```Study Advice Service
Student Support Services
Mathematics
Worksheet
Differential
Equations 2
in handling Mathematics. This worksheet contains both theory and exercises which
cover 2nd Order Differential Equations of the form
1
d2y
dx
2
 f ( x)
2
a
d2y
dx
2
b
dy
 cy  0
dx
dy
 cy  h( x)
dx
dx 2
There are often different ways of doing things in Mathematics and the methods
suggested in the worksheets may not be the ones you were taught. If you are
successful and happy with the methods you use it may not be necessary for you to
change them. If you have problems or need help in any part of the work then there
are a number of ways you can get help.
3 a
d y
2
b
For students at the University of Hull
 Contact the Study Advice Service on the ground floor of the Brynmor Jones
Library where you can access the Mathematics Tutor, or contact us by email.
 Come to a Drop-In session organised for your department
 Look at one of the many textbooks in the library.
For others
 Use any other facilities that may be available.
If you do find anything you may think is incorrect (in the text or answers) or want
Tel: 01482 466199
Email: [email protected]
dy
d2y
dy
 cy  0 a
 b  cy  hx 
dx
dx
dx 2
dx 2
dx 2
are Second Order Differential Equations. They can be a lot more complex than this
but this worksheet will only consider these three.
Equations of the form
d2y
 f x 
a
d2y
b
d2y
 f  x  can be solved by integrating twice (assuming that the function
dx 2
f x  and its integral can be integrated).
Type 1
Example Solve the equation G
d2y
dx 2
 W 1  x   0 where G and W are constants,
subject to the conditions that y(0)  0 , y '(0)  1 (ie when x  0 , y  0 and
G
d2y
dx 2
 W 1  x   0 
Integrating gives
dy
y '(0)  1 ie x  0 ,
1 
dx
giving
integrating again gives
y(0)  0 ie x  0 , y  0 
G
d2y
dx 2
dy
 1 ).
dx
 W 1  x   W  Wx
dy
Wx 2
 Wx 
A
dx
2
G 1  0  0  A  A  G
G
dy
Wx 2
G
 Wx 
G
dx
2
Wx 2 Wx 3
Gy 

 Gx  B
2
3 2
G  0 0  0  0  B  B  0
hence solution is
Gy 
or
y
Wx 2 Wx 3

 Gx
2
6
Wx 2 Wx 3

x
2G
6G
The method for type 1 is obvious but some methods used to solve Differential
Equations appear to be ‘plucked from thin air’! In many areas of life tools are
developed to tackle certain jobs. For instance in surgery there are a large number of
tools which are designed to be used in particular operations. The tools to solve
Differential Equations are often ‘designed’ to deal with a particular type. They can be
justified, some of them quite easily, others are more complex.
Type 2 Linear equations with constant coefficients of the form:
d2y
dy
a
 b  cy  0
dx
dx 2
The first order, linear equation
dy
 cy  0 has solution y  kecx so it seems logical
dx
to try y  e mx as the solution for a second order Linear Differential equation...
1
y  e mx 
dy
d2y
 memx 
 m 2e mx
2
dx
dx
substituting into the equation gives
am2emx  bmemx  cemx  0
am2  bm  c emx  0
but as e mx  0 this gives am2  bm  c  0.
This equation is often called the auxiliary equation and there are 3 cases to consider
depending on whether the roots of the quadratic equation are (i) real and different, (ii)
real and equal or (iii) complex.
Case (i) Two distinct roots
Let the roots of the auxiliary equation am2  bm  c  0 be m1  p and m2  q then
y  e px , y  eqx are solutions of the differential equation and it looks as though the
general solution could be y  Ae px  Be qx ( A and B are arbitrary constants),
y  Ae px  Be qx 
Check
dy
d2y
 pAe px  qBe qx 
 p 2 Ae px  q 2 Be qx
dx
dx 2
Substituting into the differential equation and collecting terms gives
a
d2y
dx
2
b




dy
 c  A ap 2  bp  c e px  B aq 2  bq  c e qx
dx
But as p and q are roots of the equation am2  b m  c  0 then
ap 2  bp  c  0 and aq 2  bq  c  0
hence a
d2y
dx 2
b
dy
 c  A0  B0  0
dx
y  Ae px  Be qx is the general solution of the given 2nd order differential equation as
it satisfies the differential equation and has two arbitrary constants.
