Compare the two sides

Compute the right-hand side of x' = A x

g) Compute the left-hand side of the differential system x' = A x

Therefore, the particular solution is given by

Thus,

Equate the corresponding components of the solution at 0 and the initial conditions, then solve for c1 and c2

Evaluate the solution at 0

Define the initial condition

==>

Let

Since (A - l I)3 = 0 ==> when we solve for system (A - l I)3 v3 = 0 for v3, we three free variables. to choose for the components of v3.

Compute A - l I

Define the 3 x 3 identity matrix as I

Define l

a) Define the coefficient matrix

Solution:

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Solve for the components of the e-vector

Mathcad: Reset the variables v1 and v2

Compute the e-vector

b) Choose one of the e-values

Find the roots of the char. equation

f) Evaluate the general solution at the initial point (0) and set it equal to the initial condition

e) Form the general solution

Extract the imaginary part of the complex solution to obtain the second real-valued solution

Extract the real part of the complex solution to obtain the first real-valued solution

d) Write the complex solution in terms of sine and cosine

c) Form the complex solution

Therefore, the e-vector is

==>

8. If possible, give examples of a reduced row echelon form of the augmented matrix of a linear system with the given conditions. If not possible, write "impossible" as an answer.

Impossible

g) homogeneous system with 3 equations in 4 unknowns that has no solution

f) nonhomogenous system with 3 equations in 4 unknowns that has no solution

Impossible

e) homogeneous system with two equations in 3 unknowns that has a only the trivial solution

d) nonhomogenous system with 3 equations in 3 unknowns that has infinite number of solutions

c) nonhomogeneous system with 4 equations in two unknowns that has infinite number of solutions

b) non homogeneous system with 4 equations in two unknowns that has a unique solution

a) nonhomogeneous system with 4 equations in two unknowns that has no solution

Create the three linearly independent solutions

(which is not the zero vector)

Compute v1

(which is not the zero vector)

Compute the second e-vector

==>

Let

With the above choices for the components v3, we got v2 to be the zero vector which is not acceptable since we are looking for a nonzero eigenvector. Therefore, we change our choices for the components of v3

Compute the second e-vector

Conversions

Derivatives

Substitution

Solve for the highest derivative

Mathcad: Reset the variables

Solution:

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Write the general solution

Mathcad : Reset the variables c1 and c2

Compute the rref of M

Create the augmented matrix

The vector w will be a solution if we can find constants c1 and c2 such that w = c2 v1 + c2 v2

is a solution or not. Do not use substitution.

f) Show whether the vector

Since the basis has two vectors

e) Give the dimension of the solution space

Create the augmented matrix of the system

a) Determine t so that the system is consistent

Solution:

2 . Given the linear system

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w is not a solution because the there is a leading entry in the last column of the row reduced echelon form of the augmented matrix [v1 v2 | w]

g) Give the reason for which w is or is not a solution to the system.

The rref of the augmented matrix has a leading entry in the last column. Therefore, w is not a solution.

c) Use the rank of A and the number of unknowns to compute the number of free variable in the solution of the

b) Find the rank of A.

Remark:

The small Mathcad program below defines a function R(A), where A is any matrix. R(A)i returns the ith row of matrix A. It is used, along with the command stack to illustrate the step-by-step reduction of a matrix.

A. Farhat

Final Exam Solution

Math260-022

Therefore, with the choice of s = 1 and t = 1, a basis for the solution space is:

==>

==>

==>

d) Find a basis for the solution space of the system

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Compute the general solution:

Define the general solution:

Simplify the particular solution:

Compute the particular solution:

Define the particular solution:

Compute u2

Simplify u1

Compute u1

Compute the characteristic equation

Define the 2 x 2 identity matrix as I

a) Define the coefficient matrix

Solution:

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Modified form of the particular solution

To avoid repeating the terms that appear in the complementary solution, we modify the above form by multiplying the appropriate terms by x raised to the power equal to the multiplicity of the root that produced the terms.

Unmodified form:

b) find the form of the particular solution if the differential equation is to be solved by the method of undetermined coefficients.

a) write the complementary solution

4. Given that 0, 0, -1+2i, -1-2i, -1, -1 are the roots of the characteristic polynomial of the differential equation

b) Determine t so that the system is inconsistent

This is not possible. t can not be simultaneously equal to -4 and 9. Therefore, the system is not consistent for any value of t.

==>

and

==>

For the system to be consistent, t must simultaneously satisfy the two equations

Compute the rref of the matrix M

Remark:

The Mathcad program R(A) written above does not work with non constant matrices. Instead, we define the rows separately as follows:

Compute the Wronskian:

Define f(x):

Define the two solutions:

Write the complementary solution:

Find the roots of the char. eqn.

Mathcad: Reset the variable x

Solution:

3 . Use the method of variation of parameters to find the general solution of the differential equation y'' + 4y = 12 sin2(2x)

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From part (a), the system is inconsistent for all values of t. That is, for t e (-¥, ¥)