# First Order Linear Differential Equations by hcj

VIEWS: 10 PAGES: 38

• pg 1
```									         First Order Linear Differential Equations

Any equation containing a derivative is called a differential equation.

A function which satisfies the equation is called a solution to the differential equation.

The order of the differential equation is the order of the highest derivative involved.

The degree of the differential equation is the degree of the power of the highest
derivative involved.
dy
When a differenti al equation is of the form        f ( y)
dx
1
          dy  dx      We can then integrate both sides.
f ( y)

dy
            dx   This will obtain the general solution.
f ( y)
dy
When the equation is of the form                f ( x) g ( y ), then
dx
1
dy  f ( x)dx
g ( y)

dy
          f ( x)dx
g ( y)

A first order linear differential equation is an equation of the form

dy
 P( x) y  Q( x)                1
dx

To find a method for solving this equation, lets consider the simpler equation

dy
 P( x) y  0       Which can be solved by separating the variables.
dx
(Remember: a solution is of the form y = f(x) )
dy                      dy
 P( x) y  0            P( x) y
dx                      dx
dy
     P( x) dx
y
 ln y    P( x) dx  c

 ye 
 P ( x ) dx  c

  P ( x ) dx c
 ye               e

 y  Ce 
 P ( x ) dx

ye  P ( x ) dx  C
or

Using the product rule to differentiate the LHS we get:
d  P ( x ) dx   dy  P ( x ) dx             P ( x ) dx
ye              e             yP ( x)e
dx               dx

 dy          P ( x ) dx
   P( x) y  e
 dx        

Returning to equation 1,
dy
 P( x) y  Q( x)          If we multiply both sides by     e  P ( x ) dx
dx

 dy          P ( x ) dx             P ( x ) dx
   P( x) y  e             Q ( x )e
 dx        
d  P ( x ) dx                 P ( x ) dx
ye               Q ( x )e                         Now integrate both sides.
dx
 P ( x ) dx  Q( x)e  P ( x ) dx dx
ye               
 P( x)dx and      Q( x)e 
P ( x ) dx
For this to work we need to be able to find                                                       dx
dy 2
Solve the differential equation                  yx
dx x
2
Step 1: Comparing with equation 1, we have P( x) 
x
2
 P ( x)dx   dx  2ln x  ln x 2  ln x 2
x

Step 2: e  P ( x ) dx  eln x2  x 2      e  P ( x ) dx is called the integrating factor

Step 3: Multiply both sides by the integrating factor.
 dy 2 
x2   y   x2 x
 dx x 

d 2
yx  x 3
dx
d 2
yx  x 3
dx
yx 2   x3dx

x4
yx 2   c
4
1 2
y  x  cx 2
4

Note the solution is composed of two parts.

1                                           2
The particular solution: y  x 2 and the complementary function: y  cx
4
dy 2
The particular solution satisfies the equation:      y  Q( x )
dx x
dy 2
The complementary function satisfies the equation:    y0
dx x
dy
Solve the differential equation             2 xy  3 x
dx
Step 1: Comparing with equation 1, we have P ( x)  2 x and Q( x)  3 x

 P( x)dx   2 xdx   x 2

Step 2: Integrating Factor e  P ( x ) dx  e  x2

Step 3: Multiply both sides by the integrating factor.
 x2    dy       x2
e            2 xy   e 3x
 dx     
d  x2        x2
ye  3 xe
dx
d  x2        x2
ye  3 xe
dx
ye     x2
  3xe    x2
dx   To solve this integration we need to use substitution.

Let t  x2  dt  2 xdx
3 t       3       3  x2
 3xe x dx 
2
e dt   et   e
2

2       2

3  x2
 x2
ye   e  c
2
3
y    ce  x2

2

If you look at page 113 you will see a note regarding the constant when finding the
integrating factor. We need not find it as it cancels out when we multiply both sides
of the equation.
dy 3
Solve the differential equation x     x  xy  0
2

dx
dy 1
rewriting in standard form:      yx
dx x
1
Step 1: Comparing with equation 1, we have P( x)  and Q( x)  x
x
dx  ln x
 P( x)dx  
x
Step 2: Integrating Factor e  P ( x ) dx  eln x  x
The modulus vanishes as
we will have either both
Step 3: Multiply both sides by the integrating factor.      positive on either side or
both negative. Their effect
is cancelled.
