MATH2030 Differential Equations

Tutorial #8 - Solutions

• True-False Review.

1. True. This is essentially the statement of the variation-of-parameters method

2. False. The solutions y1 , y2 , ..., yn must form a linearly independent set of solutions
to the associated homogeneous differential equation

3. False. The requirement on the functions C1 , C2 , . . . , Cn , similarly for the case of

a second-order, is that they satisfy a systems of equations with unknown C10 , C20 ,
. . . , Cn0 . Because constants of integration of these derivatives can be arbitrarily
chosen, addition of constants to the functions C1 , C2 , . . . , Cn again yields valid
solutions..

1
• Problems.

1. (a) P (t) = t2 + 4t + 4 = 0 ⇐⇒ t1,2 = −2, then yc (x) = C1 e−2x + C2 xe−2x .

Since F (x) = e−x cos x, the product of two functions e−x and cos x, we find
a particular solution in the form
yp (x) = e−x (A sin x + B cos x).
Substituting this function yp into the given equation, we have
yp00 + 4yp0 + 4yp = 2Ae−x cos x − 2Be−x sin x.
1
Equating it to F (x) = e−x cos x yields A = , B = 0, that is
2
1
yp (x) = e−x sin x.
2
Thus the general solution to the given equation is
1
y(x) = C1 e−2x + C2 xe−2x + e−x sin x,
2
where C1 , C2 are arbitrary constants.
(b) As for (a), P (t) = t2 + 4t + 4 = 0 ⇐⇒ t1,2 = −2, then yc (x) = C1 e−2x +
C2 xe−2x .
Since F (x) = 5xe−2x , is the product of two functions 5x and e−2x , and
moreover, e−2x as well as xe−2x are solutions to the associated homogeneous
differential equation, we find a particular solution in the form
yp (x) = (Ax + B)x2 e−2x = (Ax3 + Bx2 )e−2x .
Substituting this function yp into the given equation and simplifying yields
5
2B + 6Ax = 5x, which gives B = 0, A = .
6
Thus the general solution to the given equation is
5
y(x) = C1 e−2x + C2 xe−2x + x3 e−2x ,
6
where C1 , C2 are arbitrary constants.
2. (a) We have P (t) = t1 + 1 = 0 ⇐⇒ t1,2 = ±i, then yc (x) = C1 cos x + C2 sin x.
Since F (x) = 6ex , we find a particular solution in the form
yp (x) = Aex .
Substituting this function yp into the given equation and simplifying yields
A = 3, which gives yp (x) = 3ex .
Thus the general solution to the given equation is
y(x) = C1 cos x + C2 sin x + 3ex ,
where C1 , C2 are arbitrary constants.

2
 (b) Similarly, P (t) = t2 − 5t = 0 ⇐⇒ t1 = 0, t2 = 5, then yc (x) = C1 + C2 e5x .
Since F (x) = −5x2 + 2x, and t1 = 0 is a simple root of the characteristic
polynomial, we find a particular solution in the form
yp (x) = x(Ax2 + Bx + C).
Substituting this function yp into the given equation, and equating it to
1
F (x) = −5x2 + 2x yields A = , B = C = 0, that is
3
1
yp (x) = x3 .
3
Thus the general solution to the given equation is
1
y(x) = C1 + C2 e5x + x3 ,
3
where C1 , C2 are arbitrary constants.
3. (a) We have P (t) = t2 − t − 2 = 0 ⇐⇒ t1 = 2, t2 = −1, then yc = C1 e2x + C2 e−x .
Since F (x) = 5e2x is a solution to the associated homogeneous equation, we
find a particular solution in the form
yp (x) = Axe2x .
Substituting this function yp into the given equation, and equating it to
5
F (x) = −5e2x yields A = , that is
3
5
yp (x) = xe2x .
3
Thus the general solution to the given equation is
5
y(x) = C1 e2x + C2 e−x + xe2x ,
3
where C1 , C2 are arbitrary constants.
(b) We have P (t) = t2 + 16 = 0 ⇐⇒ t = ±4i, then yc = C1 cos 4x + C2 sin 4x.
Since F (x) = 4 cos x, we find a particular solution in the form
yp (x) = A cos x + B sin x.
Substituting this function yp into the given equation, and equating it to
4
F (x) = 4 cos x yields A = and B = 0, that is
15
4
yp (x) = cos x.
15
Thus the general solution to the given equation is
4
y(x) = C1 cos 4x + C2 sin 4x + cos x,
15
where C1 , C2 are arbitrary constants.

3
4. (a) We have P (t) = t2 +6t+9 = 0 ⇐⇒ t1,2 = −3, then yc (x) = C1 e−3x +C2 xe−3x .
With y1 (x) = e−3x and y2 (x) = xe−3x , we find a particular solution in the
form
yp (x) = C1 e−3x + C2 xe−3x ,
where C1 and C2 satisfy
(
C10 e−3x + C20 xe−3x =0
−3x
0 −3x 0 −3x
−3C1 e + C2 e (1 − 3x) = 2e
x2 +1
,

or equivalently (
C10 + C20 x =0
0 0
−3C1 + C2 (1 − 3x) = x22+1 ,
From this system, we have C10 and C20 and integrating each of these functions
yeilds
C1 (x) = − ln(x2 + 1) and C2 (x) = 2 arctan x.
Finally, we obtain the general solution to the given differential equation

y(x) = e−3x [C1 + C2 x + 2x arctan x − ln(x2 + 1),

where C1 , C2 are arbitrary constants.

