Download Particular Solutions of Differential Equations

Further Differential Equations
We already know how to solve first order separable differential equations.
We will now study more complex differential equations.
First Order Linear Differential Equations
A first order linear differential equation takes the form
dy
 P  x y  f  x
dx
dy
P  x   2 x, f  x   e x 
 2 xy  e x

dx
dy
P  x   sin x, f  x   x 2 
  sin x  y  x 2

dx
We may need to perform some manipulation to get the equation in this form.
dy
e.g.
x
 y  3x 2
dx
e.g.
dy y  3x 2

dx
x
dy y
  3x
dx x
dy 1
 y  3x
dx x
1


 P  x   , f  x   3x 
x


The method we use to solve this type of differential equation is called the integrating factor
method.
The integrating factor is I  x   e 
P  x  dx
.
The strategy for solving these comes from the following.
I  x   e
P  x  dx
P x  dx 
d
d
I  x     e


dx
dx 

 P  x  e
P  x  dx
(by the Chain Rule)
 P  x I  x
We now take the equation
dy
 P  x y  f  x .
dx
If we multiply both sides of our equation by I  x  we have
I  x
dy
 I  x P  x y  I  x f  x
dx
1
The LHS is
d
d
d
dy
I  x  y  by the product rule as
I  x  y    I  x   y  I  x 


dx
dx
dx
dx
dy
 P  x I  x y  I  x
dx
d
 I  x y   I  x f  x
dx
Now we can integrate both sides of this equation.



  I  x  dx  I  x  P  x  y dx    I  x  f  x  dx
dy
d

  dx  I  x  y      I  x  f  x  dx
I  x  y    I  x  f  x  dx
This can now be solved for y .
This is why the integrating factor method works. It will be much clearer how it works by
example.
Examples
① Find the general solution of
dy
 2 y  e3 x
dx
2 dx
Here P  x   2 so the integrating factor is e   e2 x .
Multiply both sides of the differential equation by the integrating factor.
 dy

e 2 x   2 y   e 2 x e3 x
 dx

d 2x
e y   e5 x

dx
Now integrate both sides.
 dx  e y dx   e
d
2x
5x
dx
1
e 2 x y  e5 x  C
5
1
y  e3 x  Ce 2 x
5
1
The general solution is y  e3 x  Ce 2 x
5
② Find the general solution of the differential equation
dy y
  4x2 .
dx x
1
dx
1

so the integrating factor is e x  eln x  x .
x
Multiply both sides of the differential equation by the integrating factor.
P  x 
2
 dy y 
x     4 x3
 dx x 
d
 xy   4 x3
dx
d
3
 dx  xy dx   4 x dx
Integrate both sides.
xy  x 4  C
C
x
y  x3 
The general solution is y  x3 
C
.
x
dy xy  4

.
dx
x2
The equation must firstly be rearranged into the correct form.
dy
x2
 xy  4
dx
dy 1
4
 y 2
dx x
x
③ Find the general solution of
1
1
1
1
 dx
P  x   so I.F.= e x  e ln x  eln x 
x
x
Multiply both sides of the differential equation by the integrating factor.
1  dy 1  4
  y 
x  dx x  x 3
d 1

dx  x
Integrate both sides.

y   4 x 3

d 1 
 dx  x y dx   4 x
3
dx
y
 2 x 2  C
x
2
y  2  Cx
x
The general solution is y  Cx 
2
.
x
④ Find the general solution of the differential equation
P  x   2 x so the I.F.  e
2xdx
dy
 2 xy  5 x .
dx
 ex .
2
Multiply both sides of the differential equation by the integrating factor.
3
2  dy
2

e x   2 xy   5 xe x
 dx

2
d x2
e y  5 xe x
dx
d x2
x2
e
y
dx

5
xe
Integrate both sides
 dx
 dx
2
5
e x y   eu du
2
2
5 2
ex y  ex  C
2
5 C
The general solution is y   x2 .
2 e
 