d
yx  x 2   (note the shortcut I have taken here)
dx
yx   x 2 dx
1 2
y  x  cx 1
3
dy
Solve the differential equation cos x  y sin x  cos 2 x
dx
dy
rewriting in standard form:       y tan x  cos x
dx
Step 1: Comparing with equation 1, we have P( x)  tan x and Q( x)  cos x

 P( x)dx   tan x dx  ln sec x
Step 2: Integrating Factor e  P ( x ) dx  eln sec x  sec x
The modulus vanishes as
we will have either both
Step 3: Multiply both sides by the integrating factor.          positive on either side or
both negative. Their effect
d                                                         is cancelled.
y sec x  1 (note the shortcut I have taken here)
dx
y sec x  1dx

y  x cos x  c cos x
So far we have looked at the general solutions only. They represent a whole family
of curves. If one point can be found which lies on the desired curve, then a unique
solution can be identified.

Such a point is often given and its coordinates are referred to as the initial
conditions or values. The unique solution is called a particular solution.
dy
Solve the differential equation x       y  x 2 given x  1 when y  0
dx
dy 1
rewriting in standard form:    yx
dx x
1
Step 1: Comparing with equation 1, we have P( x)  
x
1
 P( x)dx    x dx   ln x
Step 2: Integrating Factor e  P ( x ) dx  e  ln x  1
x
Step 3: Multiply both sides by the integrating factor.
d y
1
dx x
y
  1dx
x
y  x 2  cx
y  x 2  cx     given x  1 when y  0
0  1 c
c  1

Hence the particular solution is   y  x2  x

Page 114 and 116 Exercise 1 and 2

TJ Exercise 1
Second-order linear differential equations

d2y     dy
Differential equations of the form a 2  b  cy  Q( x)
dx      dx
are called second order linear differential equations.

When Q ( x)  0 then the equations are referred to as homogeneous,

When Q( x)  0 then the equations are non-homogeneous.

Note that the general solution to such an equation must include two arbitrary
constants to be completely general.
Theorem
If y  f ( x) and y  g ( x) are two solutions then so is y  f ( x)  g ( x)

Proof

d2 f  df             d 2g  dg
we have a 2  b  cf  0 and a 2  b  cg  0
dx    dx             dx    dx

d2 f  df       d 2g  dg
Adding:    a 2  b  cf  a 2  b  cg  0
dx    dx       dx    dx

 d 2 f d 2 g   df dg 
a 2  2   b    c f  g   0
 dx    dx   dx dx 

And so y  f ( x)  g ( x) is a solution to the differential equation.
dy
y  Ae , for A and m, is a solution to the equation b  cy  0
mx

dx

It is reasonable to consider it as a possible solution for
d2y   dy
a 2  b  cy  0
dx    dx

dy           d2y
y  Ae   mx
     Ame mx  2  Am2emx
dx           dx

If y  Aemx is a solution it must satisfy aAm 2 e mx  bAme mx  cAe mx  0

assuming Aemx  0, then by division we get am 2  bm  c  0

The solutions to this quadratic will provide two values of m which will make
y = Aemx a solution.
If we call these two values m1 and m2, then we have two solutions.

y  Ae m1x   and   y  Be m2 x

A and B are used to distinguish the two arbitrary constants.

From the theorem given previously;

y  Aem1x  Bem2 x     Is a solution.

The two arbitrary constants needed for second order differential equations ensure
all solutions are covered.

The equation am2  bm  c  0 is called the auxiliary equation.
The type of solution we get depends on the nature of the roots of this equation.
When roots are real and distinct
d2y   dy
Find the general solution of 2  5  6 y  0.
dx    dx

The auxiliary equation ism 2  5m  6  0
(m  2)( m  3)  0
m  2, or m  3

Thus the general solution is y  Ae2 x  Be3 x .

To find a particular solution we must be given enough information.
d2y   dy
Find the particular solution to 2 2  7  4 y  0,
dx    dx
dy
given y  1 and      2 when x  0.
dx

The auxiliary equation is    2m 2  7 m  4  0
(2m  1)(m  4)  0
1
m  , or m  4
2
1
x
Thus the general solution is y  Ae  Be 4 x .
