MATH2270

FW00

Solutions to Assignment 4

1. Find the general solution to the following second-order differential equations:

(i) 2x´´+ 3x´ - 2x = 0;

Characteristic equation is 2m2 + 3m - 2 = 0 with roots m = -2 and 1/2. Thus the solution is

x = C1e-2t + C2et/2

(ii) x´´ - 4x´ + 4x = 2t2 - t ;

CE is m2 - 4m + 4 = 0 with roots m = 2 repeated. Thus the complementary solution is

xc = (C1 + C2t) e2t

Take the particular solution as xp = At2 + Bt + C. Then xp´ = 2At + B and xp´´ = 2A. Substituting into the DE we have 2A - 4(2At + B) + 4(At2 + Bt + C) = 4At2 + (4B - 8A) t + (4C - 4B + 2A) = 2t2 - t

or 4A = 2; 4B - 8A = -1; 4C - 4B + 2A = 0 giving A = 1/2; B = 3/4; C = 1/2.

Thus the general solution to the DE is x = (C1 + C2t) e2t + t2/2 + 3t/4 + 1/2.

(iii) 3x´´ - 10x´ +3x = 25 cos(4t);

CE is 3m2 - 10m + 3 = 0 with roots m = 3 and 1/3. Thus the complementary solution is

xc = C1e3t + C2 et/3

Take the particular solution as xp = Acos(4t) + Bsin(4t). Then xp´ = -4Asin(4t) + 4Bcos(4t) and

xp´´ = -16Acos(4t) - 16Bsin(4t). Substituting into the DE we have

-48Acos(4t) - 48Bsin(4t) + 40Asin(4t) - 40Bcos(4t) + 3Acos(4t) + 3Bsin(4t) = 25 cos(4t)

or -45A - 40 B = 25; 40A - 45B = 0 giving A = -9/29; B = -8/29.

Thus the general solution to the DE is x = C1e3t + C2 et/3 - (9/29)cos(4t) - (8/29)sin(4t).

(iv) x´´ + 8x´ + 20x = 5te2t ;

CE is m2 + 8m + 20 = 0 with roots m = -4 + 2i and -4 - 2i. Thus the complementary solution is

xc = (C1 cos(2t) + C2 sin(2t))e-4t

Take the particular solution as xp = (At + B)e2t. Then xp´ = (2At + 2B + A)e2t and

xp´´ = (4At + 4B + 4A)e2t. Substituting into the DE we have

(4At + 4B + 4A)e2t + 8(2At + 2B + A)e2t + 20(At + B)e2t = (40At + 40B + 12A)e2t = 5te2t

or 40A = 5; 40B + 12A = 0 giving A = 1/8; B = -3/80.

Thus the general solution to the DE is x = (C1 cos(2t) + C2 sin(2t))e-4t + (t/8 - 3/80)e2t

(v) x´´ - 16x = e4t .

CE is m2 - 16 = 0 with roots m = 4 and -4. Thus the complementary solution is

xc = C1 e4t + C2 e-4t

Since the complementary solution contains a term which also appeares on the right-hand side of the DE we take the particular solution as xp = Ate4t (note the extra factort of t) . Then xp´ = (4At + A)e4t and

xp´´ = (16At + 8A)e4t . Substituting into the DE we have (16At + 8A)e4t - 16Ate4t = 8Ae4t = e4t

or 8A = 1 giving A = 1/8.

Thus the general solution to the DE is x = (C1 + t/8) e4t + C2 e-4t

2. Find the explicit solution to the following initial value problems:

(i) x´´ + 6x´ + 13x = 0, x(0) = 3, x´(0) = -3;

CE is m2 + 6m + 13 = 0 with roots m = -3 + 2i and -3 - 2i. Thus the solution is

x(t) = (C1cos(2t) + C2sin(2t)) e-3t

Then x´(t) = -3(C1cos(2t) + C2sin(2t)) e-3t + (-2C1sin(2t) + 2C2 cos(2t)) e-3t

Applying the initial conditions we have x(0) = C1 = 3 and x´(0) = -3C1 + 2C2 = -3 giving C2 = 3

Thus the solution to the DE satisfying the initial conditions is x = 3(cos(2t) + sin(2t)) e-3t.

