**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 2m^{2 }+ 3m - 2 = 0
with roots m = -2 and 1/2. Thus the solution is

x = C_{1}e^{-2t} + C_{2}e^{t/2}

(ii) x´´ - 4x´ + 4x = 2t^{2}
- t ;

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

x_{c} = (C_{1} + C_{2}t) e^{2t}

Take the particular solution as x_{p} = At^{2}
+ Bt + C. Then x_{p}´ = 2At + B and x_{p}´´
= 2A. Substituting into the DE we have 2A - 4(2At + B) + 4(At^{2}
+ Bt + C) = 4At^{2} + (4B - 8A) t + (4C - 4B + 2A) = 2t^{2}
- 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 = (C_{1}
+ C_{2}t) e^{2t } + t^{2}/2 + 3t/4 + 1/2.

(iii) 3x´´ - 10x´ +3x = 25 cos(4t);

CE is 3m^{2 }- 10m + 3 = 0 with roots m = 3 and
1/3. Thus the complementary solution is

x_{c} = C_{1}e^{3t} + C_{2}
e^{t/3}

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

x_{p}´´ = -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 = C_{1}e^{3t}
+ C_{2} e^{t/3} - (9/29)cos(4t) - (8/29)sin(4t).

(iv)
x´´
+ 8x´ + 20x = 5te^{2t}
;

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

x_{c} = (C_{1} cos(2t) + C_{2}
sin(2t))e^{-4t}

Take the particular solution as x_{p} = (At +
B)e^{2t}. Then x_{p}´ = (2At + 2B + A)e^{2t}
and

x_{p}´´ = (4At + 4B + 4A)e^{2t}.
Substituting into the DE we have

(4At + 4B + 4A)e^{2t } + 8(2At + 2B + A)e^{2t}
+ 20(At + B)e^{2t} = (40At + 40B + 12A)e^{2t} = 5te^{2t}

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

Thus the general solution to the DE is x = (C_{1}
cos(2t) + C_{2} sin(2t))e^{-4t} + (t/8 - 3/80)e^{2t}

(v) x´´ - 16x = e^{4t }.

CE is m^{2 }- 16 = 0 with roots m = 4 and -4.
Thus the complementary solution is

x_{c} = C_{1 }e^{4t} + C_{2}
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 x_{p} = Ate^{4t }(note the extra factort
of t) . Then x_{p}´ = (4At + A)e^{4t} and

x_{p}´´ = (16At + 8A)e^{4t}
. Substituting into the DE we have (16At + 8A)e^{4t} -
16Ate^{4t } = 8Ae^{4t }= e^{4t }

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

Thus the general solution to the DE is x = (C_{1 }+
t/8) e^{4t} + C_{2} 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 m^{2 }+ 6m + 13 = 0 with roots m = -3 + 2i
and -3 - 2i. Thus the solution is

x(t) = (C_{1}cos(2t) + C_{2}sin(2t))
e^{-3t}

Then x´(t) = -3(C_{1}cos(2t) + C_{2}sin(2t))
e^{-3t }+ (-2C_{1}sin(2t) + 2C_{2 }cos(2t))
e^{-3t}

Applying the initial conditions we have x(0) = C_{1}
= 3 and x´(0) = -3C_{1} + 2C_{2} = -3 giving C_{2}
= 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 m^{2} + 9 = 0 with roots m = 3i and -3i.
Thus the complementary solution is

x_{c} = C_{1} cos(3t) + C_{2}
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 x_{p} = At cos(3t) + Bt sin(3t) (note the extra
factor of t). Then

x_{p}´ = -3At sin(3t) + 3Bt cos(3t) +
Acos(3t) + Bsin(3t) and

x_{p}´´ = -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) = (C_{1} - t/3)cos(3t) + C_{2}
sin(3t) and x´(t) = -3(C_{1} - t/3)sin(3t) + (3C_{2}
-1/3)cos(3t)

Applying the initial conditions we have x(0) = C_{1}
= 0 and x´(0) = 3C_{2} -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 m^{2} - 3m - 4 = 0
with roots m = -1 and 4. Thus the solution is

x = C_{1}e^{-t} + C_{2} e^{4t}
and x´ = -C_{1}e^{-t} + 4C_{2} e^{4t}.
Thus y = (x´ - x)/2 = -C_{1}e^{-t} + (3/2)C_{2}
e^{4t}.

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´_{1}e^{-t} + C´_{2}
e^{4t} and y´ = -C´_{1}e^{-t} +
4C´_{2} e^{4t}. Now x = (y´ - 2y)/3 =
-C´_{1}e^{-t} + (2/3)C´_{2} e^{4t}

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 m^{2} - 3m - 4 = 0
with roots m = 1 and -4. Thus the complementary solution is x_{c}
= C_{1}e^{t} + C_{2} e^{-4t}.

Take the particular solution as x_{p}^{ }=
At + B. Then x_{p}´ = A and x_{p}´´
= 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 = C_{1}e^{t}
+ C_{2} e^{-4t} + 3t/4 + 9/16 and x´ = C_{1}e^{t}
- 4C_{2} e^{-4t} + 3/4.

Thus y = (x´ + 2x)/ 3 = C_{1}e^{t}
- (2/3)C_{2} 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. y_{c} = C_{1}´e^{t}
+ C_{2}´ e^{-4t}.

Take y_{p} = At + B. Then y_{p}´
= A and y_{p}´´ = 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 = C_{1}´e^{t} + C_{2}´
e^{-4t} + t/2 + 5/8 and y´ = C_{1}´e^{t}
- 4C_{2}´ e^{-4t} + 1/2.

Thus x = (y´ + y + t)/2 = C_{1}´e^{t}
- (3/2)C_{2}´ 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 m^{2} - 4m - 12 = 0
with roots m = -2 and 6. Thus the complementary solution is x =
C_{1}e^{-2t} + C_{2} e^{6t}.

For the particular solution take x_{p} = At +
B. Then x_{p}´ = A and x_{p}´´ =
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 = C_{1}e^{-2t} + C_{2}
e^{6t} + t - 1/4 with x´ = -2C_{1}e^{-2t}
+ 6C_{2} e^{6t} + 1.

Thus y = (x´ - x + t -5)/5 = -(3/5)C_{1}e^{-2t}
+ C_{2} e^{6t} - 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 = C_{1}´e^{-2t} + C_{2}´ e^{6t}.

For the particular solution take y_{p} = At +
B. Then y_{p}´ = A and y_{p}´´ =
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 = C_{1}´e^{-2t} + C_{2}´
e^{6t} - 3/4 with y´ = -2C_{1}´e^{-2t}
+ 6C_{2}´ e^{6t}.

Now x = (y´ - 3y + 3t - 3)/3 = -(5/3)C´_{1}e^{-2t}
+ C´_{2} e^{6t} + 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.