Page 317 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes k k
x
= log(k 2 ) 1 1 4 2logx dx
k 2x x
k k
= log(k 2 ) 6 2log x dx
x
k 2x x
2 k 1
x
x
= log(k 2 ).x .xdx 6x log(k 2 ) k log x 2 log .x .xdx
x
k 2x 2 x
k 2x k k
= x log(k 2 ) dx 6x log(k 2 ) k log x 2 log x 2x
x
x
x
k 2x 2
k k
x
x
x
x
= x log(k 2 ) x log(k 2 ) 6x log(k 2 ) k log x 2 log x 2x
2 2
k k
x
= x log y x log y 6x log y k log x 2 log x 2x (putting back y =k + 2x)
2 2
= x log y x 6x k logx 2 logx 2x
x
x
= x log y 3x (y 2 )log x 2 log x
x
x
= x log y 3x y log .
Hence the complete solution is
x
x
x
z = 1 (y 2 ) (y 2 ) x log y 3x y log .
3
3
x
Example 3: Solve: r t tan x tany tan tan y
Solution: The given equation is
2
(D 2 D 2 )z = tan tan (tanx y 2 x tan y ).
2
= tan tan (secx y 2 x sec y )
C.F. = (y x ) (y x ).
1 2 2
y
x
P.I. = tan tan (sec x sec y )
(D D )(D D )
1 2 2
x
= tan tan(c x ){sec x sec (c x )}dx [where c x = y]
D D
1 2 2
x
x
= tan tan(c x )sec x dx tan tan(c x )sec (c x )dx
D D
1 1 2 1 2 2
= tan x tan(c x ) tan x sec (c x )dx
D D 2 2
1 2 1 2 2
tan tan (c x ) tan (c x )sec xdx
x
2 2

310 LOVELY PROFESSIONAL UNIVERSITY