2.7 HW 7
2.7.1 Problems listing
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2.7.2 Problem 3 section 5.5
In Problems 1 through 20, find a particular solution of the given equation.
Solution
The first step is to find the homogeneous solution in order to determine the basis solutions to
check for any duplication with basis solutions for the particular solution. The characteristic
equation is
Hence the roots are . Therefore the basis solutions are
Which implies Now that we found the basis solution, we turn our attention to finding . The
RHS is . Looking at this function and all possible derivatives gives
Notice that we ignore any leading coefficients when doing this. Now we compare the above to the
basis of the homogeneous solution found in (1) to check if there are duplication in basis or not.
There is no duplication. Therefore we assume that particular solution is a linear combination of
the functions in (2). This implies that
Substituting the above back in original ODE (A) gives
Comparing coefficients gives
Multiplying first equation by and adding result to second equation gives
From (3)
Hence the particular solution is
Therefore the general solution is
2.7.3 Problem 9 section 5.5
In Problems 1 through 20, find a particular solution of the given equation.
Solution
The first step is to find the homogeneous solution in order to determine the basis solutions to
check for any duplication with basis solutions for the particular solution. The characteristic
equation is
Hence the roots are . Therefore the basis solutions are
Which implies Now that we found the basis solution, we turn our attention to finding . The
RHS is . Hence it basis functions are taking derivatives of each basis gives
Where we used to group all basis generated from same one.
Now we compare the above to the basis of the homogeneous solution found in (1) to check if
there are duplication in basis or not. We see duplication since is basis in both (1)
and (2). Therefore we multiply the group which generated by . The the above now
becomes
We compare again (1) against (2A) and now we see no duplication. Therefore we assume that
particular solution is a linear combination of the functions in (2A). This implies that
Substituting the above back in original ODE (A) gives
Comparing coefficients
Hence and from second equation , or and . Therefore the particular solution is
Therefore the general solution is
2.7.4 Problem 11 section 5.5
In Problems 1 through 20, find a particular solution of the given equation.
Solution
The first step is to find the homogeneous solution in order to determine the basis solutions to
check for any duplication with basis solutions for the particular solution. The characteristic
equation is
Hence the roots are . Therefore the basis solutions are
Which implies Now that we found the basis solution, we turn our attention to finding . The
RHS is . Hence it basis functions are
Taking derivatives does not add any new basis. Now we compare the above to the basis of the
homogeneous solution found in (1) to check if there are duplication in basis or not. We see
duplication the constant is in both (1) and (2). Therefore we multiply the group by . We took
the whole basis as one group, since the constant above is generated by taking derivative of , so it
is really in the same group. The above now becomes, after multiplying everything by
We compare again (1) against (2A) and now we see no duplication. Therefore we assume that
particular solution is a linear combination of the functions in (2A). This implies that
Substituting the above back in original ODE (A) gives
Comparing coefficients
Hence . Therefore the particular solution is
Therefore the general solution is
2.7.5 Problem 23 section 5.5
In Problems 21 through 30, set up the appropriate form of a particular solution , but do not
determine the values of the coefficients.
Solution
The first step is to find the homogeneous solution in order to determine the basis
solutions to check for any duplication with basis solutions for the particular solution.
The characteristic equation is Hence the roots are . Therefore the basis solutions are
Which implies Now that we found the basis solution, we turn our attention to finding . The
RHS is . Hence it basis functions are
Taking all possible derivatives of the above gives
Where in the above all signs and coefficients were ignored.
Now we compare the above to the basis of the homogeneous solution found in (1) to check if
there are duplication in basis or not. We see duplication as are in both. Therefore
we multiply the group by . We took the whole basis as one group since everything
above was generated from (2). The above now becomes, after multiplying each term by
Now we compare (2B) again with (1) and see no duplication. Hence
2.7.6 Problem 32 section 5.5
Solve the initial value problems in Problems 31 through 40.
Solution
The first step is to find the homogeneous solution in order to determine the basis solutions to
check for any duplication with basis solutions for the particular solution. The characteristic
equation is
Hence the roots are . Therefore the basis solutions are
Which implies Now that we found the basis solution, we turn our attention to finding . The
RHS is . Hence it basis functions are
Taking all derivatives does not any terms. We also see no duplication between (2) and (1). Hence
let
Substituting these into (A) gives
Hence
Therefore Therefore the complete solution is
We are now ready to apply the initial conditions. . Applying first IC to (3) gives
Taking derivative of (3) gives Applying second IC to the above gives
We now need to solve (4,5) for . Adding (4,5) gives
From (4)
Therefore the complete solution (3) becomes
2.7.7 Problem 7 section 6.1
In Problems 1 through 26, find the (real) eigenvalues and associated eigenvectors of the given
matrix . Find a basis for each eigenspace of dimension 2 or larger. Solution
We first need to find the eigenvalues. These are found by solving . Hence
Hence . For each eigenvalue we find its associated eigenvectors.
We nee to solve . This becomes . Therefore
Augmented matrix Therefore (1) becomes
is free variable. Let . Then from first row or . Hence
We nee to solve . This becomes . Therefore
Augmented matrix Therefore (1) becomes
is free variable. Let . Then from first row or . Hence This table gives summary of the
result
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2.7.8 Problem 17 section 6.1
In Problems 1 through 26, find the (real) eigenvalues and associated eigenvectors of the given
matrix . Find a basis for each eigenspace of dimension 2 or larger. Solution
We first need to find the eigenvalues. These are found by solving . Hence Expanding along the
first columns.