Example
Find the general solution of the equation
d2y
dx
2
5
dy
 6y  0
dx
The auxiliary equation is m2  5m  6  0 which factorises to give m  1m  6  0 with
roots m1  1, m2 6
General solution is y  Ae x  Be 6 x
2
Case (ii) Roots real and equal
Experience shows that if the auxiliary equation am2  bm  c  0 has equal
roots m1  p and m2  p then the general solution is y   Ax  B e px
[This can be checked, as above, by substituting into the differential equation.]
Example
d2y
Find the general solution of the equation
dx
2
6
dy
 9y  0
dx
The auxiliary equation is m  6m  9  0 which gives m1  3 , m2  3 and general
2
solution y   Ax  B  e3x . Check this by substitution.
Case (iii) complex roots
If the auxiliary equation am2  bm  c  0 has complex roots of the form
m    j where j   1 , the solution is y  ex  A cos x  B sin x  .
[This can be checked, as above, by substituting into the differential equation.]
Example Solve the equation
d2y
2
2
dy
 4y  0
dx
dx
The auxiliary equation is m  2m  4  0 which gives
2
m
  2 
 22  414 2 

21
 12 2  2 j 3

 1 j 3
2
2




The general solution is y  e x  A cos x 3  B sin x 3  . Check this by substitution.

 
 
 
NOTE: we write cos x 3 rather than cos 3 x as it can be misread as cos 3 x !
Exercise 1
Solve the equations
d 2 y dy
1.

0
dx 2 dx
4.
d2y
dx
2
2
7.
10.
d y
2
dx
d2y
dx 2
d2y

dy
 2y  0
dx
y0
3
dy
 10 y  0
dx
dy
13.

y0
2
dx
dx
d2y
dy
15.
 2  4y  0
dx
dx 2
2. 2
5.
d2y
dx 2
d2y
dx

2
2
dy
0
dx
3.
dy
y0
dx
6.
d2y
dx
2
d y
dy
8. 3
2  y 0
2
dx
dx
2
11.
d y
dx 2
14.
 9y  0
d2y
dx
16. 3
2
 10
d2y
dx 2
4
9. 4
5
2
d2y
dx
12.
 4y  0
dx 2
d2y
2
d2y
dx 2
dy
 6y  0
dx
4
dy
y0
dx
 9y  0
dy
 25 y  0
dx
dy
y0
dx
3
17. a 2
d2y
dx 2
 2a
dy
y0
dx
18.
d2y
dx 2
6
dy
 13 y  0
dx
d 2x
dx k 2
dx
 u, when t  0.
k

x  0 , given x  0,
19. Solve the equation
2
dt
dt
2
dt
20. Solve the equation
d 2s
dt
2
4
ds
ds
 u, when t  0.
 13s  0 , given s  a,
dt
dt
Type 3 Linear equations with constant coefficients of the form
a
d2y
dx
2
b
dy
 cy  f  x 
dx
This is an extension of type 2. We can get the solution of a
d2y
2
b
dy
 cy  0 in the
dx
dx
form y  Au1  Bu2 where u1 and u2 are functions of x .
If we can find a particular function y  g ( x ) which satisfies the equation
d2y
dy
 cy  f  x 
dx
dx
then we can write the general solution as y  Au1  Bu2  g ( x)
The main problem is finding the function y  g ( x ) !
a
2
b
y  Au1  Bu2 is often called the complementary function
and y  g ( x ) the particular integral.
