User:Egm4313.s12.team1.stewart

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Reports[edit]

Report 1
Report 2

6.6[edit]

From Lecture 10 Pg. 9


(6.1)

To find the particular solution of an ODE we can use the method of undetermined coefficients. In report 3 we used the table 2.1 below to find the standard form of the particular solution given the form of the excitation .

Table 2.1

Table2.1.JPG


So the general form of the particular solution is:

(6.2)

Task 1 asks to simplify they first term of the given ODE which is . This is done by taking the second derivative of and plugging it into equation 6.1

(6.3)

simplifies to:

(6.4)

Now we must do the same for the second term .

(6.5)

(6.6)

Add the second term to the first to continue building the original ODE (equation 6.1).

(6.7)

(6.8)


Finally add the third term to the other two to have the lhs of the ODE 6.1

(6.9)

(6.10)


This simplifies to the particular solution below:

(6.11)


R5[edit]

See R5.7 Lect. 8b pg. 11:

The oblique basis vectors b1,b2 are:


1. Find the components c1, c2 using the Gramian matrix.
2. Verify the result found above.

R4.3[edit]

4.3eqn.JPG


(3.30)

R4[edit]

Statement[edit]

4.1 from lecture notes R4.1 Lect. 7c pgs. 19-22

Given the general form of polynomial excitation.

(1.1)

The particular solution that satisfies:

(1.2)

The first and second derivative of the particular solution that solves the original polynomial excitation equation.

(1.3)

(1.4)

The particular solutions are put into the polynomial excitation equation to give the general summation form:

(1.5)

Obtain the equations associated with , coefficients of ; , coefficients of ; , coefficients of ; , coefficients of ; , coefficients of . Five total equations for coefficients.

Also set up the matrix that satisfies .

Solution[edit]

The given equation associated with taking

(1.6)

Taking

(1.7)


(1.8)

The equation associated with , coefficients of

(1.9)

Taking

(1.10)

(1.11)

The equation associated with , coefficients of

(1.12)

Taking

(1.13)

(1.14)

The equation associated with , coefficients of

(1.15)

Taking

(1.16)

(1.17)

The equation associated with , coefficients of

(1.18)

Taking .

(1.19)


The equation associated with , coefficients of

(1.20)

Now set up the equation:

(1.21)

(1.22)

Therefore the matrix that satisfies the matrix equation is:

(1.23)

Author[edit]

Solved and Typed By - Chris Stewart User:Egm4313.s12.team1.stewart -- 21:12, 11 March 2012 (UTC)
Reviewed By -




R3[edit]

Use the Basic Rule 1 and the Sum Rule to show that the appropriate particular solution for

(4.1)

is of the form

(4.2)

i.e.,

(4.3)

Basic Rule: Select from the table and determine the coefficients by substituting in

(4.4)

Sum Rule: If is the sum of the terms in the 1st column of table 2.1 then is the sum of the corresponding terms in the 2nd column of this table.

Table 2.1

Table2.1.JPG


Solution[edit]

According to the table the Homogeneous equation has two values of the form . Using the Basic Rule this means that there is a particular solution of the form for each .

(4.5)

So the particular solution for this should be:

(4.6)

Which simplifies to:

(4.7)

For the second :

(4.8)

So the particular solution for this should be:

(4.9)

Which simplifies to:

(4.10)

Now using the Sum Rule which just states if there are two values in any form on the left side of the table, then the particular solution is the sum of the solutions for on the right side of the table.

So the particular solution for is:

(4.11)

Report 2 Progress[edit]

Statement[edit]

From K 2011 pg. 59 #3,4

Find the General Solution to the two ODEs below. (See Sec.5 Pg.5-6 notes)

Solution[edit]

Given:

The Homogeneous equation is given:

(3.0)

The first step is to change the equation into the characteristic equation.

(3.1)

Next use either completing the square or the quadratic formula to solve for (the roots of the characteristic equation). In this case completing the square is used.

(3.3)

(3.4)

The roots of the characteristic equation:

(3.5)

Now that the roots are known we can use the Homogeneous solutions in this equation:

(3.6)

Finally plug in the known roots and this is the final answer and general solution:

(3.7)

The Homogeneous equation is given:

(3.8)

Again, the first step is to convert to the characteristic equation.

(3.9)

In order to solve for the roots () either use the quadratic formula or completing the square. Once again in this case completing the was used to solve for .

(3.10)


(3.11)


(3.12)

The root takes the square root of a negative one, therefore the complex number is in the solution. The roots of the characteristic equation is shown:

(3.13)

For complex roots the Homogeneous solution is different and the root must broken into components and plugged into the complex Homogeneous solution equation below.

(3.14)


(3.15)

When plugging in the final answer and general solution is:

(3.16)

Author[edit]

Solved and typed by - Egm4313.s12.team1.stewart 02:03, 4 February 2012 (UTC)
Reviewed By -
Edited By -