MATH2270

FW99

Solutions to Term Test 2

1. (12 marks) Find the general solution to the following second-order differential equations:

(a) 2x´´ + 6x´ + 5x = 0

CE: 2r2 + 6r + 5 = 0; r = -3/2 ± i/2

x = (C1sin(t/2) + C2cos(t/2)e-3t/2

(b) x´´ + 9x´ + 14x = 3e-2t

CE: r2 + 9r + 14 = 0; r = -2, -7

xh = C1e-2t + C2e-7t

xp = Ate-2t

xp´ = (A - 2At) e-2t
xp´´ = (-4A + 4At) e-2t

xp´´ + 9xp´ + 14xp = (-4A + 9A)e-2t + (4A -18A + 14A)te-2t = 3e-2t

Thus 5A = 3 or A = 3/5 and the general solution is

x = C1e-2t + C2e-7t + 3/5 e-2t

2. (14 marks) Solve the following initial value problem

x´´ + 6x´ + 9x = 5e3t; x(0) = 1, x´(0) = 2

CE: r2 + 6r + 9 = 0; r = -3 (repeated)

xh = (C1 + C2t) e-3t

xp = Ae3t

xp´ = 3Ae3t
xp´´ = 9Ae3t

xp´´ + 6xp´ + 9xp = (9A + 18A + 9A)e3t = 5e3t

Thus A = 5/36 and the general solution is

x = (C1 + C2t) e-3t + 5e3t/36

x´ = (C2 -3C1 - 3C2t) e3t + 5e3t/12

Applying the initial conditions we get

x(0) = C1 + 5/36 = 1 or C1 = 31/36

x´(0) = C2 -3C1 + 5/12 = 2 or C2 = 50/12

so that the solution is

x = (31/36 + 25t/6)e-3t + 5/36 e3t

3. (6 marks) The second-order differential equation x´´ + 3x´ + 2x = f(t) has a homogeneous solution

xh = C1e-t + C2 e-2t

Write down the form of the particular solution xp to the differential equation for the various right hand sides f(t) given below. You are not required to find values for the undetermined coefficients.

(a) f(t) = 2t2 - 5

At2 + Bt + C

(b) f(t) = 3 sin(2t)

Asin(2t) + Bcos(2t)

(c) f(t) = 2t cos(3t)

(At + B)sin(3t) + (Ct + D)cos(3t)

(d) f(t) = 7te3t

(At + B)e3t

(e) f(t) = 4te-2t

t(At + B)e-2t

(f) f(t) = 4 cos(5t) e-t

(Asin(5t) + Bcos(5t))e-t

4. (2 marks) Convert the second-order differential equation

3x´´-2x´ + 4x = e-2t

into a system of two first-order differential equations. You are NOT required to solve this equation.

x´ = y and 3y´ = 2y - 4x + e-2t

5. (9 marks) Solve the differential equation (sin(2y) + 4x) dx + (2x cos(2y) + 3) dy = 0

Let M = sin(2y) + 4x and N = 2x cos(2y) + 3. Then

so the DE is exact. Then

Thus f´(y) = 3 or f(y) = 3y and the solution is

H = x sin(2y) + 2x2 + 3y = C

Alternately we can get the solution as

Thus g´(x) = 4x and g(x) = 2x2 giving the solution above.

6. (9 marks) Find the solution to the first-order differential equation

This is a Bernoulli DE with n = 3. Thus take u = y-1/3 and thus u´ = -y-4/3y´/3. Substituting this into the DE yields -3xu´ + 6u = 3x or u´ - 2u/x = -1. This is a linear equation with p = -2/x so the integrating factor is x-2. Thus we have

or u = x + Cx2. Thus the solution is y = u-3 = (x + Cx2)-3.

7 (4 marks) Convert the system of first-order differential equations

x´ = 2x + y + t; y´ = x - 3y - 4

into a second-order differential equation in one of the dependent variables.

Differentiating the first equation, substituting y´ from the second equation and then substituting y from the first equation yields:

x´´ = 2x´ + y´ + 1 = 2x´ + x - 3y - 4 + 1 = 2x´ + x -3(x´ - 2x - t) - 3

or x´´ = x´ - 7x = 3t - 3

Alternately, we can differentiate the second equation, substitute for x´ from the first equation and then substitute for x from the second equation yielding:

y´´ = x´ - 3y´ = 2x + y + t - 3y´ = 2(y´ + 3y + 4) + y + t - 3y´ or y´´ + y´ - 7y = t + 8

8. (5 marks) The following plot represents the orbit of the system of differential equations

x´ = y

y´ = -25x - 2y

representing the motion of a damped pendulum. The variable x measures the distance of the pendulum from its equilibrium position (its lowest point) and y is the velocity of the pendulum

(a) Mark on the orbit the point where the pendulum has the greatest speed (greatest magnitude of the velocity).

It is the lowest point on the orbit

(b) What is the approximate value of the velocity of the pendulum as it passes through the equilibrium position for the second time?

1.8 - it is the highest point where the orbit crosses the positive y-axis

(c) What is the approximate position of the pendulum after it has completed one complete swing (back and forth)?

0.28 - it is the point where the orbit crosses the positive x-axis for first time after the initial point.

(d) Draw the slope field for this autonomous system at the point x = 0.2, y = -2.

Since this is an autonomous system we can write

which has the value 0.5 for x = 0.2, y = 2. This is plotted on the posted diagram.

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