Study Advice Service
Mathematics
Worksheet
Differential
Equations 1
This is one of a series of worksheets designed to help you increase your confidence
in handling Mathematics. This worksheet contains both theory and exercises which
cover 1st Order Differential Equations. It deals with 4 types:1. Exact
2. Variables Separable
3. Homogeneous
4. Linear
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.
For students at the University of Hull
 Ask your lecturers
 Contact the Study Advice Service in 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
 Ask your lecturers
 Access your Study Advice or Maths Help Service
 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
further help please contact us by email.
Web: www.hull.ac.uk/studyadvice
Email: studyadvice@hull.ac.uk
Tel: 01482 466199
What Are Differential Equations?
Any equation containing terms of the form
dny
is a differential equation.
dx n
The following are all differential equations of varying order and degree
dy
x
 y 2  0,
dx
Sometimes
dy
,
dx
d2y
dx 2
2
 y sin x  0,
dx 2
d3y
are written as y ' ,
dx3
xy
,
d2y
Order: If the highest derivative appearing is
If the highest is
2
d3y
 dy 
   3.
 dx 
dx
y' ',
y ' ' ' to save space!
dy
then the equation is of order 1.
dx
d2y
then it is of order 2 etc. Not all such equations can be solved in
dx 2
terms of the unknown, some have to be solved by numerical methods giving
approximate results.
Type 1 Exact
If you are asked to solve the Differential Equation
dy
 cos x  3x 2 you can integrate
dx
immediately giving y  sin x  x 3  A . In the same way the following second order
Differential Equation can be solved directly by integration:
d2y
2
integrating
integrate again
 3e 3x  4 x  sin x
dx
dy
 e 3x  2 x 2  cos x  A
dx
e 3x 2 x 3
y

 sin x  Ax  B
3
3
Note that integrating twice has introduced the 2 arbitrary constants A and B.
dny
 f x  then you can, in theory, integrate n times to
dx n
find the solution. This assumes that you can integrate at each step (remember that
you cannot integrate every function algebraically). Your solution will contain n
arbitrary constants.
If you have a DE of the form
Type 2 Variables Separable
dy
Equations of the form f  x   g  y   0 can be solved by ‘separating the variables’
dx
which means all the x -terms along with dx can be put on one side of the equation
and all the y -terms along with dy can be put on the other side.
page 1
Examples

dydx  xy  0
1 : Find the general solution of the equation x 2  1
This gives
dy
xy

2
dx
x 1



Integrating
dy
x

dx
2
y
x 1
dy
x

dx
y
x2  1




ln y  1 ln x 2  1  A
2
and simplify, putting ln B  A
ln y  ln x 2  1  ln B  ln B x 2  1 


y  B x 2  1 is the general solution
hence
2 : Solve the equation
4 x2  1
dy
 xy given y(1)  0 (ie y  0 when x  1)
y 1
dx
This can be written as
Simplify and Integrate
4 x2  1
dx  y y  1 dy
x
1

2
  4 x  x  dx   y  y dy




y3 y2
2 x  ln x 

A
3
2
2  ln1  0  0  A  A  2
2
When x  1, y  0
y3 y2
2 x  ln x 

2
3
2
2
Solution
Exercise 1
Solve the following
dy
1. x  y 2
dx


dy
2. x x 2  1 
dx
2
3.
2 xy
2

dy
dx
x 1
dy
6. e x  y
1
dx
dy x  x
dy

5. cos 2 x
 cos 2 y
2
dx y  y
dx
dy y  1
dy
7. x

y
8. ( x  1) y  x , where y(1)  1
dx y  1
dx
dy
dy

9. tan x
 cot y where y(0) 
10. sin x  cos x 
 cos x  sin x
dx
dx
2
4.
11. Find the equation of the curve which has gradient function
dy
y

2
dx x  1
and passes through the point (2, -1).
page 2
Type 3 Homogeneous Equations
A homogeneous equation of the first order is one that can be written in the form
dy
 y
 f   . To test whether a function can be written in that form substitute y  vx
dx
 x
and the function should reduce to one containing only v .
For instance
dy x3 y  x 2 y 2

dx 4 xy3  x3 y
becomes
dy
vx4  v 2 x 4
v  v2 x4
v  v2



.
dx 4v3 x 4  vx4
4v3  v x 4 4v3  v


 
 


