram Alpha) If we set $$k_1=1$$ and $$k_2=0$$, we get

Which is the 2nd homogeneous solution.

Part three
The 2nd homogeneous solution $$u_2(x)=x$$ compare to $$e^{xr_2(x)}$$.

we can easily find $$r_2(x)=\frac{\ln(-x)}{x}$$

Given
Consider the following integrals: Here $$z(t):=v_{y}(t)$$ And n=2, a=k/m=2, b=mg=10

Do the same with n=3.

Find
For each value fo time t, solve for velocity z(t), then plot z(t) versus t. Also plot the altitudey(t) vs time and find the time the projectile hits the ground.

Solution
For n=3, We can solve for altitude z(t) in Matlab:

Assume the initial velocity of the projectile is 50m/s



Given
(From lecture notes sec 37)

Legengre differential equation:

For n=2

First homogeneous solution:

Find
Show the second homogeneous solution:

Using the Method of Variation of Parameters.

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. We can rewritte the ($$) as follows:

Let $$y(x)=U(x)u_1(x) $$ and $$u_1(x)=P_2(x) $$ Then we can get:

Where $$y=Uu_1 $$ $$y'=U'u_1+U{u_1}' $$ $$y=U{u_1}+2U'{u_1}'+U''u_1 $$ To reduce the order of ($$), we set $$ Z=U' $$

Using the integrating factor method, we first finding the integrating factor, h(x), as follows:

Since we already get $$a_1=\frac{-2x}{1-x^2} $$ and $$ u_1=P_2 $$, we can slove h(x) as follows:

(From lecture notes sec 34) we can get that

So

Further simplied, we can get:

Hence proved.

Comments
Intial solved by Enze

Given
(From lecture notes sec 37) Given a non-homogeneous linear 2nd order ordinary differential equation with variable coefficients

and a homogeneous solution, $$u_1(x)$$, the method of variation of parameters can be used to solve for the complete solution. The process is outlined below. First assume that the solution is of the form:

where $$U(x)$$ is the remaining parts of the solution that is being sought. For the given trial solution, the first and second derivatives are:

Substituting $$y(x)$$ and its derivatives into ($$) results in

where the bracketed term multiplying $$U$$ goes to zero since $$u_1$$ is a homogeneous solution, and '(x)' has been omitted for brevity. Making the substitution

into ($$) gives,

rearranging ($$)

which can be solved using the integrating factor method from lecture notes sec 11. Using the method outlined in sec 11, the integrating factor, $$h(x)$$, and the solution, $$Z(x)$$, are determined to be

Given the relation in ($$), $$U(x)$$ is given by

and $$y(x)$$ is obtained by substituting ($$) into ($$), which leads to

Find
Find the solution for the Legendre equation

given that

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. First rearrange ($$) into the form in ($$)

From ($$) we can see that

Using ($$), the integrating factor, $$h(x)$$, is found to be

and using ($$) $$Z(x)$$ is found to be

$$U(x)$$ can be found by plugging in ($$) into ($$)

Partial fractions method can be used to expand the fractions so that they can be integrated.

Using the values of A, B, and C, ($$) becomes

Similarily,

Plugging in the values for A and B into ($$)

Substituting ($$) and ($$) into ($$) and integrating

Now that $$U(x)$$ is known, $$y(x)$$ is determined by substituting ($$) into ($$)

Comments
initial soln by --Egm6321.f12.team5.kim (talk) 04:54, 11 November 2012 (UTC)

= Contributors = Report Leader: Enze Zhu Report Co-Leaders : Yi Zhongwen,Vishnu