(ii) x´´ + 9x = 2 sin(3t), x(0) = 0, x´(0) = 1;

CE is m2 + 9 = 0 with roots m = 3i and -3i. Thus the complementary solution is

xc = C1 cos(3t) + C2 sin(3t)

Since the complementary solution contains a term which also appeares on the right-hand side of the DE we take the particular solution as xp = At cos(3t) + Bt sin(3t) (note the extra factor of t). Then

xp´ = -3At sin(3t) + 3Bt cos(3t) + Acos(3t) + Bsin(3t) and

xp´´ = -9At cos(3t) - 9Bsin(3t) - 6Asin(3t) + 6Bcos(3t). Substituting into the DE we have

-9At cos(3t) - 9Bsin(3t) - 6Asin(3t) + 6Bcos(3t) + 9(At cos(3t) + Bt sin(3t)) = - 6Asin(3t) + 6Bcos(3t) = 2 sin(3t). Thus we have -6A = 2 and 6B = 0 giving the general solution

x(t) = (C1 - t/3)cos(3t) + C2 sin(3t) and x´(t) = -3(C1 - t/3)sin(3t) + (3C2 -1/3)cos(3t)

Applying the initial conditions we have x(0) = C1 = 0 and x´(0) = 3C2 -1/3 = 1.

Thus the general solution to the DE is x = -(t/3) cos(3t) + (4/9)sin(3t).

3. An Electrical RLC circuit has an inductance L = 1 henry, a resistance R = 0.1 ohms and a capacitance of C farads.

(a) If the derivative of the applied voltage E is sin(t) + 4*sin(5*t) volts use MAPLE to plot the current I (in amperes) in the circuit if I(0) = 1 and I´(0) = 0 coulombs. Take C = 1, 1/25 and 1/81 (i.e. produce three diferent plots). Use the t range of 80 to 100 to ensure the transients have died down.

(b) What is the amplitude of the current in each of the three cases?

(c) What is the frequency of the current in each of the three cases? The frequency can be estimated by counting the number of complete oscillations in a given time interval, multiplying by 2p and dividing by the length of the time interval.

See MAPLE output.

4. Convert the following systems of first-order differential equations into a second-order differential equation in one of the dependent variables where the prime indicates differentiation with respect to the variable t. Then find the general solution to the system:

(i) x´ = x + 2y

y´ = 3x + 2y

Differentiating the first equation we get x´´ = x´ + 2y´

= x´ + 2( 3x + 2y) using the second equation

= x´ + 6x + 2(x´ - x) using the first equation.

Thus we have the DE x´´ - 3x´ - 4x = 0.

Solving this we get the CE m2 - 3m - 4 = 0 with roots m = -1 and 4. Thus the solution is

x = C1e-t + C2 e4t and x´ = -C1e-t + 4C2 e4t. Thus y = (x´ - x)/2 = -C1e-t + (3/2)C2 e4t.

If we eliminated x from the equations we would get y´´ = 3x´ + 2y´ = 3(x + 2y) + 2y´ =

y´ - 2y + 6y + 2y´ giving the DE y´´ - 3y´ - 4y = 0. This is the same DE as above and hence

y = C´1e-t + C´2 e4t and y´ = -C´1e-t + 4C´2 e4t. Now x = (y´ - 2y)/3 = -C´1e-t + (2/3)C´2 e4t

which is equivalent to the previous result.

(ii) x´ = -2x + 3y

y´ = 2x - y - t

Differentiating the first equation we get x´´ = -2x´ + 3y´

= -2x´ + 3( 2x - y -t) using the second equation

= -2x´ + 6x -2x - x´ using the first equation.

Thus we have the DE x´´ + 3x´ - 4x = -3t.