Hence roots (eigenvalues) are . For each eigenvalue we find its associated eigenvectors.
We nee to solve . This becomes . Therefore
Free variable is . Let . Last row gives . Second row gives . Hence the eigenvector is
We nee to solve . This becomes . Therefore
Swap (for clarify only) Free variable is . Let . From second row . Hence . First row gives . Hence .
Hence the eigenvector is
We nee to solve . This becomes . Therefore
Free variable is . Let . From second row . First row gives . Hence . Hence the eigenvector is
This table gives summary of the result
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eigenvalue |
associated eigenvector |
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2.7.9 Problem 21 section 6.1
In Problems 1 through 26, find the (real) eigenvalues and associated eigenvectors of the given
matrix . Find a basis for each eigenspace of dimension 2 or larger. Solution
We first need to find the eigenvalues. These are found by solving . Hence Expanding along the
last row
Hence the eigenvalues are of algebraic multiplicity and . For each eigenvalue we find its
associated eigenvectors.
We nee to solve . This becomes . Therefore
Free variables are . This means this is a complete eigenvalue. Since it has algebraic
multiplicity of and have a geometric multiplicity of as well. This means we can find two
linearly independent eigenvectors from it. Let . First row gives or . Hence the solution is
Therefore the basis (eigenvectors) are Now that we found the eigenvectors associated with , we
will do the same for second eigenvalue.
We nee to solve . This becomes . Therefore
Free variable is , leading variables are . Let . From second row, First row gives . Hence . Hence
the eigenvector is This table gives summary of the result
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eigenvalue |
associated eigenvector |
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2.7.10 Problem 25 section 6.1
In Problems 1 through 26, find the (real) eigenvalues and associated eigenvectors of the given
matrix . Find a basis for each eigenspace of dimension 2 or larger. Solution
We first need to find the eigenvalues. These are found by solving . Hence Since this is an upper
triangle matrix, then the determinant is the product of the diagonal. Hence the above reduces to
Therefore the eigenvalues are of algebraic multiplicity 2 and also of algebraic multiplicity 2. For
each eigenvalue we find its associated eigenvectors.
We nee to solve . This becomes . Therefore
Swapping (for clarify) Swapping (for clarify) Hence leading variables are free variables are .
Let . Second row gives . First row gives . Therefore the solution is Therefore the two eigenvectors
associated with this eigenvalues are
We nee to solve . This becomes . Therefore
Hence leading variables are free variables are Let . Second row gives or . First row gives or .
Hence the solution is Therefore the two eigenvectors associated with this eigenvalues are This
table gives summary of the result
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eigenvalue |
associated eigenvector |
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2.7.11 Problem 29 section 6.1
Find the complex conjugate eigenvalues and corresponding eigenvectors of the matrices given in
Problems 27 through 32 Solution
We first need to find the eigenvalues. These are found by solving . Hence
Hence . For each eigenvalue we find its associated eigenvectors.
We nee to solve . This becomes . Therefore
Leading variable is , free variable is . Let . From first row or . Hence the eigenvector is
We nee to solve . This becomes . Therefore
Leading variable is , free variable is . Let . From first row or . Hence the eigenvector is This
table gives summary of the result
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eigenvalue |
associated eigenvector |
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2.7.12 Additional problem 1
Find particular solution to
Solution
From HW6, we found as Therefore, we see the basis functions for are
Looking at RHS of (1), we see the basis functions for are Taking derivative does not generate
new basis. Taking derivative of generates . And taking derivative of generates . Hence the above
becomes
There are 3 groups. Comparing (2,3) we see there is one duplication, which is the constant term.
Hence we need to multiply that one group by . The above becomes
Now we again compare (3A) and (2). Now there is no duplication. Therefore the particular
solution is
2.7.13 Additional problem 2
Let where are distinct real numbers. Find the eigenvalues of and the corresponding
eigenvectors.
Solution
We first need to find the eigenvalues. These are found by solving . Hence Since this is
a diagonal matrix, then the determinant is the product of the diagonal. Hence the
above reduces to Hence the eigenvalues are . For each eigenvalue we find its associated
eigenvectors.
We nee to solve . This becomes . Therefore
Leading variables are and free variables is . Let . Second and third rows give , this is because are
distinct real numbers therefore . Therefore the solution is For this gives the eigenvector
We nee to solve . This becomes . Therefore
Leading variables are and free variable is . Let . First and third rows give , this is because are
distinct real numbers therefore . Therefore the solution is For this gives the eigenvector
We nee to solve . This becomes . Therefore
Leading variables are and free variable is . Let . First and third rows give , this is because are
distinct real numbers therefore . Therefore the solution is For this gives the eigenvector This
table gives summary of the result
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eigenvalue |
associated eigenvector |
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2.7.14 Additional optional problem 3
Extend the result in problem 2 to the case of matrices. That is, let be a matrix with entries on
the main diagonal and 0s everywhere else, where the are distinct real numbers. Find the
eigenvalues and corresponding eigenvectors.
Solution
This follows immediately from the last problem. Therefore each eigenvalue will be . And
corresponding eigenvectors are (each eigenvector is . These eigenvectors are the standard basis
for .
2.7.15 key solution for HW 7
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