ENGI 9420 Engineering Analysis
Mid Term Test Solutions

2007 Fall
  1. Find the complete solution to the initial value problem

    x" + 10 x' + 21 x = 8 e^(-3t) , x(0) = 0, x'(0) = 2

    A.E.: lamda = -3, -7
    C.F.: xC = A e^(-3t) + B e^(-7t)
    P.S. by the method of undetermined coefficients:
    try xP = c t e^(-3t)
    xP = 2t e^(-3t)

    OR
    P.S. by the method of variation of parameters:
    Wronskian = -4 e^(-10t)
    W1 = -8 e^(-10t)
    W2 = +8 e^(-6t)
    u' = W1/W = 2, u = 2t
    v' = W2/W = -2 e^(4t), v = -e^(4t) / 2
    xP = u x1 + v x2
    But the C.F. contains   e-3t
    xP = 2t e^(-3t)

    G.S.: x(t) = (A + 2t) e^(-3t) + B e^(-7t)
    x'(t) = (-3A - 6t + 2) e^(-3t) - 7B e^(-7t)
    Imposing the initial conditions:
    x(0) = 0 ==> B = -A
    x'(0) = 2 leads to B = A = 0
    The complete solution is therefore

    x(t) = 2t e^(-3t)

    OR

    Laplace Transform Method:

    Let X(s) = Laplace transform of x(t) then
    x" + 10 x' + 21 x = 8 e^(-3t) , x(0) = 0, x'(0) = 2
    (s^2 X - 0 - 2) + 10(sX - 0) + 21X = 8 / (s+3)
    (s^2 + 10s + 21)X = (2s + 6 + 8) / (s+3)
    X = 2 / (s+3)^2
    The complete solution is therefore

    x(t) = 2t e^(-3t)

  1. Either

    Use the identities for Laplace transforms
    L{(sin wt - wt cos wt) / (2w^3)} = 1/(s^2 + w^2)^2 and L{f'(t) = s L{f(t)} - f(0)
    to find L{t sin wt}

    or

    Use the identities for Laplace transforms
    L{t f(t)} = -d/ds L{f(t)} and L{sin wt} = w / (s^2 + w^2)
    to find L{t sin wt}.

    Let f(t) = (sin wt - wt cos wt) / (2w^3) then
    f'(t) = t sin wt / (2w) and f(0) = 0
    L{t sin wt} = 2ws/(s^2 + w^2)^2

    OR

    L{t sin wt} = -d/ds L{sin wt}
    From either method,

    L{t sin wt} = 2ws/(s^2 + w^2)^2

  1. Find the MacLaurin series solution of Airy’s equation

    y" + xy = 0

    with arbitrary initial conditions   y(0) = a,   y' (0) = b , as far as the term in x6.

    Assume a power series, with an = y^(n)(0) / n!
    then   a0 = y(0) = a ,         a1 = y'(0) = b
    y^(2)(x) = -x y(x) a2 = 0
    y^(3)(x) = -y - x y' a3 = -a/6
    y^(4)(x) = -2y' - x y" a4 = -b/12
    y^(5)(x) = -3y" - x y^(3)     a5 = 0
    y^(6)(x) = -4y^(3) - x y^(4) a6 = a/180

    Therefore the MacLaurin series is

    y = a(1 - x^3/6 + x^6/180 + ...) + b(x - x^4/12 + ...)

    OR

    Assume a power series solution and substitute it into the ODE:
    y(x) = Sum {an x^n} ; y' and y"
    y" + xy = ...
    [manipulate indices]
    [bring series together]
    a_(n+2) = -a_(n-1) / ((n+2)(n+1))
    [values of a3, a4, a5, a6]
    Therefore the MacLaurin series is

    y = a(1 - x^3/6 + x^6/180 + ...) + b(x - x^4/12 + ...)

  1. The cubic equation
                  x^3 - 1.01x + 0.01 = 0
    is a special case (e = 0.01) of the perturbation
                  x^3 - (1+e)x + e = 0
    of the simple cubic equation
                  x^3 - x = 0
    Find the perturbation series, as far as the term in e2, for the root near x = –1 and hence find the solution of x^3 - 1.01x + 0.01 = 0
    nearest to x = –1, correct to four decimal places.

    Let x(e) = x0 + x1 e + x2 e^2 + ... then
    (x(e))^3 - (1+e)*x(e) + e = 0
    Table of coefficients:
    e0 e1 e2
    x03     3x02 x1         3x02 x2         first column leads to x0 = ±1 or 0
    3x0 x12     Here we require the negative root
        –x0     –x1 –x2 x0 = -1 and 3x0^2 - 1 = 2
    –x0 –x1
    1 second column leads to x1 = -1
    		second column leads to x1 = -1
    000
    third column leads to x2 = 1
    Therefore the required perturbation series is

    x(e) = -1 - e + e^2 + ...

    x(0.01) = -1.0099
    [This is actually correct to five decimal places!]

    Additional Notes:
    [The exact solutions to the perturbed equation are found easily, upon noticing
    factorisation into (x-1)(x^2 + x - e) = 0
    binomial expansion is also -1 - e + e^2 + ...
    which matches the perturbation series.]

    [With   e = 0.01, the three solutions evaluate to   1 (exactly), 0.00990195... and –1.00990195...]

  1. BONUS QUESTION:

    An over-damped mass-spring system is modelled by the ordinary differential equation

    x" + 3x' + 2x = 0

    where   x(t)   is the displacement of the centre of mass of the spring from its equilibrium position at time t.   The spring is released with velocity   vo   from location   x = +1   at time   t = 0.

    Find the condition on the initial velocity   vo   in order for the centre of mass of the spring to pass through its equilibrium position at some finite positive time   tc   and determine an exact expression for   tc   in terms of   vo.

    Solving the homogeneous ODE:
    auxiliary equation leads to lambda = -2 or -1
    complementary function = general solution x(t) = A exp(-2t) + B exp(-t)
    x'(t) = -2A exp(-2t) - B exp(-t)
    The initial conditions are     x(0) = 1,   x'(0) = vo
    A + B = 1 and -2A - B = vo
    A = -(vo + 1) and B = vo + 2
    The complete solution is therefore
    x(t) = (vo + 2) exp(-t) - (vo + 1) exp(-2t)

    OR   using Laplace transforms,
    (s^2 X - s - vo) + 3(sX - 1) + 2X = 0
    (s^2 + 3s + 2) X = s + 3 + vo
    X(s) = A/(s+1) + B/(s+2)
    Using the cover-up rule,
    A = vo + 2, B = -(vo + 1)
    X(s) = (vo + 2)/(s+1) - (vo + 1)/(s+2)
    The complete solution is therefore
    x(t) = (vo + 2) exp(-t) - (vo + 1) exp(-2t)

    The centre of mass passes through the equilibrium position   x = 0   at time   tc
    Solving x(tc) = 0 , exp(tc) = (vo + 1) / (vo + 2)
    In order for   tc   to be a positive real number, we must have
    exp(tc) > 1
    The upper branch has no solution, while the lower branch is true for all   vo < –2.
    Therefore the centre of mass of the spring will pass through its equilibrium position at the finite positive time

    tc = ln((vo + 1)/(vo + 2))

    if and only if     vo < -2

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