MA2001N Differential Equations
Lecture Notes for Week 6
[6. Euler’s Equation]
6.
Euler’s Equation
Euler’s equation takes the following form
x 2 y    x y    y  0 ,
x0 ,
(76)
where  and  are constants.
Equation (76) suggests a solution of the form
y  xr ,
(77)
where the form of r will become clear shortly. The reason for writing (77) in this
way is that then all three terms on the left-hand side of (76), that is,  y ,  x y 
and x 2 y  , will all be of the same degree, that is, degree x r .
To confirm this we need to differentiate (77) and then substitute into (76). So,
differentiating (77) gives
y   r x r 1
y   r (r  1) x r 2 .
and
(78)
Substituting (77) and (78) into (76) now gives
x 2 (r (r  1) x r 2 )   x (r x r 1 )   ( x r )  0
or, after combining terms and taking out a common factor,
[ r (r  1)   r   ] x r  0 .
(79)
For (79) to be valid for all x in its domain, we must therefore have
r (r  1)   r    0
or
r 2  (  1) r    0 .
(80)
Equation (80) is a quadratic equation in r and can be solved to give
r1 , r2 
 (  1)  (  1) 2  4 
2
1
.
(81)
MA2001N: Lecture Notes for Week 6 (continued)
6.
Euler’s Equation (continued)
As usual with quadratic equations, there are three possible cases to consider:
Case 1: Roots of (81), r1 , r2 , are real and distinct;
Case 2: Roots of (81), r1 , r2 , are both complex;
Case 3: Roots of (81), r1 , r2 , are real and equal.
Each of these cases will now be dealt with in turn.
6.1
Case 1: Roots of (81): real and distinct
When (  1) 2  4  , equation (81) will give real and distinct values for r1 and
r2 . Equation (77) then gives the two solutions:
y1  x r1
and
y 2  x r2 ,
corresponding to the two roots r1 , r2 . The general solution of (76) is then
the sum of k1 y1 and k 2 y 2 , which gives
y  k1 x r1  k 2 x r2 ,
(82)
with r1 , r2 from (81).
6.2
Case 2: Roots of (81): both complex
When (  1) 2  4  , equation (81) will give complex number values for r1 and
r2 . To develop this further, write equation (81) as follows:
r1  
4  (  1) 2
(  1)
i
2
2
and
4  (  1) 2
(  1)
,
r2  
i
2
2
where i is the unit imaginary number, i   1 , and where it should be
noticed that, by reversing signs, the term
take the term to be strictly positive.
2
4  (  1) 2 is now real. We also
MA2001N: Lecture Notes for Week 6 (continued)
6.2
Case 2: Roots of (81): both complex (continued)
The roots, r1 , r2 , may be abbreviated as
r1    i 
and
r2    i  ,
(83)
where
(  1)

2
and

4  (  1) 2
.
2
(84)
The general solution can now be found by combining (82), which still applies
even though the roots are now complex, and (83). This gives:
y  c1 x (  i )  c2 x (  i ) ,
(85)
where c1 and c2 have been used to indicate complex constants.
Taking out the common factor, x  , (85) becomes
y  x  (c1 x i   c2 x  i  ) .
(86)
It is important to realise that the solution, y , as given in (86), is a complex number.
We are, however, looking for a real solution. To find such a solution, we need to
develop the terms in (86), using the theory of complex numbers.
To extract all the real parts of (86), consider the complex number, z , where
z  xi  .
From the definition of z , it follows that
ln z  ln ( x i  )  i  ln x .
Taking the exponential of both sides now gives
e ln z  e i  ln x
or
z  e i  ln x .
But this can be written in trigonometrical form and so
z  cos (  ln x)  i sin (  ln x) .
(continued overleaf)
3
MA2001N: Lecture Notes for Week 6 (continued)
6.2
Case 2: Roots of (81): both complex (continued)
Using the definition of z and extending to include the negative case,   , gives
x  i   cos (  ln x)  i sin (  ln x) .
(87)
After substituting (87) back into (86), and then remembering that c1 and c2 are
complex constants, it is possible to select all possible real quantities from the
resulting expression. Having done that, it becomes possible to write the real
solution for y as
y  x  (k1 cos (  ln x)  k 2 sin ( ln x) ) ,
(88)
where k1 and k 2 are now real constants and  and  are given in (84).
6.3
Case 3: Roots of (81): real and equal
When (  1) 2  4  , equation (81) will give real and equal values for r1 and
r2 . It follows from equation (81) that r1 (  r2 ) is given by:
r1  
(  1)
.
2
(89)
Since r1 in (89) is now the one and only root of (81), it follows that equation (77)
will give one (and only one) solution, which will be:
y1  x  ( 1) / 2 .
(90)
To find the second solution, y 2 , we can apply the method of Reduction of Order.
The method of Reduction of Order has been derived in section 5.2 (of the notes for
weeks 3, 4 and 5) and restating equation (72) gives
y 2  y1 
W
dx ,
y12
(91)
where
W  exp(   p ( x) dx) .
To apply (91), we need to identify p (x ) . In this case, from (76), after division by x 2 ,
the equation we are attempting to solve can be written:
(continued overleaf)
4
MA2001N: Lecture Notes for Week 6 (continued)
6.3
Case 3: Roots of (81): real and equal (continued)
y  

x
y 

x2
y0 ,
x0 ,
(92)
with  constrained to be (  1) 2 / 4 for the equal roots case. So, from (92)
we can now identify the expression for p (x ) to be used in this case as:
p ( x) 

x
.
We can now evaluate the Wronskian, W , which we need to use in (91), as:
W  exp(  

x
dx) .
Integrating now gives
W  exp(   ln x)
or
W  x  .
(93)
We can now apply the Reduction of Order formula for y 2 . Using (91), with y1
from (90) and W from (93), we can write
y 2  x  ( 1) / 2 
x 
( x  ( 1) / 2 ) 2
dx .
Noting that the square in the denominator simplifies to x  ( 1) , y 2 becomes
y 2  x  ( 1) / 2 
x 
x  ( 1)
dx ,
which again can be simplified, by combining the powers of x , to give
y 2  x  ( 1) / 2 
1
dx .
x
Upon integrating, the final form for y 2 can therefore be written
y 2  x  ( 1) / 2 ln x .
(94)
(continued overleaf)
5
MA2001N: Lecture Notes for Week 6 (continued)
6.3
Case 3: Roots of (81): real and equal (continued)
To obtain the general solution, it is now necessary to combine y1 from (90) and
y 2 from (94) with the constants k1 and k 2 respectively. This gives the general
solution for this case, the case of real and equal roots, as
y  ( k1  k 2 ln x ) x  ( 1) / 2 .
Now try Example Sheet 6.
That concludes the notes for Week 6.
6
(95)