The substitution y  vx simplifies the function from two variables to one so it seems
dy
logical to try the same substitution for
. This is another technique which may have
dx
come about by ‘trial and error’, but helps us in this case, to solve equations.
Differentiating y  vx (using product rule) gives
dy d vx dv
dx
dv


xv
 x v.
dx
dx
dx
dx
dx
So by substituting y  vx the differential equation
dv
dy
 y
 f   becomes x  v  f v 
dx
dx
 x
which can be solved by separating the variables.
dy 4 y

dx
x
dv
4vx
dv
x v 
 4v  x
 3v
dx
x
dx
dv 3dx

v
x
Example 1 Solve the differential equation
Putting y = vx this becomes
Separating the variables gives
Integrate ln v  3ln x  c  ln Ax3 where c  ln A
giving v  Ax3 
y
 Ax3  y  Ax4
x
4
dy
4y
3 4 Ax
Check y  Ax 
which is where we started!
 4 Ax 

dx
x
x
4
page 3
Example 2 : Solve the equation
Substitute y  vx
dy y  x  y 

given y(1)  1 (ie y  1 when x  1).
dx x y  x 
dv
vxx  vx v1  v x 2
x
v 

dx
xvx  x  v  1x 2
dv
v1  v 

v
v  1
dx
dv v1  v   vv  1 2v
x


dx
v 1
v 1
x
separate variables
2 dx
 v  1

 dv 
x
 v 


 1  v  dv   x  2 x


1
integrate
2 dx
dx
v  ln v  2 ln x  c
put c  ln A
But y  vx , hence
or

v  2 ln x  ln v  ln A  ln Ax 2 v

y
y

 ln Ax 2 
x
x

 ln Axy
y  x ln Axy
But y(1)  1; substituting in gives 1  ln A  A  e
The solution can be written as y  x ln  exy  or y  x ln  exy   x ln e  ln xy   x(1 ln xy )
It is important to realise that there are often many different ways of writing the final
solution. It depends on how it is to be used as to which is the most useful form.
page 4
Example 3 : Solve the equation x
dy
y
dx
x2  y 2
dy

dx
this can be written as
substitute y  vx
x
dv
v 
dx
x2  y2  y
x
x 2  v 2 x 2  vx
x


x 1  v 2  vx

x
x
simplify
separate the variables

1  v2  v
1  v2
dx
x
x
dx
sin 1 v  ln x  c
integrate
put v 
dv
v 
dx
dv
x 
dx
dv

1  v2
dv

2
1 v
sin 1
y
and c  ln A
x
y
 ln x  c  ln Ax
x
y  x sinln Ax 
dy
 x y
dx
dy x  y

dx
x
dv
x  vx
x
v 
 1 v
dx
x
dv
x
1
dx
dx
dv 
x
1
v   dx  ln x  A
x
v  ln x  ln B  ln Bx
Example 4: Solve the equation x
Rearrange equation
putting y  vx gives
simplify
separate the variables
integrate
& put A  ln B
y
put v 
x
giving answer
y
 ln Bx
x
y  x ln Bx
page 5
Exercise 2
Solve the following
dy x  2 y
1.

dx
x
2.
3. 2 x 2
dy
 x2  y 2
dx
dy
dy
6. y
 x  2 y
 y  x2  y 2
dx
dx
dy
dy
7. x x 2  y 2
 y y 2  x 2 8. x 3  xy 2
 x2 y  y3
dx
dx
4. xy2

dy
 x3  y 3
dx
dy x 2  y 2

dx
xy

5. x






Type 4 - Linear Equations
dy
 P x  y  Q x  . The method used to solve
dx
such equations depends, again, on noticing a particular pattern.
A linear equation is one of the form
If z  ye f x  then differentiating (using the product rule and the chain rule
dz dy f  x 
 dy


e
 yf ' x e f  x   e f  x   f '  x  y 
dx dx
 dx

When compared to the linear equation above, this suggests that if we can find an
e f  x  to multiply both sides of the linear equation by then we’ll be able to integrate
the left hand side immediately.
Multiplying the linear equation
dy
 P x  y  Q x  by e f  x  gives
dx
 dy

e f  x   Px  y   e f x Qx 
 dx

 dy

comparing left hand side with e f  x   f ' x  y  (which can be integrated
 dx

immediately) shows that we want to find f ( x ) where

f ' x   Px   f x   Px dx
i.e. multiply the equation by e f x   e 
Example 1 Solve the equation
P x dx
dy
 3 y  e2 x
dx
From above the Integrating Factor is e
Multiplying through gives
which is called the Integrating Factor.
e 3 x
3dx
 e3x
dy
 3e 3 x y  e 3 x e 2 x
dx
The left hand side is the differential of ye 3 x
Check, using product rule,



d ye 3 x
dy 3 x

e
 y  3e 3 x
dx
dx
dy
 e 3 x
 3e 3 x y
dx

page 6
From the original equation
dy
 3 y  e2 x
dx
multiply through by e 3 x
e 3 x
dy
 3e 3 x y  e 3 x e 2 x
dx


d ye 3 x
 e x
dx
this can be written as

ye 3 x  e  x dx
integrating
 e  x  A
y  e 2 x  Ae3 x
or, multiplying through by e3 x
dy
 4 y sin x  sin 2 x find y( x ) (ie y as a function of x ),
dx
(ie y  0 when x   ).
Example 2 Given 2 cos x
given that y(0)  
3
3
dy
 Px  y  Qx 
dx
dy 2 sin x
sin 2 x 2 sin x cos x

y

dx cos x
2 cos x
2 cos x
dy
 2 tan x  y  sin x
dx
2 tan x dx
The equation is linear, integrating factor e 
Divide through by 2cos x to write it in the form
 2 tan xdx 2 ln sec x  ln sec
2
x
ln sec 2 x 
2 tan x dx