 
Let u  x 2 ,
du
 2x
dx
5
du  5 xdx
2
Questions
Find the general solution of the following differential equations.
①
dy
 2y 1
dx
②
dy
 2 y  1 2x
dx
③ x
dy
 2 y  6x4
dx
④
dy
 4 xy  10 x
dx
Particular Solutions of Differential Equations
If we are given initial conditions for x and y we can find the particular solution of a
differential equation.
Examples
① Find the unique solution of the differential equation x 2
dy
 x 3  x 2 y  0 if when
dx
x  1 , y  1.
Rearrange into correct form
P  x   1 so I.F.  e
1dx
dy
x y 0
dx
dy
yx
dx
 ex
Multiply both sides of the differential equation by the integrating factor.
 dy

e x   y   xe x
 dx

d x
e y   xe x

dx
4
d x
x
 dx  e y  dx   xe dx
Integrate both sides
ux
v '  ex
u' 1
v  ex
e x y  xe x   e x dx
e x y  xe x  e x  C
y  x  1  Ce x
1  1  1  Ce 1
C e
y  x  1  ee x
Apply the initial conditions
So the unique solution is y  x  1  e1 x
② Find the particular solution of the differential equation x
dy
 y  x 2 given that x  1
dx
when y  0 .
Rearrange into correct form
dy y
 x
dx x
1
1
1
1
 dx
P  x 
so I.F.  e x  e ln x  eln x 
x
x
Multiply both sides of the differential equation by the integrating factor.
1  dy y 
   1
x  dx x 
d 1 
 y 1
dx  x 
Integrate both sides
d 1 
 dx  x y dx   1dx
1
y  xC
x
y  x 2  Cx
Apply the initial conditions
0  12  C
C  1
So the particular solution is y  x 2  x .
Questions
Find the particular solutions of the following differential equations.
①
dy
 3y  x
dx
given x  0 when y  1
5
②
dy
 y tan x  sec x given y  0  1
dx
3
dy 3
 2 y  e x given x  3 when y  5e
dx x
dy
 y cos x  2 xesin x given y  0 when x  
④
dx
③
Second Order Linear Differential Equations
A second order differential equation takes the form
d2y
dy
a 2  b  cy  f  x  where a , b and c are constants and a  0 .
dx
dx
There are 2 different cases.
If f  x   0 , this is a homogenous equation.
If f  x   0 , this is a non-homogenous equation.
Homogenous Equations
As these equations involve a second derivative the general solution will contain two
arbitrary constants.
The method we use to find the general solution is to consider a solution of the form
y  Aemx for some values of m (to be calculated) and A (to be calculated if given initial
conditions).
y  Aemx
dy
 mAe mx
dx
d2y
 m2 Aemx
2
dx
d2y
dy
 b  cy  0
2
dx
dx
am2 Aemx  bmAe mx  cAe mx  0
Substitute these into a
divide through by Ae mx to get
am 2  bm  c  0
This is known as the auxiliary equation. It can be factorised to solve for m .
There are 3 possible cases when solving the auxiliary equationb 2  4ac  0
two real distinct roots
b  4ac  0
repeated real roots
2
6
b 2  4ac  0 complex roots.
Example - two real distinct roots
d2y
dy
 4  3y  0 .
Find the general solution of the differential equation
2
dx
dx
2
The auxiliary equation is m  4m  3  0
 m 1 m  3  0
m  1 or m  3
The general solution is y  Ae x  Be3 x
Example - real repeated roots
If the roots are equal we must consider y  Aemx as one solution and y  Bxemx as the
other.
Find the general solution of the differential equation
d2y
dy
 4  4y  0 .
2
dx
dx
The auxiliary equation is m 2  4m  4  0
 m  2
2
0
m  2 (twice)
The general solution is y  Ae2 x  Bxe2 x .
Example - complex roots
If the roots are complex we will need to use the quadratic formula and the roots will be in
the form p  qi . The general solution is given by y  e px  A cos qx  B sin qx  .
d2y
dy
 2  5y  0 .
Find the general solution of the differential equation
2
dx
dx
2
The auxiliary equation is m  2m  5  0 .
2  22  4 1 5
m
2
2  16
2
1
m  1 
16i 2
2
m
m  1  2i
The general solution is y  e x  A cos 2x  B sin 2 x  .
7
Questions
Find the general solutions of the following differential equations.
①
d 2 y dy
  6y  0
dx 2 dx
②
d2y
dy
 6  9y  0
2
dx
dx
③
d2y
dy
 4  13 y  0
2
dx
dx
④
d2y
dy
 6  5y  0
2
dx
dx
⑤
d2y
dy
 2  17 y  0
2
dx
dx
⑥2
d2y
dy
 5  3y
2
dx
dx
Particular Solutions to Second Order Differential Equations
In a similar way to before we can apply initial conditions to find the values of the constants
A and B . We have to differentiate the general solution to be able to do this.
Examples
① Find the particular solution of
d2y
dy
 3  2 y  0 given that when x  0 , y  0 and
2
dx
dx
dy
 2.
dx
Auxiliary equation is m2  3m  2  0 .
 m 1 m  2  0
m  1 or m  2
General solution is y  Ae x  Be2 x .
Apply initial conditions : y  Ae x  Be2 x
 0  A B
dy
 Ae x  2 Be2 x  2  A  2B  A  2, B  2
dx
Particular solution is y  2e2 x  2e x
② Find the particular solution of
d2y
dy
 10  25 y  0 given that when x  0 , y  1 and
2
dx
dx
dy
 2.
dx
8
Auxiliary equation is m2  10m  25  0
 m  5
2
0
m  5 (twice)
General solution is y  Ae  Bxe5 x .
5x
1  A
Apply initial conditions: y  Ae5 x  Bxe5 x
dy
 5 Ae5 x  Be5 x  5Bxe5 x
dx
 2  5 B
B  3
Particular solution is y  e5 x  3xe5 x .
d2y
dy
③ Find the particular solution of the differential equation 4 2  4  9 y  0 given that
dx
dx
dy
 1.
when x  0 , y  6 and
dx
Auxiliary equation is 4m 2  4m  9  0
4  42  4  4  9
8
1 1
m