2

Using the initial conditions.   1  Ae0  Be0  1  A  B

dy 1 1 x           1 0             1
 Ae  4 Be  2  Ae  4 Be  2  A  4 B
4 x
2                      0
Now
dx 2               2               2
1 A B                                   4      1
Solving gives   A , B
1                                     3      3
2     A  4B
2

4 1 x 1 4 x
Thus the particular solution is   y  e2  e
3     3

Page 119 Exercise 3 Questions 1(a), (b) 2(a), (b)

But we will now look at the solution when the roots are real and coincident.
Roots are real and coincident
When the roots of the auxiliary equation are both real and equal to m, then the
solution would appear to be y = Aemx + Bemx = (A+B)emx

A + B however is equivalent to a single constant and second order equations need two.

With a little further searching we find that y = Bxemx is a solution. Proof on p119

So a general solution is

y  Aemx  Bxemx
Find the general solution of y''  6 y ' 9 y  0.

The auxiliary equation is   m 2  6m  9  0
(m  3)( m  3)  0
m  3 (twice)

Thus the general solution is y  Ae3 x  Bxe3 x .
Find the particular solution to 9 y ''- 6 y ' y  0,
given y  1 when x  0 and y  4e when x  3.

The auxiliary equation is       9m 2  6m  1  0
(3m  1)(3m  1)  0
1
m  (twice)
3
1       1
x       x
Thus the general solution is y  Ae  Bxe .    3       3

Using the initial conditions.       1  Ae0  B 0e 0  1  A

Now 4e  Ae1  3Be1  4e  e  3Be  1  B

1        1          1
x        x
1  x 
x
y  e  xe
3        3
e   3
Page 120 Exercise 4 Questions 1(a), (b) 2(a), (b)

But we will now look at the solution when the roots are complex conjugates.
Roots are complex conjugates
When the roots of the auxiliary equation are complex, they will be of the form
m1 = p + iq and m2 = p – iq. Hence the general equation will be

y  Ae( piq ) x  Be( piq ) x
 Ae px eiqx  Be px e  iqx
 e px  Aeiqx  Beiqx        We know that ei  cos   i sin 
 e px  A  cos qx  i sin qx   B  cos( qx)  i sin( qx)  
 e px  A  cos qx  i sin qx   B  cos qx  i sin qx  
 e px   A  B  cos qx   A  B  i sin qx 
 e px  C cos qx  D sin qx 

Where C  A  B and D  ( A  B )i
d2y     dy
Find the solution to    2
 6  13 y  0.
dx      dx
The auxiliary equation is   m 2  6m  13  0
6  36  52   6  16
m              
2         2
m  3  2i

Thus the general solution is y  e3 x  C cos 2 x  D sin 2 x 
d2y     dy
Find the particular solution to    2
 4  13 y  0,
dx      dx
dy
given that y  2 and      0 when x  0.
dx

The auxiliary equation is   m 2  4m  13  0
4  16  52 4  36
m              
2         2
m  2  3i

Thus the general solution is y  e 2 x  C cos3x  D sin 3 x 

Substituting gives:   2  C cos0  D sin 0  C  2
dy
 2e 2 x  C cos3x  D sin 3 x   e 2 x  3D cos3x  3C sin 3x 
dx
dy
 2e 2 x  C cos3x  D sin 3 x   e 2 x  3D cos3x  3C sin 3x 
dx

Substituting gives:   0  2  2cos 0  D sin 0    3D cos 0  6sin 0 
0  4  3D
4
D
3

2 x            4        
The particular solution is   ye           2cos3x  sin 3 x 
         3        

Page 122 Exercise 5A Questions 1(a), (b) 2(a), (b)

TJ Exercise 2
Non homogeneous
second order differential equations
d2y   dy
Non homogeneous equations take the form      a 2  b  cy  Q( x)
dx    dx

Suppose g(x) is a particular solution to this equation. Then

d 2g  dg
a 2  b  cg  Q( x)
dx    dx
Now suppose that g(x) + k(x) is another solution. Then

d 2 (g  k)    d (g  k)
a        2
b            c( g  k )  Q( x)
dx            dx
d 2g  d 2k  dg  dk
Giving    a 2  a 2  b  b  cg  ck  Q( x)
dx    dx    dx  dx

 d 2g   dg      d 2k     dk    
  a 2  b  cg    a 2  b  ck   Q( x)
 dx     dx      dx       dx    

 d 2k   dk    
 Q( x)   a 2  b  ck   Q( x)
 dx     dx    
d 2k   dk
 a 2  b  ck  0
dx    dx

From the work in previous exercises we know how to find k(x).

