Page 334 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 20: Higher Order Equations with Constant Coefficients and Monge’s Method

or p/q = k = f(c) Notes
p qf(c) = 0.

dx dy dz
= ,
c
1 f ( ) 0
c
y xf ( ) = K F ( )
c
z
y xf ( ) = F ( ). ...(3)
z
The integral of the differential equation is the surface (3) which is the locus of the straight lines
c
c
given by the intersections of planes y xf ( ) F ( ), and z = c. These lines are all parallel to the
plane z = 0 as they lie on the plane z = c for varying values of c.

Example 6: Solve by Monge’s method
2
ab
r a t 2 (p qa ) = 0.
Solution: Putting

dp s dy dq s dx
r = and t , we get
dx dy
2
2
dp dy a dq dx 2ab (p aq )dx dy = s (dy 2 a dx 2 )
The subsidiary equations are

2
dy 2 a dx 2 = 0 ...(1)
2
dp dy a dq dx 2ab (p qa )dx dy = 0 ...(2)
From (1),
y + ax = , ...(3)
y ax = . ...(4)
From (3) and (2)

)
dp a dq 2ab p qa dx = 0
(
dp a dq
or = 2ab dx
p aq

log(p qa ) = 2abx log ,
c
p aq (p aq ) 2abx
= e
c f ( )
or p + qa = f ( ) e 2abx ...(5)

dx dy dz
=
1 a f ( )e 2abx

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