It quite logical to assume that the function y  g ( x ) will be the same sort of function
as f ( x ) . For example if f ( x ) is of the form e x then g ( x ) must be of the form ke x or
kxe x ... as there are no combinations of other functions which will give the correct
f ( x ) . There are a number of different possibilities three of which are discussed here.
case (i) f ( x ) of the form ke mx
Examples
d 2 y dy
d 2 y dy

 2 y  e3 x (b)

 2 y  e2 x
Solve the equations (a)
2
2
dx
dx
dx
dx
d 2 y dy
  2y  0
In both the complementary function is the solution of
dx 2 dx
This has auxiliary equation m2  m  2  0 which gives m1  2 , m2  1.
Hence complementary function is y  Ae2 x  Be  x .
4
(a)
d2y

dy
 2 y  e3 x has Complementary function y  Ae2 x  Be  x
dx
dx 2
To find a particular integral try y  ke3 x
y  ke3 x 
So, for y  ke
3x
dy
d2y
 3ke3 x 
 9ke3 x
2
dx
dx
to satisfy
d2y
dx
then
2

dy
 2 y  e3 x
dx
9ke3 x  3ke3 x  2ke3 x  e3 x
4ke3 x  e3 x  k  1
Hence the general solution is y  Ae
(b)
d2y

2x
 Be
x
4
1
 e3 x
4
dy
 2 y  e 2 x has Complementary Function y  Ae2 x  Be  x
dx
dx 2
There is a term Ae2x in the complementary function so the particular integral cannot
be of the form y  ke2x as y  Ae2 x  Be  x  ke2 x  A1e2 x  Be  x which will satisfy
d 2 y dy
d 2 y dy

 2 y  0 but not

 2 y  e2 x
2
2
dx
dx
dx
dx
so we try y  kxe2 x
y  kxe2 x

dy
d2y
 ke2 x  2kxe2 x 
 2ke2 x  2ke2 x  4kxe2 x  4ke2k  4kxe2 x
2
dx
dx
So, for y  kxe2 x to satisfy
d2y
dx
e
2

dy
 2 y  e2 x
dx
4k  4kx  k  2kx  2kx  e2 x
e2 x 4k  4kx  k  2kx  2kx  e2 x
2x
3ke2 x  e  k  1
3
Hence the general solution is y  Ae2 x  Be  x  13 xe2 x
case (ii) f ( x ) is a polynomial of degree n
A particular integral can be found by substituting
y  a0 x n  a1x n 1  a2 x n  2  a3 x n 3  .........  an
The highest power in the particular function cannot be higher than the highest power
in the function f ( x ) . You can check this by taking such a function, substituting into
the left hand side and showing that all the coefficients of terms with higher powers
must be zero as there are no similar powers on the right hand side.
5
Example Solve the equation
d 2s
2
 4s  t 2  2
dt
The auxiliary equation is m  4  0 which gives m  2 j . Hence the
complementary function is s  A cos 2t  B sin 2t
2
To find a particular integral try s  a0t 2  a1t  a2
s  a0t 2  a1t  a2 
ds
d 2s
 2a0t  a1 
 2a0
dt
dt 2
d 2s
To satisfy the equation
dt

2
 4s  t 2  2

2a0  4 a0t 2  a1t  a2  t 2  2
4a0t 2  4at1  2a0  4a2  t 2  2
Equating coefficients
1
4
4a1  0
 a1  0
2a0  4a2  2
 a2  38
t term
constant term
 a0 
4a0  1
of t 2 term


Particular Integral is s  14 t 2  83  18 2t 2  3


General solution s  A cos 2t  B sin 2t  18 2t 2  3
case (iii) f ( x ) is of the form k1 cos cx  k2 sin cx (where k1 or k2 may be zero).
