Need 5 questions answered with written explanations and intermediate steps when

Question

Need 5 questions answered with written explanations and intermediate steps when appropriate. Answers also need to be in terms of rational numbers and constants like e, ?, ?2, etc instead of decimal approximations. First person to answer all 5 in this manner gets the points. Will choose best answer as soon as I can. Thanks!

6. Does pure or practical resonance occur when the position of a particle is governed by the equation x'' + 9x = cos(3t)? Explain, and ?nd a particular solution

7. Solve the system of equations

8. Which sets of vectors below constitute a basis for R^2? Give reasons

9. Let D be the linear di?erential operator de?ned on the vector space V of in?nitely di?erentiable functions

f : R ? R by D(f) = f '. Describe the kernel of D^3, and determine the nullity

10. Find a basis for the kernel of the linear transformation T : R^4 ? R^2 de?ned by

Explanation / Answer

6.Your complimentary solution will be of the form e^mx.

in particular, you will solve m^2 + 9 = 0. So m = 3i

so your solution is e^(3it), but remember that e^aix = cos ax + i sin (ax)

So xc = c1 cos (3t) + c2 sin(3t) are your complimentary solutions to the differential equation.

Now, since your complimentary solution contains cos (3t), your particular solution cannot contain
cos (3t), so
xp = At cos (3t).
x'p = -At sin(3t) + 3Acos(3t)
x''p = -Atcos(3t) - 3Asin(3t) + 9Acos(3t) (I hope I did those derivatives correct, lol)

Now plug these in for your x'' , x', and x (don't forget the coefficients!) and solve for A.

Then your solution is x = xc + xp

7.x^1 + 2x^3 + 4x^4 = 8
x^2 - x^4 = 0
x(4x^3 + 2x^2 + 1) = 8
x = 1

9.The kernel, or otherwise known as the nullspace, is the set of vectors which, when mapped by the linear transformation in question, goes the zero to the target vector space. D^3 implies differentiating an infinitely differentiable function 3 times, so any polynomial of degree 2 or less (including constants and 0 itself) will be mapped to 0. I can't think of any other functions.

The nullity is the dimension of the nullspace, otherwise the number of basis vectors that generates the nullspace. A basis of a vector space is a set of linearly independent vectors that spans said vector space. Since we have previously established that any polynomial of degree 2 or less is part of the nullspace, we can use the standard basis for polynomials of degree 2, that is B = {1, x, x^2}, which consists of 3 vectors, thus the nullity = dim(nullspace of D^3) = 3.

10.We want to find all [x, y, z, w] such that T[x, y, z, w] = [0, 0].

This leads to the augmented matrix
[1 2 1 -1|0]
[2 3 -1 1|0]

Applying -2 * R1 + R2 --> R2 (elementary row operation) yields
[1 2 1 -1|0]
[0 -1 -3 3|0]

Next, apply -R2 --> R2:
[1 2 1 -1|0]
[0 1 3 -3|0]

Finally, -2R2 + R1 --> R1 yields
[1 0 -5 5|0]
[0 1 3 -3|0].

So, we have x - 5z + 5w = 0 and y + 3z - 3w = 0.

Solving for the pivots (with w, z being free variables):
x = 5z - 5w
y = -3z + 3w
z = z
w = w.

In vector form (letting z = a and w = b):
[x, y, z, w] = a [5, -3, 1, 0] + b [-5, 3, 0, 1] for some scalars a and b.

Hence, ker T has basis {[5, -3, 1, 0], [-5, 3, 0, 1]}.

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