Page 202 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 12: Sturm Comparison and Separation Theorems

This gives the second solution of (2) as Notes

x 1 1 1
u 2 ( ) x 2 dt
x
t 2(1 t ) 2(1 t )
x
1 1 1 t x 1 x
x log log 1,
t 2 1 t 2 1 x
and hence the general solution is

x 1 x
y = Ax + B log 1 .
2 1 x

Self Assessment

1. Use the reduction of order method to find the second independent solution of the equation

2
d y 2 dy y 0
dx 2 x dx
1
with the solution u (x) = x sin x
1
12.3 The Method of Variation of Parameters

Let’s now consider how to find the particular integral given the complementary function,
comprising u (x) and u (x). As the name of this technique suggests, we take the constants in the
1 2
complementary function to be variable, and assume that
y = c (x)u (x) + c (x)u (x)
1 1 2 2
Differentiating, we find that

dy du 1 dc 1 du 2 dc 2
c 1 u 1 c 2 u 2
dx dx dx dx dx
We will choose to impose the condition
dc dc
u 1 1 u 2 2 0, ...(4)
dx dx
and thus have

dy du du
c 1 1 c 2 2 ,
dx dx dx
which, when differentiated again, yields
2
2
2
d y c d u 1 du dc 1 c d u 2 du dc 2
1
2
dx 2 1 dx 2 dx dx 2 dx 2 dx dx
This form can then be substituted into the original differential equation to give
2
2
d u du dc d u du dc du du
c 1 1 1 1 c 2 2 2 2 a c 1 1 c 2 2 a 0 (c u c u ) f .
2 2
1 1
1
dx 2 dx dx dx 2 dx dx dx dx
LOVELY PROFESSIONAL UNIVERSITY 195