Page 338 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

Notes
R 1
Or (U dy T 1 dx 1 U dp ) ( 1 Rdy U dp 1 U dq ).
U R
1
Similarly if is a root of (6), the same is,
2
R 1
(U dy T 2 dx 2 U dp ) ( 2 Rdy U dx 2 U dq ).
U 2 R

Now we may obtain two integrals u 1 a 1 , 1 b 1 of the equations
U dy 1 T dx 1 U dp 0
and ...(7)
U dx Rdy U dq 0
2 2
or we may obtain two integrals u 2 a 2 , 2 b 2 of the equations

U dy 2 T dx 2 U dp 0
...(8)
U dx 1 Rdy 1 U dq 0
Sets of equations (7) and (8), when written down, constitute the second important step in the
solution of the given equation.
)
Thus we get two intermediate integrals u 1 f 1 ( 1 and u 2 f 2 ( 2 and substituting in dz = p dz
)
+ q dy, the values of p and q obtained from the two intermediate integrals, and we get the
solution after integrating.
In case the two roots of the equation (6) are equal, we shall get only intermediate integral
u f ( 1 which together with one of the integrals u = a and b 1 will give values of p and
)
1 1 1 1 1
q suitable to solve dz p dx q dy .
If it is not possible to obtain the values of p and q from the two intermediate integrals u 1 f 1 ( 1 )
)
and u 2 f 2 ( 2 , suitable for integration in dz pdx q dy , we may take one of the intermediate
)
integrals say u 1 f 1 ( 1 and one of the integrals from u = a and = b .
2
2
2
2
The values of p and q obtained from these and substituted in dz p dx q dy will give the
solution of the given equation.
Illustrative Examples

Example 1: Solve:

ar bs ct ( e rt s 2 ) h where , , ,a b c e and h are constants.
Solution: Here R , a S , b T , c U , e V h

The equation in is

2 2
(ac eh ) be e = 0. ...(1)
Putting = e/m, ...(2)
(1) becomes

2
e 2 (ac eh ) e b e 2 = 0
m 2 m

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