User:EGM6321.F10.TEAM1.HW5

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

Solution
Author Egm6321.f10.team1.allison 17:08, 21 October 2010 (UTC), EGM6321.f10.team1.russo 16:31, 2 November 2010 (UTC)

Given
and boundary conditions defined as:

Find
A. Solve Eq2.1 using Method 2 (i.e. trial solution $$ y=x^r\ $$) with boundary conditions Eqs 2.2 and 2.3 B. Find coefficients $$ c_1\ $$ and $$ c_2\  $$ and compare with results from (Method 1) C. plot the solution

Solution
A. For two roots, use the form:

Then:

Sustitute eqs 2.4 thru 2.6 into eq2.1:

reduce reduce divide thru by $$ 2x^r\ $$ to get: separate and solve for r:

B) Solve for $$ C_1\ $$ and $$ C_2\  $$ Using andswer from eq2.13 and plugging into eq2.4: Using the given in eq2.2 gives: Using the given in eq2.3 gives: Solving for $$ C_1\  $$ and $$ C_2\  $$ reults in the following: Substituting eq2.17 into eq2.4 yields:

Comparing results to direct method yields the same solution.

C) graph solution from x = -10 to x = 10:



Author EGM6321.F10.TEAM1.HW5 02:43, 25 October 2010 (UTC)EGM6321,F10.TEAM1.WILKS

Given
Consider the characteristic equation: and given:

Find
For an Euler L2_ODE_VC 1.1) Find $$ a_2\ $$, $$ a_1\ $$ and $$ a_0\ $$ such that eq.(3.1) is a characteristic equation of eq.(3.2) 1.2) Find the first homogeneous solution: $$ y_1=x^2\ $$ 1.3) Find the second homogeneous solution: $$ y_2=c(x)\ $$ 1.4) Find the general homegenous solution

For an Euler L2_ODE_CC 2.1) Find $$ b_2\ $$, $$ b_1\ $$ and $$ b_0\ $$ such that eq.(3.1) is a characteristic equation of eq.(3.3) 2.2) Find the first homogeneous solution: $$ y_1=x^2\ $$ 2.3) Find the second homogeneous solution: $$ y_2=c(x)\ $$ 2.4) Find the general homegenous solution

Solution
1.1) Do trial solution method 

Substitute eq(3.4),(3.5) and (3.6) into eq(3.2).

Compare eq(3.10) to eq(3.3). So,

Author team1. Taeho Kim

1.2) Substitute 1st homogeneous solution into eq(3.2) 

The left side hand of eq(3.19) is zero.

So, eq(3.14) is a homogeneous solution of eq(3.2).

Author team1. Taeho Kim

1.3) Euler variation of parameter 

Substitute eq(3.21), (3.22) and (3.23) into eq(3.2).

Since,

So, u(x) will be missing.

Substitute eq(3.13), (3.17) and (3.18) into (3.27).

Let,

Then, we get reduced order equation of L1-ODE-VC.

The eq(3.32) is exact.

So,

Author team1. Taeho Kim

1.4) General homogeneous equation 

Author team1. Taeho Kim

2.1) Trial solution method  Let,

Substitue eq(3.44), (3.45) and (3.46) into eq(3.3).

Author team1. Taeho Kim

2.2) 1st homogeneous solution  Let,

Substitute eq(3.52) into eq(3.3).

The left side hand of eq(3.53) is zero.

So, eq(3.52) is a homogeneous solution of eq(3.3).

Author team1. Taeho Kim

2.3) 2nd homogeneous solution 

Substitute eq(3.55), (3.56) and (3.57) into eq(3.3).

Since,

Substitute eq(3.51) and (3.52) into (3.61)

Author team1. Taeho Kim

2.4) General homogeneous solution 

Author team1. Taeho Kim

Find
Use the idea of variation of parameters/constants to find the particular solution $$ y_P\ $$ after knowing the homogeneous solution $$ y_H\ $$

Solution
First let us motivate the problem by consider under what type circumstance would we find ourselves knowing only the homogeneous solution and needing to search for the particular solution. Consider the following general non-homogeneous first order differential equation with vary coefficients.

Now consider the homogeneous case of this equation. The homogeneous form of this equation may be directly integrated by separation of variables.

