Page 158 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 8: Total Differential Equations, Simultaneous Equations

Notes
t
or (D 2 9D 14)x = 1 5t 3e ...(6)
or (D 7)(D 2)x = 1 5t 3e t ...(7)

C.F. is C e 7t C e 2t
1
2
The particular integral, P.I. is given by
1 t
P.I. = 2 {1 5t 3 }
e
[14 9D D ]
1
1 9D D 2
= 1 1 5t 3e t
14 14
1 9D 3e t
= 1 (1 5 )
t
14 14 14 9(1) (1) 2
1 45 3e t
= 1 5t
14 14 24

1 31 e t
= 5t ...(8)
14 14 8
So the complete solution is

5 31 e t
7t
2t
C e + C e t ...(9)
2
1
14 196 8
Self Assessment
dx
11. Solve 7x y 0
dt
dy
2x 5y 0
dt

dx dy t
12. Solve 2 2x 2y 3e
dt dt
dx dy 3t
3 2x y 4e
dt dt
The equation of the type
P dx Q dy R dz = 0
1
1
1
...(1)
P dx Q dy R dz = 0
2
2
2
,
,
,
Where P P Q Q and R 1 , R 2 are functions of , ,x y z
1
2
1
2
We can write these equations as
dx dy
P 1 Q 1 R 1 = 0
dz dz
LOVELY PROFESSIONAL UNIVERSITY 151