128
2 8
1
m
 2i
2
m
General solution is y  e
x
2
 A cos
x
2 x  B sin 2 x


Apply initial conditions: y  e 2 A cos 2 x  B sin 2 x

 6  e0  A cos 0  B sin 0
A6
x
2
dy e

dx
2


x

A cos 2 x  B sin 2 x  e 2  2 A sin 2 x  2 B cos 2 x
1 
1
6  B  0   2 A 0  2B
2
1  3  2B
4
B
2 2
2
x

Particular solution is y  e 2 6cos 2 x  2 2 sin 2 x
9




Questions
Find the particular solution of the following differential equations.
①2
②
d2y
dy
dy
 5  3 y  0 given that when x  0 , y  1 and
3
2
dx
dx
dx
d2y
dy
dy
 4  13 y  0 given that when x  0 , y  2 and
0
2
dx
dx
dx
③4
d2y
dy
 20  25 y  0 given that when x  0 , y  2 and when x  2 , y  3e5
2
dx
dx
④ 16
⑤
d2y
dy
 9 y  0 given that when x  0 , y  1 and
3
2
dx
dx
d 2 y dy
dy
  y  0 given that when x  0 , y  1 and
4
2
dx
dx
dx
Non-Homogenous Equations
When we are dealing with a non-homogenous equation we initially deal with it as if it were
homogenous.
We set the RHS=0 to obtain the auxiliary equation which we then solve for m . We then use
these values of m to obtain the complementary function.
To deal with the RHS we need something called a particular integral which should take the
form (or similar form) to the RHS.
Finallygeneral solution = complementary function + particular integral
Some examples will make this clearer.
Examples
① Find the general solution of the differential equation
The corresponding homogenous equation is
The auxiliary equation is m2  m  2  0
10
d 2 y dy
  2 y  4e2 x .
2
dx
dx
d 2 y dy
  2y  0 .
dx 2 dx
 m 1 m  2  0
m  1 or m  2
The complementary function is y  Ae x  Be2 x .
The particular integral is of the form y  ke2 x .
dy
 2ke 2 x
dx
d2y
 4ke2 x
2
dx
d 2 y dy
  2 y  4e2 x
dx 2 dx
4ke2 x  2ke2 x  2ke2 x  4e2 x
Substitute these into the differential equation
4ke2 x  4e2 x
k 1
So the particular integral is y  e2 x .
The general solution is y  Ae x  Be2 x  e2 x .
② Find the general solution of the differential equation
d2y
dy
 3  2 y  2x .
2
dx
dx
d2y
dy
3  2y  0 .
The corresponding homogenous equation is
2
dx
dx
2
The auxiliary equation is m  3m  2  0
 m 1 m  2  0
m  1 or m  2
The complementary function is y  Ae2x  Be x
The particular integral is of the form y  ax  b
dy
a
dx
d2y
0
dx 2
d2y
dy
 3  2 y  2x
2
dx
dx
0  3a  2  ax  b   2 x
Substitute these into the differential equation
2ax  2b  3a  2x
Equating coeffiecients:
2a  2,
a 1
2b  3a  0
2b  3  0
11
b
3
2
3
So the particular integral is y  x  .
2
The general solution is y  Ae2 x  Be x  x 
3
.
2
d2y
dy
 5  4 y  8x2  3 .
③ Find the general solution of the differential equation
2
dx
dx
The corresponding homogenous equation is
d2y
dy
5  4y  0
2
dx
dx
The auxiliary equation is m2  5m  4  0
 m 1 m  4  0
m  1 or m  4