(b) We have P (t) = t2 − 4 = 0 ⇐⇒ t = ±2, then yc (x) = C1 e2x + C2 e−2x .
With y1 = e2x and y2 = e−2x , we find a particular solution in the form

yp (x) = C1 e2x + C2 e−2x ,

where C1 and C2 satisfy

(
C10 e2x + C20 e−2x =0
2C10 e2x − 2C20 e−2x = e2x8+1 ,

From this system, we have C10 and C20 and integrating each of these functions
yeilds

C1 (x) = ln(e2x + 1) − 2x − e−2x and C2 (x) = − ln(e2x + 1).

Finally, we obtain the general solution to the given differential equation

y(x) = C1 e2x + C2 e−2x + [ln(e2x + 1) − 2x − e−2x ]e2x − ln(e2x + 1)e−2x ,

where C1 , C2 are arbitrary constants.

5. (a) We have P (t) = t2 − 4t + 5 = 0 ⇐⇒ t = 2 + ±i, then yc = e2x (C1 cos x +

C2 sin x).
With y1 = e2x cos x and y2 = e2x sin x, we find a particular solution in the
form
yp (x) = C1 e2x cos x + C2 e2x sin x,

4
 where C1 and C2 satisfy
(
C10 e2x cos x + C20 e2x sin x =0
0 2x 0 2x
C1 e (2 cos x − sin x) + C2 e (2 sin x + cos x) = e2x tan x.

From this system, we have C10 and C20 and integrating each of these functions
yeilds

C1 (x) = sin x − ln | sec x + tan x| and C2 (x) = − cos x.

Finally, we obtain the general solution to the given differential equation

y(x) = (C1 cos x + C2 sin x)e2x + e2x [cos x − ln | sec x + tan x|],

where C1 , C2 are arbitrary constants.

(b) Similarly, we have P (t) = t2 − 6t + 9 = 0 ⇐⇒ t1,2 = 3, then yc = C1 e3x +
C2 xe3x .
With y1 = e3x and y2 = xe3x , we find a particular solution in the form

yp (x) = C1 e3x + C2 xe3x ,

where C1 and C2 satisfy

(
C10 e3x + C20 xe3x =0
C10 3e3x + C20 e3x (3x + 1) = 4e3x ln x.

From this system, we have C10 and C20 and integrating each of these functions
yeilds
C1 (x) = x2 (1 − 2 ln x) and C2 (x) = 4x(ln x − 1).
Finally, we obtain the general solution to the given differential equation

y(x) = C1 e3x + C2 xe3x + x2 e3x (2 ln x − 3),

where C1 , C2 are arbitrary constants.

6. (a) P (t) = t2 + 1 ⇐⇒ t1,2 = ±i, two linearly independent solutions to a homo-

geneous equation is

y1 (x) = cos x and y2 (x) = sin x.

Then we find a particular solution in the form

yp (x) = C1 (x)y1 (x) + C2 (x)y2 (x),

where C1 and C2 satisfy

(
C10 cos x + C20 sin x = 0
−C10 sin x + C20 cos x = sec x + 4ex

5
 The solution to this system is

 0 sin x
C1 = − − 4 sin xex
cos x
C 0 = 1 + 4 cos xex
2

Hence, (
C1 = ln(cos x) − 2(sin x − cos x)ex
.
C2 = x + 2(sin x + cos x)ex
R R
(here we use the integration by parts to compute sin xex dx and cos xex dx,
π
and also note that ln | cos x| = ln(cos x), because |x| < )
2
Thus the general solution is
y(x) = C1 cos x + C2 sin x + 2ex + cos x ln(cos x) + x sin x,
where C1 , C2 are arbitrary constants.
Note. We can also separate F into the sum of sec x and 4ex , and find
particular solutions to each of equations y 00 + y = sec x and y 00 + y = 4ex ,
and then add the two solutions together.
(b) This time, let’s follow the way given in Note of (a).
We first find particular solutions to the differential equations y 00 + y = csc x
and y 00 + y = 2x2 + 5x + 1, and then add the two solutions together.
By variation-of-parameters, a particular solution to y 00 +y = csc x is yp1 (x) =
C1 (x) cos x + C2 (x) sin x, where
(
cos x C10 + sin x C20 =0
− sin x C10 + cos x C20 = csc x
which give a solution
C1 = −1 and C20 = cot x.
Hence, we can choose C( x) = −x and C2 (x) = ln(sin x) to get
yp1 (x) = −x cos x + ln(sin x) sin x.

Next, to find a particular solution to y 00 + y = 2x2 + 5x + 1, we can use

the method of undetermined coefficients to choose yp2 = Ax2 + Bx + C.
Substituting it into the preceding equation and equating it yields A = 2,
B = 5, and C = −3. Hence,
yp2 (x) = 2x2 + 5x − 3.
Thus we obtain a particular solution to a given differential equation
yp (x) = yp1 (x) + yp2 (x) = −x cos x + ln(sin x) sin x + 2x2 + 5x − 3.

Finally, we have the general solution to the given differential equation

y(x) = C1 cos x + C2 sin x − x cos x + ln(sin x) sin x + 2x2 + 5x − 3,
where C1 , C2 are arbitrary constants.

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7. We have P (t) = t2 + 16 ⇐⇒ t = ±4i, then yc (t) = cos 4t + sin 4t = A cos(4t − φ).
Since F (t) = 130e−t cos t, we we find a particular solution in the form

yp (t) = e−t (A sin t + B cos t).

Substituting this expression into the differential equation, and equating it to

F (t) = 130e−t cos t, we obtain
(
B + 8A = 0
8B − A = 65

which gives A = −1 and B = 8.

Consequently,
yp (t) = e−t (8 cos t − sin t).

Transient part: e−t (8 cos t − sin t)

Steady-state part: A cos(4t − φ)