Now we see that we indeed have a homogeneous solution to the above general first order non-homogeneous differential equation. In considering the search for the particular solution, we make a guess that the total solution of y(x), which will contain the particular solution will be the product of our homogeneous solution and some function for x.

This type of relationship may seem like a reasonable guess when thinking about the relationship between a general first order non-homogeneous differential operator with constant coefficients and an associated particular solution to that operator. We know that the families of input functions are defined by the characteristic values of the particular solution and the polynomial factor of the general exponential function solution is determined by the nature of the differential operator. Therefore it's a natural guess to think that our total solution is a product of some function x and the homogeneous solution, and push ahead to explore the characteristics of such a solution under the differential operator. Now plug the assumed solution and it's derivative back into the differential operator. Because we are looking for some details on how the differential operator operates on our proposed solution, we want to rearrange the equation into proper form with respect to A(x), i.e. we already know the image of homogeneous solution under the differential operator is zero.

We can see if we look closely that the factor multiplying A(x) is the same form as our originally posed homogeneous problem. We already know the image of the homogeneous solution under the differential operator is zero. This can now be solved directly by integration. Now this is our A(x) that we originally guessed was a factor of the total solution of y(x) times the homogeneous solution of y(x). Let's plug it in and see what form y(x) will take. Therefore, we notice that our y(x) solution is a sum of 2 terms (i.e a linear combination). With comparison of the total solution to the general first order non-homogeneous differential operator with varying coefficients in the literature (such as King et al. p. 512) it's easily seen that the following is true and our guess at the form of the total y(x) solution was correct.

Author Team1. Paul Lang

Given
Equation of Motion

Find
1) Find the PDE's for the integrating factor $$ h(t,y)\ $$ 2) Find the trial solution for the integrating factor $$ h(t)=e^{ \alpha\ t}\ $$ 3) Find $$ \bar a \ _1\ $$ and $$ \bar a \ _0\ $$ in terms of $$ a_2\ $$, $$ a_1\ $$ and $$ a_0\ $$ 4) Find the quadratic equation for $$ \alpha\ \ $$ 5) Solve the reduced order equation equation: 6) Use the integrated factor mathed to find y, for general excitation $$ f(t)\ $$ 7) Show that and

8) Deduce the expression for a particular solution $$ y_P\ $$ for general excitation $$ f(t)\ $$ 9) Verify with table of particular solutions (consider $$ f(t)=t^2\ $$ ) 10) Solve L2_ODE_CC, using:

5.1 Find the PDE's for the integrating factor
First, put the equation into the standard form

From here, y' is defined as p to reduce the order. Next the terms are defined as:

Then the two pde's to test for exactness are used. These are:

and

Directly substituting into this will yield:

But since all of the a's are given to be constants there derivatives will be zero. Similarly,in this case, derivatives with respect to p and y all are constants hence the second derivative is zero. Therefore, equation 5.12 and 5.13 can be simplified to:

and the trivial

respectively.

5.2 Find the trial solution for the integrating factor
Using these, the original ODE can be reduced to:

Dividing to get in the right form then the equation:

is acquired as:

This can be solved with the given integrating factor as:

This can then be integrated again to yield the original variable, y

This is not solvable for a generic f(t), but given a specific case it could be. Solving only the homogenous solution will yield:

where b and c are additional constants dependent on initial conditions.

5.3 Find $$ \bar a \ _1\ $$ and $$ \bar a \ _0\ $$
assume $$\bar a\_2 $$ is zero and differentiate.

compare this to the given equation

will give the combined equation

From here, the coefficients can be solved easily.

5.4 Find the quadratic equation for $$ \alpha\ \ $$
using equation 5.27

rearrange this equation as:

5.6 Use the integrated factor mathed to find y
Dividing by the leading coefficient will put the equation in the form of 5.17 as:

for which the solution is:

5.7 prove $$ \alpha \beta $$ relations
combine 5.26, 5.27 and 5.33 then multiply by $$ \alpha $$

start with 5.27.

now combine 5.26, 5.27, and 5.33 seperate and add $$ \alpha $$

Author Egm6321.f10.team1.allison 17:52, 5 November 2010 (UTC)

Given
and

Find
Show that eq 6.1 agrees with with expression in King, page 8, eq 1.6:

Where:

Solution
Equation (6.2) may be rearranged as follows:

Taking the derivative of both sides of equation (6.5) yields:

Carrying out the derivative of the left side of equation (6.6) yields:

Upon inspection, one can notice a relationship between equations (6.7) and (6.4), which may be written as:

It is now possible to develop an expression for $$ h(x)\ $$ by setting equations (6.6) and (6.8) equal to one another and rearranging:

King's expression (equation (6.3)) may be rearranged, and its integration variables substituted, yielding: Where the two homogeneous solutions, namely $$ Au_1(x)\ $$ and $$ Bu_2(x)\ $$ have been omitted so that only the particular solution is considered.