  sec 2 x
e 
hence integrating factor is e
multiply by I. F.
sec 2 x


dy
 2 sec 2 x tan x y  sec 2 x sin x
dx


d y sec 2 x
sin x
 sec 2 x sin x 
dx
cos2 x
y sec 2 x  
Integrate
Multiply by cos2 x
sin x
cos 2 x
dx 
1
A
cos x
y  cos x  A cos 2 x
But y  0 when x   hence 0  cos   A cos2 
3
3
0
1 1
2 4
2
3
A  A  2
Solution y( x)  cos x  2 cos x
page 7
Example 3
Solve the equation
dy
yx
dx
1dx
The I. F. is e 
 e x giving
e x
dy
 ye  x  xe x
dx
 
d ye x
 xe x
dx
hence
giving

ye  x  xe x dx

  xe x   e  x dx
Integrating the right side by parts
  xe x  e  x  A
giving
y  Ae x  x  1
Exercise 3
Solve the following
dy
dy
1.
 y  e x
2.
 xy  0
dx
dx
dy
dy
4. x
 y  x2
5.
 y x
dx
dx
dy
dy
8. x  2 y 
7.
 y cot x  cos x
dx
dx
dy
 y tan x  cos 2 x
dx
dy
11. 1  x 2
 xy  1
dx
9.



dy y 1
 
dx x x
dy
6.
 2y  x
dx
3.
x3
dydx  2 xy  x
10. x 2  1
12.
dy
 3 y  e2 x
dx
Exercise 4
Miscellaneous Differential Equations
Solve the following
dx
dy
1. tan t
x
2. x
 y  x 2 ln x
dt
dx
dy
dy
4
4. 3 x
 x 2  2 5. 2 x
 8y 
dx
dx
x2
3. cos   sin
6. 3
d2y
dx
2
d
dr
 12 x 2  2 sin x
dy
 xy ln x which satisfies
dx
the conditions x  1, y  1. Give ln y in terms of x .
dy
8. Solve the differential equation x  2 y  e x  x  0  given that
dx
y  1 when x  1.
dy
9. Find the solution of the differential equation
 y cot x  sin x
dx
for which y  1 when x   .
7. Find the solution of the differential equation
page 8
dy
 x which satisfies the
dx
boundary condition y  2 when x  0 . Give y in terms of x .
10. Find the solution of the differential equation ye 2 x
ANSWERS
(note many of the answers can be put in slightly different forms, depending whether
the constant is put as A or ln B or  ln C etc.)
Exercise 1
1. 3x 2  2 y 2  A
2. x 4  2 x 2  4 y  A
4. 2 y 3  3 y 2  2 x 3  3x 2  A
5. tan y  tan x  A
6. e y  e  x  A
7. ln  x y 2  1  tan 1 y  A


10. y  lnsin x  cos x   A
9. sin x cos y  1
Exercise 2
1. x3  3x 2 y  A
2. y 2  2 x 2 ln Ax 

3. ln x 2  1  ln y  A or


y  A x2  1
8. x  ln x  1  ln y
11. y 2 x  1  3x  1
3. 2 x   y  xln Ax

1 x3  2 y3
4. ln x  ln
 A or Bx 3 x 3  2 y 3  1
3
6
x
y
y
5. sin1  ln x  A or sin1  ln Ax
x
x
yx
x
x
6. ln

  ln Ax
or ln y  x 
A
x
yx
yx
7. 2 x 2 ln Axy  y 2  0
8. 2 y 2 ln xy  x 2  Cy 2
page 9
Exercise 3
x2
1. ye x  x  A or y  x  Ae x
2. y  Ae 2
3. xy  x  A or y  1  Ax1
4. y  x 2  Ax
6. 4 y  2 x  1  Ae2 x
7. y  1 sin x  A or 4 y sin x  cos 2 x  A
8. y  x3  Ax2
9. 3 y cos x  3 sin x  sin3 x  A

1
sin x
2

5. y  x  1  Ae x
10. 2 y x 2  1  x 2  A 11. y 1  x 2  sin1 x  A 12. 5 ye 3x  e 5x  A
Exercise 4
1. x  B sin t
2. y  x 2 ln x  x 2  Ax
4. y  1 x 2  2 ln Ax
6
3

2
5. y 
 4
7. ln y  1 x 2 ln x  1 1  x 2
9. y sin x  1 x  1 sin 2 x  
2
4
2

3. r  ln sec   c
x2  A
x 4  2 sin x
6. y 
 Ax  B
3
x4
8. y 
xe x  e x  1

x2
10. y 2  1 9  e  2 x  2 xe  2 x
2

We would appreciate your comments on this worksheet, especially if
you’ve found any errors, so that we can improve it for future use.
Please contact the Maths tutor by email at studyadvice@hull.ac.uk
updated 29th November 2004
The information in this leaflet can be made available in an
alternative format on request. Telephone 01482 466199
© 2009
page 10