In general a particular integral can be found by trying y  p cos cx +q sin cx
Example Solve the equation
d 2r
d 2
4
dr
 5r  12 sin 2
d
Auxiliary equation m 2  4m  5  0 gives complementary function
r  Ae5  Be
Try particular integral r  p cos 2 +q sin 2
dr
 2 p sin 2 +2q cos 2
d
d 2r
d 2
To satisfy the equation
 4 p cos 2 4q sin 2
d 2r
d
2
4
dr
 5r  12 sin 2
d
6
 4 p cos 2  4q sin 2  4 2 p sin 2  2q cos 2   5 p cos 2  q sin 2   12 sin
 4 p cos 2  8q cos 2  5 p cos 2  4q sin 2  8 p sin 2  5q sin 2  12 sin
cos 2  9 p  8q  sin 2  9q  8 p  12 sin
Equating terms in
cos 2 :
sin 2 :
 9 p  8q  0
 9q  8 p  12
Solving
 9 p  8q  0 
 9 p  8q  0 
 81 p  72q  0 



 9q  8 p  12 
 8 p  9q  12 
 64 p  72q  96 
Adding gives  145 p  96  p   96
145
Substituting in gives q 
 108
145
96 cos 2  108 sin 2
General solution r  Ae 5  Be   145
145
This can, of course, be checked in the usual way!
Further examples
(a) Solve the equation
d 2r
d 2
4
dr
 5r  7  34  15 2
d
From the previous example the complementary function is r  Ae5  Be
Try particular integral r  a 2  b  c
dr
 2a  b;
d
d 2r
d 2
To satisfy the differential equation:
 2a
d 2r
d

2
4
dr
 5r  7  34  15 2
d

2a  42a  b   5 a 2  b  c  7  34  15 2
2a  4b  5c   8a  5b  5a 2  7  34  15 2
Equating
 2 terms
 terms
constant terms
 5a  15
8a  5b  34
2a + 4b - 5c = -7
Solving gives a  3 , b  2 , c  1
General Solution r  Ae5  Be  3 2  2  1
7
(b) Solve the equation
d2y
dx
2
 16 y  2 cos 4 x  20e 2 x (a more complex equation!)
Auxiliary Equation m2  16  0  m  4 j
Complementary Function y  A cos 4 x  B sin 4 x
The Particular Integral must have terms to give both the 2cos 4 x term and the
20e
2x
term when put into
d2y
dx 2
 16 y .
We would normally go for y  p cos 4 x  q sin 4 x  ke2 x but we already have
a cos4x term in the CF so we need to use
y  x p cos 4 x  q sin 4 x   ke2 x
This gives
dy
 x 4 p sin 4 x  4q cos 4 x    p cos 4 x  q sin 4 x   2ke 2 x
dx
d2y
dx 2
 x 16 p cos 4 x  16q sin 4 x    4 p sin 4 x  4q cos 4 x 
  4 p sin 4 x  4q cos 4 x   4ke 2 x
 16 x p cos 4 x  q sin 4 x   8 p sin 4 x  q cos 4 x   4ke 2 x
To satisfy the differential equation
d2y
2
 16 y  2 cos 4 x  20e 2 x
dx
 16 x p cos 4 x  q sin 4 x   8 p sin 4 x  q cos 4 x   4ke2 x
 16 x p cos 4 x  q sin 4 x   16ke2 x  2 cos 4 x  20e 2 x
Equating terms in
 8 p sin 4 x  8q cos 4 x  20ke2 x  2 cos 4 x  20e 2 x
sin 4 x :  8 p  0  p  0
cos 4x :
8q  2  q  1
e
2x
4
:
20k  20  k  1
Hence General Solution is y  A cos 4 x  B sin 4 x  14 x sin 4 x  e 2 x
8
Exercise 2 Solve the following
1.
3.
d2y
dx 2
d2y
dx 2
5. 4
 9 y  2e 2 x
2. 4
 y  3 cos 2 x
4.
d2y
dx 2
2
 y  x  3x
6.
dy
x3
7. 2

 3y 
2
dx 2 dx
d y
9.
d y
dx 2
 4 y  4e
8.