Solving this we get the CE m2 - 3m - 4 = 0 with roots m = 1 and -4. Thus the complementary solution is xc = C1et + C2 e-4t.

Take the particular solution as xp = At + B. Then xp´ = A and xp´´ = 0. Substituting into the DE we get 3A - 4(At + B) = -3t giving -4A = -3 and 3A - 4B = 0 or A = 3/4, B = 9/16.

Thus the general solution is x = C1et + C2 e-4t + 3t/4 + 9/16 and x´ = C1et - 4C2 e-4t + 3/4.

Thus y = (x´ + 2x)/ 3 = C1et - (2/3)C2 e-4t + t/2 + 5/8

If we eliminated x from the equations we would get y´´ = 2x´ - y´ - 1 = 2(-2x + 3y) - y´ - 1 =

-2(y´ + y + t) + 6y - y´ - 1 giving the DE y´´ + 3y´ - 4y = -2t - 1. This DE has the same complementary solution as above i.e. yc = C1´et + C2´ e-4t.

Take yp = At + B. Then yp´ = A and yp´´ = 0. Substituting into the DE we get 3A - 4(At + B) = -2t - 1 giving -4A = -2 and 3A - 4B = -1 or A = 1/2, B = 5/8. Thus the general solution is

y = C1´et + C2´ e-4t + t/2 + 5/8 and y´ = C1´et - 4C2´ e-4t + 1/2.

Thus x = (y´ + y + t)/2 = C1´et - (3/2)C2´ e-4t + 3t/4 + 9/16.

(iii) 3x´ - y´ - 12y = 12

3x´ - 5 y´ + 12x = 12t

We must first solve this system for x´ and y´. This yields the system

x´ = x + 5y - t + 5

y´ = 3x + 3y - 3t + 3

Differentiating the first equation we get x´´ = x´ + 5y´ - 1

= x´ + 5( 3x + 3y - 3t + 3) - 1 using the second equation

= x´ + 15x + 3(x´ - x + t - 5) -15t + 14 using the first equation.

Thus we have the DE x´´ - 4x´ - 12x = -12t - 1.

Solving this we get the CE m2 - 4m - 12 = 0 with roots m = -2 and 6. Thus the complementary solution is x = C1e-2t + C2 e6t.

For the particular solution take xp = At + B. Then xp´ = A and xp´´ = 0. Substituting into the DE we get -4A - 12(At + B) = -12t - 1 giving -12A = -12 and -4A - 12B = -1 or A = 1, B = -1/4. Thus the general solution is x = C1e-2t + C2 e6t + t - 1/4 with x´ = -2C1e-2t + 6C2 e6t + 1.

Thus y = (x´ - x + t -5)/5 = -(3/5)C1e-2t + C2 e6t - 3/4.

If we eliminated x from the equations we would get y´´ = 3x´ + 3y´ - 3 = 3(x + 5y - t + 5) + 3y´ - 3 =

y´ - 3y + 3t - 3 + 15y - 3t + 3y´ + 12 giving the DE y´´ - 4y´ - 12y = 9. This has the same right-hand side as above and hence the complementary solution is y = C1´e-2t + C2´ e6t.

For the particular solution take yp = At + B. Then yp´ = A and yp´´ = 0. Substituting into the DE we get -4A - 12(At + B) = 9 giving -12A = 0 and -4A - 12B = 9 or A = 0, B = -3/4. Thus the general solution is y = C1´e-2t + C2´ e6t - 3/4 with y´ = -2C1´e-2t + 6C2´ e6t.

Now x = (y´ - 3y + 3t - 3)/3 = -(5/3)C´1e-2t + C´2 e6t + t - 1/4

which is equivalent to the previous result.

5. (a) Use MAPLE to illustrate the concept of beats by plotting solutions of the differential equation

x´´ + 25x = cos(wt)

for w = 4, 4.5 and 4.9. Use the initial conditions x(0) = 0, x´(0) = 1. Choose the plot range to contain at least two beats.

(b) What are the beat frequencies in each case?

See MAPLE output.