Page 201 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes which can be integrated directly to yield
x
log|Z| + 2log|u | + a ( )ds C ,
s
1 1
where s is a dummy variable, for some constant C. Thus

c x dU
s
Z exp a 1 ( )ds
u 2 1 dx
C
where c = e . This can then be integrated to give
z c t
x
U ( ) 2 exp a 1 ( )ds dt c ,
s
t
u ( )
1
for some constant c . The solution is therefore
x c t
x
x
y ( ) u 1 ( ) 2 exp a 1 ( )ds dt u 1 ( ).
x
s
t
u 1 ( )
We can recognize u (x) as the part of the complementary function that we knew to start with,
1
and
x 1 t
x
u 2 ( ) u 1 ( ) exp a 1 ( )ds dt ...(3)
x
s
u 2 1 ( )
t
as the second part of the complementary function. This result is called the reduction of order
formula.
Example: Let us try to determine the full solution of the differential equation
2
2 d y dy
(1 x ) 2x 2y 0
dx 2 dx
given that y u 1 ( ) x is a solution. We firstly write the equation in standard form as
x
2
d y 2x dy 2 y 0
dx 2 1 x 2 dx 1 x 2
2
Comparing this with (2), we have a (x) = 2x/(1 x ). After noting that
1
t t 2s 2
a 1 ( )ds 2 ds log(1 t ),
s
1 s
the reduction of order formula gives

x 1 x dt
x
u 2 ( ) x 2 exp{ log(1 t 2 )}dt x 2 2
t t (1 t )
We can express the integrand in terms of its partial fractions as

1 1 1 1 1 1
t 2 (1 t 2 ) t 2 1 t 2 t 2 2(1 t ) 2(1 t )

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