The objective now is to use rearrangement and substitution to put the particular solution in a form equivalent to equation (6.10). The rule of Integration by Parts may be written as follows:

Rearranging this rule into an applicable form yields:

It can easily be seen that the left side of equation (6.12) is of nearly the same form as the particular solution, equation (6.1).

To apply the Integration by Parts rule to the particular solution, the following substitutions may be made: and

Substituting equations (6.13),(6.14) and (6.15) into equation (6.12) yields:

Multiplying both sides of equation (6.15) by $$ u_1(x)\ $$ and substituting equation (6.9) for $$ h(x)\ $$ in the right side of (6.16) yields:

By inspection, the left side of equation (6.17) is exactly the particular solution from equation (6.1). Thus, after rearranging the right side of equation (6.17) and canceling out like terms, the particular solution may be written as:

It is my belief that the given particular solution from lecture 30-1 equation 2 is a misprint. If the particular equation had been given as:

Then equation (6.18) would instead have become:

Which is equivalent to King's expression for y(x) as stated in equation (6.10).

Author EGM6321.f10.team1.russo 00:58, 3 November 2010 (UTC)

Given
Trial Solution:

Find
$$ u_1\ $$ and $$ u_2\ $$ of eq 7.1 using trial solution eq 7.2

Solution
First, solve r to satisfy this set of equations.

substitute 7.2-7.4 into 7.1 cancel out exponentials

factor in terms of x

for non-trivial solution satisfy two equations

Put 7.1 in the proper form

such that

Then use the trial solution method to find another solution.

Specifically,

The first integral is carried through to yield: which simplifies to:

The second integral is carried through to yield: which simplifies to:

Author Egm6321.f10.team1.allison 00:33, 1 November 2010 (UTC) 173.24.228.78 05:47, 1 November 2010 (UTC)EGM6321,F10.TEAM1.WILKS 01:57, 2 November 2010 (UTC) Egm6321.f10.team1.allen 02:03, 2 November 2010 (UTC)

Given
Trial Solution: and characteristic equation:

Find
An L2_ODE_VC such that the trial solution is eq 8.1 and the characteristic equation is eq 8.2.

Solution
First, differentiate the given solution.

Now to solve the characteristic equation, multiply 8.2 termwise by exponential function

Assume an exponential solution

Equation 8.6 can be differentiated to yield

Equations 8.6-8.8 are then substituted into 8.5 to get the DE

Substituting back in the given trial solution (equations 8.1 and 8.4) gives Multiplying the entire equation termwise by $$ x^2 $$ and combining in terms of r

Now again using the solution style in 8.6-8.8 will give the final equation

Author Egm6321.f10.team1.allison 15:17, 31 October 2010 (UTC)

Signatures
Solved problem 1 -- EGM6321.F10.TEAM1.WILKS 05:32, 30 October 2010 (UTC)Egm6321.f10.team1.allison, EGM6321.f10.team1.russo 01:17, 3 November 2010 (UTC)

Solved problem 2 -- EGM6321.F10.TEAM1.WILKS 05:17, 30 October 2010 (UTC)EGM6321,F10.TEAM1.WILKS

Solved problem 3 -- EGM6321.F10.TEAM1.Taeho Kim

Solved problem 4 -- 70.185.108.64 07:53, 1 November 2010 (UTC)EGM6321.F10.TEAM1.Lang

Solved problem 5 -- XXXXXXXXXX

Solved problem 6 -- EGM6321.f10.team1.russo 01:17, 3 November 2010 (UTC)

Solved problem 7 -- EGM6321.F10.TEAM1.ALLISON 11:17, 31 October 2010 (UTC)EGM6321,F10.TEAM1.allison

Solved problem 8 -- EGM6321.F10.TEAM1.ALLISON 20:40, 31 October 2010 (UTC)EGM6321,F10.TEAM1.allison