2 x
10.
dy
11.
2
 5 y  80e 3 x
2
dx
dx
15.
d2y
dx 2
d2y
dx 2
6
d2y
12.
 y  ex  x
dx 2
d2y
d 2
dt 2
16.
2
d2y
2
d2y
dx 2
d2y
dx 2
14. 2
dy
 14 y  6 cos x  13 sin x
dx
17. Solve the equation
18. If
d2y
dx
d2y
13. 4
dx 2
dx
2
2
d2y
dy
 16 y  e  x
dx
2
dy
 y  100 sin 3x
dx
2
dy
 5 y  15
dx
8
dy
 25 y  50
dx
2
dy
 2y  1  x  x2
dx
dx 2
dx 2
dy
y2
dx
6
d2y
d2y
3

dy
 4x  1
dx
 9 y  3 cos 2 x  sin 2 x
 4  sin t given that  

6
,
d
 0 when t  0 .
dt
dy
 2 y  4  x , find y in terms of x , given that, when x  0 ,
dx
dx
dy
y  4,
 2.
dx
2
3
19. Obtain the solutions of the equation
when x  0 , y  0,
d2y
dx 2
 9 y  2 sin 3x given that
dy
0.
dx
Exercise 1
 2x
1. y  Ae x  B
2. y  Ae
4. y  Ae2 x  Be  x
5. y   Ax  B e x
7. y  A cos x  B sin x
8. y  Ae 3  Be  x
10. y  Ae5 x  Be 2 x
B
x
11. y  Ae3x  Be 3x
3. y  Ae2 x  Be 2 x
6. y  Ae3x  Be 2 x
x
9. y   Ax  B e 2
12. y  A cos 3x  B sin 3x
9
13. y  e
x
2

14. y   Ax  B e 5 x

 3 
 3 
 A cos 2 x   B sin 2 x 
 
 
x
16. y  Ae x  Be 3
15. y  e x A cos x 3  B sin x 3
x
18. y  e 3 x  A cos 2 x  B sin 2 x 
2 
20. s  e 2t a cos 3t  u  2a sin 3t
17. y   Ax  B e a


19. x  2u e 2 sin kt
kt
k
3

Exercise 2
1. y  Ae3 x  Be 3 x  2 e 2 x
2. y  Ae
3. y  A cos x  B sin x  cos 2 x
4. y  Ae8 x  Be 2 x  1 e  x
x
5
4
 Be x  2
9
5. y 
x
Ae 2

x
Be 2
7. y  Ae x  Be
 3x
2
6. y   Ax  B e
 x 2  3x  8

 1 9 x3  9 x 2  42 x  26
54
8. y  e  x  A cos 2 x  B sin 2 x   3
10. y  e 4 x  A cos 3x  B sin 3x   2

x
 23 cos 3x  4 sin 3x 

9. y  Ae 2 x  Be 2 x  xe2 x
11. y  e x  A cos 2 x  B sin 2 x   10e 3 x

12. y  e x  A cos x  B sin x   1 x 2  x  1
2 
2
2 
13. y  A cos x  B sin x  x  1 e x

 
5
 
x
14. y  A  Be 2  2 x2  9 x
15. y  e 3 x A cos x 5  B sin x 5  sin x
16. y  A cos 3x  B sin 3x  1 3 cos 2 x  sin 2 x 
5
17.  
1
6
 cos 2t  sin 2t  2 sin t 


19. y  1 (e 3 x  e 3 x  2 sin 3x )
18
18. y  1 4e x  e 2 x  2 x  11
4
you’ve found any errors, so that we can improve it for future use.
updated 29th November 2004
The information in this leaflet can be made available in an
alternative format on request. Telephone 01482 466199
10
```