The complementary function is y  Ae4x  Be x .
The particular integral is of the form y  ax 2  bx  c
dy
 2ax  b
dx
d2y
 2a
dx 2
d2y
dy
 5  4 y  8x2  3
Substitute these into the differential equation
2
dx
dx
2a  5  2ax  b   4  ax 2  bx  c   8 x 2  3
4ax2   4b 10a  x   2a  5b  4c   8x2  3
Equating coefficients: 4a  8
a2
4b 10a  0
4b  20
b5
2a  5b  4c  3
4  25  4c  3
c6
So the particular integral is y  2 x 2  5x  6 .
The general solution is y  Ae4 x  Be x  2 x 2  5x  6
d2y
dy
 2  2 y  5sin 2 x
④ Find the general solution of the differential equation
2
dx
dx
d2y
dy
 2  2y  0 .
The corresponding homogenous equation is
2
dx
dx
2
The auxiliary equation is m  2m  2  0
12
2  22  4 1 2
2
1
m  1 
4i 2
2
m  1  i
m
The complementary function is y  e x  A cos x  B sin x 
The particular integral is of the form y  p sin 2 x  q cos 2 x
dy
 2 p cos 2 x  2q sin 2 x
dx
d2y
 4 p sin 2 x  4q cos 2 x
dx 2
d2y
dy
 2  2 y  5sin 2 x
Substitute these into the differential equation
2
dx
dx
4 p sin 2 x  4q cos 2 x  2  2 p cos 2 x  2q sin 2 x   2  p sin 2 x  q cos 2 x   5sin 2 x
 4q  2 p  sin 2x   4 p  2q  cos 2x  5sin 2x
Equating coefficients:
4 p  2q  0
4 q  2 p  5
2  4 p   2 p  5
2q  4 p
10 p  5
p
1
, q  1
2
1
sin 2 x  cos 2 x
2
1
The general solution is y  e  x  A cos x  B sin x  sin 2 x  cos 2 x
2
So the particular integral is y 
⑤ Find the general solution of the differential equation
The corresponding homogenous equation is
d2y
dy
 3  2 y  e4 x  cos x .
2
dx
dx
d2y
dy
3  2y  0 .
2
dx
dx
The auxiliary equation is m2  3m  2  0 .
 m 1 m  2  0
m  1 or m  2
The complementary function is y  Ae2x  Be x .
The particular integral is of the form y  ae4 x  p cos x  q sin x
13
dy
 4ae4 x  p sin x  q cos x
dx
d2y
 16ae4 x  p cos x  q sin x
2
dx
Substitute these into the differential equation
d2y
dy
 3  2 y  e4 x  cos x
2
dx
dx
16ae4 x  p cos x  q sin x  3  4ae 4 x  p sin x  q cos x   2  ae 4 x  p cos x  q sin x   e 4 x  cos x
6ae4 x   p  3q  cos x   3 p  q  sin x  e4 x  cos x
Equating coefficients: 6a  1
a
3p  q  0
1
6
q  3 p
p  3q  1
p 9p 1
p
1
,
10
q
3
10
1
1
3
So the particular integral is y  e 4 x  cos x  sin x .
6
10
10
1
1
3
The general solution is y  Ae2 x  Be x  e 4 x  cos x  sin x
6
10
10
Questions
Find the general solution of each of the following
①
d 2 y dy
  6 y  sin x
dx 2 dx
d2y
dy
 2  3 y  22e2 x
②
2
dx
dx
d2y
dy
 2  3 y  9 x 2  3x
③
2
dx
dx
④
d 2 y dy
  12 y  x
dx 2 dx
When the ‘Normal’ Particular Integral is in the Complementary Function
The particular integral cannot have the same form as either term of the complementary
function. If it is, we should multiply the particular integral by x .
If this is also part of the complementary function (e.g. in the case of a repeated root) we