This function is referred to as the Complimentary Function. (CF)

The function g(x) is referred to as the Particular Integral. (PI)

General Solution = CF + PI
d2y      dy
Find the general solution to      2
 5  6 y  15 x  7,
dx       dx
given that the PI is of the form k ( x)  Px  Q

Finding the (CF): the auxiliary equation is    m 2  5m  6  0
 (m  3)( m  2)  0
m  2 or m  3
Thus the CF is y  Ae  Be 2x       3x

dy     d2y
Finding the PI: y  Px  Q     P  2 0
dx     dx
Substituting into the original equation     0  5P  6  Px  Q   15 x  7
 6 Px  6Q  5 P  15 x  7
5
 6P  15  p 
2
11
Q
12
5    11
Thus the PI  x 
2 12

Hence the general solution is                       5    11
y  Ae 2 x  Be3 x  x 
2 12
d2y      dy
Find the general solution to         4  6 x  2,
dx 2     dx
given that the PI is of the form k ( x)  Px 2  Qx

Finding the (CF): the auxiliary equation is     m 2  4m  0
 m( m  4)  0
m  0 or m  4
Thus the CF is y  A  Be4 x
dy             d2y
Finding the PI: y  Px 2  Qx      2 Px  Q  2  2 P
dx             dx
Substituting into the original equation       2 P  4(2 Px  Q )  6 x  2
 8 Px  2 P  4Q  6 x  2
3
 8P  6  p  
4
7
Q
8
3     7
Thus the PI   x 2  x
4     8

3 2 7
Hence the general solution is   y  A  Be  x  x
4x

4   8
Finding the form of the PI
d2y   dy
a 2  b  cy  Q( x) is non homogeneous because Q( x)  0
dx    dx

The individual terms of the CF make the LHS zero
The PI makes the LHS equal to Q(x). Since Q(x) ≠ 0, it stands to reason that the
PI cannot have the same form as the CF.
When choosing the form of the PI, we usually select the same form as Q(x).

This reasoning leads us to select the PI according to the following steps.
•try the same form as Q(x)
•If this is the same form as a term of the CF, then try xQ(x)
•If this is the same form as a term of the CF, then try x2Q(x)
CF: y  Ae3 x  Be2 x ; Q( x)  2 x, for PI try y  Cx  D a linear function

CF: y  Ae3 x  Be2 x ; Q( x)  2e4 x , for PI try y  2e4 x

CF: y  Ae3 x  Be2 x ; Q( x)  2e3 x , for PI try y  Cxe3 x (note extra x factor)

CF: y  A  Be4 x ; Q( x)  6 x  2, for PI try y  x(Cx  D)
(note extra x factor since A is a linear function)

CF: y  A sin 3x  B cos3x; Q( x)  3x 2  1, for PI try
y  Cx 2  Dx  E a quadratic function

CF: y  Ae3 x  Be3 x ; Q( x)  3cos x, for PI try y  C sin x  D cos x, a wave function

CF: y  A sin 3 x  B cos3 x; Q( x)  3cos x, for PI try y  Cx sin x  Dx cos x
d2y     dy
Find the general solution to the equation       5  6 y  8e2 x
dx 2    dx
Finding the (CF): the auxiliary equation is  m 2  5m  6  0
 (m  2)( m  3)  0
m  2 or m  3
Thus the CF is y  Ae2 x  Be3 x

Finding the PI. Q( x)  8e2 x so we might try Ce2 x
but this is the same form as the Ae2 x in the CF. So we try Cxe2 x
dy             d2y
y  Cxe   Ce  2Cxe  2  4Ce2 x  4Cxe2 x
2x   2x     2x

dx             dx
Substituting into the original equation

4Ce2 x  4Cxe2 x  5(Ce2 x  2Cxe2 x )  6Cxe2 x  8e2 x
Equating Coefficients, we get C  8
Hence the PI is y  8 xe2 x

Thus the general solution is y  Ae2 x  Be3 x  8 xe2 x

If the wrong selection of PI is made, you will generally be alerted to this by the
occurrence of some contradiction in later work.

Page 126 – Exercise 7A as many as possible.

TJ Exercise 3. This assesses beyond poly and trig work.

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