should multiply the particular integral by x 2 .
14
Examples
d2y
dy
 5  6 y  8e2 x .
① Find the general solution of the equation
2
dx
dx
The corresponding homogenous equation is
d2y
dy
5  6y  0
2
dx
dx
The auxiliary equation is m2  5m  6  0
 m  2 m  3  0
m  2 or m  3
The complementary function is y  Ae2 x  Be3 x
For the particular integral we cannot use y  ke2 x as this is already part of the
complementary function so we use y  kxe2 x .
dy
 ke 2 x  2kxe 2 x
dx
d2y
 2ke2 x  2  ke2 x  2kxe2 x 
dx 2
 4ke 2 x  4kxe 2 x
d2y
dy
 5  6 y  8e2 x .
Substitute these into the differential equation
2
dx
dx
4ke 2 x  4kxe 2 x  5  ke 2 x  2kxe 2 x   6kxe 2 x  8e 2 x
ke2 x  8e2 x
k  8
So the particular integral is y  8 xe2 x .
The general solution is y  Ae2 x  Be3 x  8xe2 x .
NB If you pick the wrong particular integral a contradiction will occur at some stage in the
calculation and this should flag it up.
d2y
dy
 8  16 y  6e4 x .
② Find the general solution of the differential equation
2
dx
dx
d2y
dy
 8  16 y  0 .
The corresponding homogenous equation is
2
dx
dx
The auxiliary equation is m2  8m  16  0 .
 m  4
2
0
15
m  4 (twice)
The complementary function is y  Ae4 x  Bxe4 x .
For the particular integral we cannot use y  ke4 x or y  kxe4 x as these are both already
in the complementary function so use y  kx 2e4x .
dy
 2kxe4 x  4kx 2 e4 x
dx
d2y
 2ke4 x  8kxe4 x  8kxe4 x  16kx 2e4 x
dx 2
 2ke4 x  16kxe4 x  16kx 2e4 x
d2y
dy
 8  16 y  6e4 x .
Substitute these into the differential equation
2
dx
dx
2ke4 x  16kxe4 x  16kx 2e 4 x  8  2kxe 4 x  4kx 2e 4 x   16kx 2e 4 x  6e 4 x
2ke4 x  6e4 x
2k  6
k 3
So the particular integral is y  3x 2e4 x .
The general solution is y  Ae4 x  Bxe4 x  3x 2e4 x .
③ Find the general solution of the differential equation 4
The corresponding homogenous equation is 4
d2y
 36 y  sin 3x .
dx 2
d2y
 36 y  0 .
dx 2
The auxiliary equation is 4m2  36  0 .
m 2  9
m  3i
The complementary function is y  e0  A cos3x  B sin 3x 
y  A cos 3x  B sin 3x
For the particular integral try y  x  p cos3x  q sin 3x  .
dy
 p cos 3x  q sin 3x  x  3 p sin 3x  3q cos 3x 
dx
d2y
 6 p sin 3x  6q cos 3x  9 xp cos 3x  9 xq sin 3x
dx 2
d2y
Substitute these into the equation 4 2  36 y  sin 3 x .
dx
4  6 p sin 3x  6q cos3x  9 xp cos3x  9 xq sin 3x   36 x  p cos3x  q sin 3x   sin 3x
24 p sin 3x  24q cos 3x  sin 3x
16
Equating terms: 24 p  1
p
1
24
So the particular integral is y 
24q  0
q0
x
cos 3x .
24
The general solution is y  A cos 3 x  B sin 3 x
x
cos 3 x .
24
Past Paper Questions
2001
① Find the solution of the following differential equation:
dy y
  x,
x0
dx x
(4 marks)
② Find the general solution of the following differential equation:
d2y
dy
 2  3 y  6x 1
2
dx
dx
(5 marks)
2002
Find the general solution of the differential equation
d2y
dy
 2  5 y  4cos x
2
dx
dx
Hence determine the solution which satisfies y  0  0 and y '  0  1 .
(6, 4 marks)
2003
Solve the differential equation
d2y
dy
 4  4 y  ex
2
dx
dx
given that y  2 and
dy
 1 , when x  0 .
dx
(10 marks)
2004
(a) A mathematical biologist believes that the differential equation x
process. Find the general solution of the differential equation.
17
dy
 3 y  x 4 models a
dx
Given that y  2 when x  1 , find the particular solution , expressing y in terms of x .
(5, 2 marks)
(b) The biologist subsequently decides that a better model is given by the equation
dy
y  3x  x 4 .
dx
Given that y  2 when x  1 , obtain y in terms of x .
(4 marks)
2005
Obtain the general solution of the differential equation
d2y
dy
 3  2 y  20sin x
2
dx
dx
Hence find the particular solution for which y  0 and
dy
 0 when x  0 .
dx
(7, 3
marks)
2006
Solve the differential equation
d2y
dy
 2  2y  0
2
dx
dx
dy
 2.
given that when x  0 , y  0 and
dx
(6 marks)
2007
d2y
dy
 6  9 y  e2 x
Obtain the general solution of the equation
2
dx
dx
(6 marks)
2008
Obtain the general solution of the differential equation
d2y
dy
 3  2 y  2 x2
2
dx
dx
Given that y 
1
dy
 1 , when x  0 , find the particular solution.
and
2
dx
2009
(a) Solve the differential equation
18
(7, 3 marks)
dy
4
 3 y   x  1
dx
given that y  16 when x  1 , expressing the answer in the form y  f  x  .
 x  1
(b) Hence find the area enclosed by the graphs of y  f  x  , y  1  x  and the x -axis.
4
(6, 4 marks)
2010
Obtain the general solution of the equation
d2y
dy
 4  5y  0 .
2
dx
dx
Hence obtain the solution for which y  3 when x  0 and y  e when x 

2
.
(4, 3 marks)
2011
Find the general solution of the differential equation
d 2 y dy
  2 y  e x  12
2
dx
dx
Find the particular solution for which y 
3
dy 1
 when x  0 .
and
2
dx 2
(7, 3 marks)
2012
(a) Express
1
 x  1 x  2 
2
in partial fractions.
(b) Obtain the general solution of the differential equation
dy
x 1
 x  1  y 
2
dx
 x  2
expressing your answer in the form y  f  x  .
2013
Solve the differential equation
19
(4, 7 marks)
d2y
dy
dy
 6  9 y  4e3 x , given that y  1 and
 1 when x  0 .
2
dx
dx
dx
(11 marks)
2014
Find the solution y  f ( x) to the differential equation
4
d2y
dy
dy
 4  y  0, given that y  4 and
 3 when x  0 .
2
dx
dx
dx
(6 marks)
2015
Solve the second order differential equation
d2y
dy
 2  10 y  3e2 x
2
dx
dx
dy
 0.
given that when x  0 , y  1 and
dx
(10 marks)
2015
Solve the differential equation
d2y
dy
 5  6 y  12 x 2  2 x  5
2
dx
dx
given y  6 and
dy
 3, when x  0.
dx
(10 marks)
20